CH08 Question Bank
Introduction to Trigonometry • 50 Questions
Question 1
Hint available
In the given right triangle, $\sin\theta$ equals:
$\dfrac35$
$\dfrac45$
$\dfrac34$
$\dfrac43$
$\dfrac35$
$\dfrac45$
$\dfrac34$
$\dfrac43$
Key Idea
$\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}$.
Step-by-Step Solution
$\sin\theta=\dfrac{AB}{AC}=\dfrac35$. [1.0 Mark]
Question 2
Hint available
In the given right triangle, $\cos\theta$ equals:
$\dfrac45$
$\dfrac35$
$\dfrac43$
$\dfrac34$
$\dfrac45$
$\dfrac35$
$\dfrac43$
$\dfrac34$
Key Idea
$\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}$.
Step-by-Step Solution
$\cos\theta=\dfrac{BC}{AC}=\dfrac45$. [1.0 Mark]
Question 3
Hint available
In the given right triangle, $\tan\theta$ equals:
$\dfrac{5}{12}$
$\dfrac{12}{5}$
$\dfrac{5}{13}$
$\dfrac{12}{13}$
$\dfrac{5}{12}$
$\dfrac{12}{5}$
$\dfrac{5}{13}$
$\dfrac{12}{13}$
Key Idea
$\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}$.
Step-by-Step Solution
$\tan\theta=\dfrac{AB}{BC}=\dfrac{5}{12}$. [1.0 Mark]
Question 4
Hint available
The value of $\sin 30^\circ$ is:
$\dfrac12$
$\dfrac{\sqrt3}{2}$
$\dfrac{1}{\sqrt2}$
$1$
$\dfrac12$
$\dfrac{\sqrt3}{2}$
$\dfrac{1}{\sqrt2}$
$1$
Key Idea
Direct recall of the standard angle value table.
Step-by-Step Solution
$\sin30^\circ=\dfrac12$. [1.0 Mark]
Question 5
Hint available
The value of $\cos 60^\circ$ is:
$\dfrac12$
$\dfrac{\sqrt3}{2}$
$\dfrac{1}{\sqrt2}$
$0$
$\dfrac12$
$\dfrac{\sqrt3}{2}$
$\dfrac{1}{\sqrt2}$
$0$
Key Idea
Direct recall of the standard angle value table.
Step-by-Step Solution
$\cos60^\circ=\dfrac12$. [1.0 Mark]
Question 6
Hint available
The value of $\tan 45^\circ$ is:
$0$
$1$
$\sqrt3$
Not defined
$0$
$1$
$\sqrt3$
Not defined
Key Idea
Direct recall of the standard angle value table.
Step-by-Step Solution
$\tan45^\circ=1$. [1.0 Mark]
Question 7
Hint available
The value of $\sin 90^\circ$ is:
$0$
$\dfrac12$
$1$
Not defined
$0$
$\dfrac12$
$1$
Not defined
Key Idea
Direct recall of the standard angle value table.
Step-by-Step Solution
$\sin90^\circ=1$. [1.0 Mark]
Question 8
Hint available
The value of $\cos 0^\circ$ is:
$0$
$\dfrac12$
$1$
Not defined
$0$
$\dfrac12$
$1$
Not defined
Key Idea
Direct recall of the standard angle value table.
Step-by-Step Solution
$\cos0^\circ=1$. [1.0 Mark]
Question 9
Hint available
The value of $\tan 0^\circ$ is:
$0$
$1$
$\dfrac{1}{2}$
Not defined
$0$
$1$
$\dfrac{1}{2}$
Not defined
Key Idea
Direct recall of the standard angle value table.
Step-by-Step Solution
$\tan0^\circ=0$. [1.0 Mark]
Question 10
Hint available
$\text{cosec}\,\theta$ is defined as:
$\dfrac{1}{\sin\theta}$
$\dfrac{1}{\cos\theta}$
$\dfrac{1}{\tan\theta}$
$\sin\theta$
$\dfrac{1}{\sin\theta}$
$\dfrac{1}{\cos\theta}$
$\dfrac{1}{\tan\theta}$
$\sin\theta$
Key Idea
Recall the reciprocal relationships between the trigonometric ratios.
Step-by-Step Solution
By definition, $\text{cosec}\,\theta=\dfrac{1}{\sin\theta}$. [1.0 Mark]
Question 11
Hint available
If $\sin\theta=\cos\theta$ for an acute angle $\theta$, then $\theta$ equals:
$0^\circ$
$30^\circ$
$45^\circ$
$60^\circ$
$0^\circ$
$30^\circ$
$45^\circ$
$60^\circ$
Key Idea
Recall that $\sin45^\circ=\cos45^\circ=\dfrac{1}{\sqrt2}$, the only acute angle where the two are equal.
Step-by-Step Solution
$\sin45^\circ=\cos45^\circ=\dfrac{1}{\sqrt2}$, so $\theta=45^\circ$. [1.0 Mark]
Question 12
Hint available
The value of $\sin^2 30^\circ+\cos^2 30^\circ$ is:
$0$
$\dfrac12$
$1$
$2$
$0$
$\dfrac12$
$1$
$2$
Key Idea
Direct application of the identity $\sin^2A+\cos^2A=1$.
Step-by-Step Solution
By the identity $\sin^2A+\cos^2A=1$, this equals $1$ for any angle $A$, including $30^\circ$. [1.0 Mark]
Question 13
Hint available
$1+\tan^2A$ is equal to:
$\sin^2A$
$\cos^2A$
$\sec^2A$
$\text{cosec}^2A$
$\sin^2A$
$\cos^2A$
$\sec^2A$
$\text{cosec}^2A$
Key Idea
Standard derived identity from $\sin^2A+\cos^2A=1$, dividing throughout by $\cos^2A$.
Step-by-Step Solution
$1+\tan^2A=\sec^2A$ is a standard trigonometric identity. [1.0 Mark]
Question 14
Hint available
If $\cot\theta=\dfrac{5}{12}$, the value of $\text{cosec}^2\theta$ is:
$\dfrac{25}{144}$
$\dfrac{169}{144}$
$\dfrac{144}{169}$
$\dfrac{13}{12}$
$\dfrac{25}{144}$
$\dfrac{169}{144}$
$\dfrac{144}{169}$
$\dfrac{13}{12}$
Key Idea
Apply the identity $1+\cot^2\theta=\text{cosec}^2\theta$.
Step-by-Step Solution
$\text{cosec}^2\theta=1+\cot^2\theta=1+\dfrac{25}{144}=\dfrac{169}{144}$. [1.0 Mark]
Question 15
Hint available
For any acute angle $\theta$, the value of $\tan\theta\times\cot\theta$ is:
$0$
$\dfrac12$
$1$
$2$
$0$
$\dfrac12$
$1$
$2$
Key Idea
$\cot\theta$ is the reciprocal of $\tan\theta$.
Step-by-Step Solution
Since $\cot\theta=\dfrac{1}{\tan\theta}$, we get $\tan\theta\times\cot\theta=1$. [1.0 Mark]
Question 16
Hint available
Assertion (A): For any acute angle $A$, the value of $\sin A$ can never exceed $1$.
Reason (R): In a right triangle, the hypotenuse is always the longest side, so $\dfrac{\text{opposite}}{\text{hypotenuse}}\leq1$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): In a right triangle, the hypotenuse is always the longest side, so $\dfrac{\text{opposite}}{\text{hypotenuse}}\leq1$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
Direct geometric reasoning about right-triangle side lengths.
Step-by-Step Solution
Since $\sin A=\dfrac{\text{opposite}}{\text{hypotenuse}}$ and the opposite side can never be longer than the hypotenuse, $\sin A\leq1$ always — A is true. [0.5 Mark]
R correctly identifies why: the hypotenuse is the longest side of a right triangle, which is exactly the geometric fact that bounds $\sin A$. [0.5 Mark]
R correctly identifies why: the hypotenuse is the longest side of a right triangle, which is exactly the geometric fact that bounds $\sin A$. [0.5 Mark]
Question 17
Hint available
Assertion (A): $\sin 30^\circ + \cos 30^\circ = 1$.
Reason (R): $\sin 30^\circ = \dfrac12$ and $\cos 30^\circ = \dfrac{\sqrt3}{2}$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): $\sin 30^\circ = \dfrac12$ and $\cos 30^\circ = \dfrac{\sqrt3}{2}$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
R is correct, but adding these two values gives $\dfrac{1+\sqrt3}{2}
eq1$, so A is false.
eq1$, so A is false.
Step-by-Step Solution
$\sin30^\circ+\cos30^\circ=\dfrac12+\dfrac{\sqrt3}{2}=\dfrac{1+\sqrt3}{2}\approx1.37
eq1$ — so A is false. [0.5 Mark]
R correctly states the standard values of $\sin30^\circ$ and $\cos30^\circ$, so R is true. [0.5 Mark]
eq1$ — so A is false. [0.5 Mark]
R correctly states the standard values of $\sin30^\circ$ and $\cos30^\circ$, so R is true. [0.5 Mark]
Question 18
Hint available
Assertion (A): For an acute angle $A$, $\sec^2A-\tan^2A=1$.
Reason (R): This follows directly by dividing the identity $\sin^2A+\cos^2A=1$ throughout by $\cos^2A$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): This follows directly by dividing the identity $\sin^2A+\cos^2A=1$ throughout by $\cos^2A$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
Standard derivation of a secondary trigonometric identity from the primary one.
Step-by-Step Solution
$\sec^2A-\tan^2A=(1+\tan^2A)-\tan^2A=1$, confirming A is true. [0.5 Mark]
R correctly describes exactly how this identity is derived (dividing $\sin^2A+\cos^2A=1$ by $\cos^2A$ gives $\tan^2A+1=\sec^2A$, which rearranges to A), so R correctly explains A. [0.5 Mark]
R correctly describes exactly how this identity is derived (dividing $\sin^2A+\cos^2A=1$ by $\cos^2A$ gives $\tan^2A+1=\sec^2A$, which rearranges to A), so R correctly explains A. [0.5 Mark]
Question 19
Hint available
Assertion (A): The value of $\tan 90^\circ$ is not defined.
Reason (R): $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, and $\cos 90^\circ=0$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, and $\cos 90^\circ=0$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
Division by zero makes the ratio undefined at this specific angle.
Step-by-Step Solution
Since $\tan90^\circ$ would require dividing by $\cos90^\circ=0$, it is indeed not defined — A is true. [0.5 Mark]
R correctly identifies both the quotient relationship and the reason ($\cos90^\circ=0$) that causes the expression to be undefined, so R correctly explains A. [0.5 Mark]
R correctly identifies both the quotient relationship and the reason ($\cos90^\circ=0$) that causes the expression to be undefined, so R correctly explains A. [0.5 Mark]
Question 20
Hint available
Assertion (A): If $\sin A=\dfrac35$ for an acute angle $A$, then $\cos A=\dfrac45$.
Reason (R): For any acute angle $A$, $\cos A=1-\sin A$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): For any acute angle $A$, $\cos A=1-\sin A$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
A is correctly derived using the Pythagorean identity, but R states an incorrect (non-existent) relationship — the correct identity is $\sin^2A+\cos^2A=1$, not $\cos A=1-\sin A$.
Step-by-Step Solution
Using $\sin^2A+\cos^2A=1$: $\cos^2A=1-\dfrac{9}{25}=\dfrac{16}{25}\Rightarrow\cos A=\dfrac45$ (taking the positive root for an acute angle) — A is true. [0.5 Mark]
R's claimed relationship $\cos A=1-\sin A$ is not a valid trigonometric identity (e.g. it fails for $A=30^\circ$: $1-\sin30^\circ=0.5
eq\cos30^\circ\approx0.866$), so R is false. [0.5 Mark]
R's claimed relationship $\cos A=1-\sin A$ is not a valid trigonometric identity (e.g. it fails for $A=30^\circ$: $1-\sin30^\circ=0.5
eq\cos30^\circ\approx0.866$), so R is false. [0.5 Mark]
Question 21
Hint available
In the given right triangle, find $\sin\theta$ and $\cos\theta$.
Key Idea
Apply the ratio definitions directly using the given side lengths.
Step-by-Step Solution
$\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{8}{17}$. [1.0 Mark]
$\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{15}{17}$. [1.0 Mark]
$\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{15}{17}$. [1.0 Mark]
Question 22
Hint available
If $\sin A=\dfrac{3}{5}$, find $\cos A$ and $\tan A$ (assuming $A$ is acute).
Key Idea
Use the identity $\sin^2A+\cos^2A=1$ to find $\cos A$, then compute $\tan A=\sin A/\cos A$.
Step-by-Step Solution
$\cos^2A=1-\sin^2A=1-\dfrac{9}{25}=\dfrac{16}{25}\Rightarrow\cos A=\dfrac45$. [1.0 Mark]
$\tan A=\dfrac{\sin A}{\cos A}=\dfrac{3/5}{4/5}=\dfrac34$. [1.0 Mark]
$\tan A=\dfrac{\sin A}{\cos A}=\dfrac{3/5}{4/5}=\dfrac34$. [1.0 Mark]
Question 23
Hint available
Evaluate: $\sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ$.
Key Idea
Substitute the standard values and simplify.
Step-by-Step Solution
$=\dfrac{\sqrt3}{2}\times\dfrac{\sqrt3}{2}+\dfrac12\times\dfrac12=\dfrac34+\dfrac14$. [1.0 Mark]
$=1$. [1.0 Mark]
$=1$. [1.0 Mark]
Question 24
Hint available
If $\tan A=1$, find the value of $A$ and hence find $\sin A + \cos A$.
Key Idea
Recall the standard angle for which tangent equals 1, then use the standard values.
Step-by-Step Solution
$\tan A=1\Rightarrow A=45^\circ$. [1.0 Mark]
$\sin45^\circ+\cos45^\circ=\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt2}=\dfrac{2}{\sqrt2}=\sqrt2$. [1.0 Mark]
$\sin45^\circ+\cos45^\circ=\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt2}=\dfrac{2}{\sqrt2}=\sqrt2$. [1.0 Mark]
Question 25
Hint available
Verify that $\sin^2 60^\circ + \cos^2 60^\circ = 1$.
Key Idea
Substitute standard values and simplify.
Step-by-Step Solution
$\sin^260^\circ+\cos^260^\circ=\left(\dfrac{\sqrt3}{2}\right)^2+\left(\dfrac12\right)^2=\dfrac34+\dfrac14$. [1.0 Mark]
$=1$, verified. [1.0 Mark]
$=1$, verified. [1.0 Mark]
Question 26
Hint available
If $\text{cosec}\,\theta=\dfrac{13}{12}$, find $\sin\theta$ and hence find $\cos\theta$.
Key Idea
Use the reciprocal relationship to find $\sin\theta$, then apply $\sin^2\theta+\cos^2\theta=1$.
Step-by-Step Solution
$\sin\theta=\dfrac{1}{\text{cosec}\,\theta}=\dfrac{12}{13}$. [1.0 Mark]
$\cos^2\theta=1-\dfrac{144}{169}=\dfrac{25}{169}\Rightarrow\cos\theta=\dfrac{5}{13}$. [1.0 Mark]
$\cos^2\theta=1-\dfrac{144}{169}=\dfrac{25}{169}\Rightarrow\cos\theta=\dfrac{5}{13}$. [1.0 Mark]
Question 27
Hint available
Simplify: $\dfrac{\sin^2 45^\circ + \cos^2 45^\circ}{\tan^2 45^\circ}$.
Key Idea
Substitute standard values.
Step-by-Step Solution
Numerator $=\sin^245^\circ+\cos^245^\circ=1$ (by the Pythagorean identity). [1.0 Mark]
Denominator $=\tan^245^\circ=1^2=1$. So the expression $=\dfrac11=1$. [1.0 Mark]
Denominator $=\tan^245^\circ=1^2=1$. So the expression $=\dfrac11=1$. [1.0 Mark]
Question 28
Hint available
If $\cos A=\dfrac{1}{2}$, find the value of $3\cos A - 4\cos^3A$.
Key Idea
Substitute the given value and evaluate directly.
Step-by-Step Solution
$3\cos A-4\cos^3A=3\left(\dfrac12\right)-4\left(\dfrac12\right)^3=\dfrac32-\dfrac{4}{8}$. [1.0 Mark]
$=\dfrac32-\dfrac12=1$. [1.0 Mark]
$=\dfrac32-\dfrac12=1$. [1.0 Mark]
Question 29
Hint available
Express $\sec\theta$ and $\cot\theta$ in terms of $\sin\theta$ and $\cos\theta$.
Key Idea
Recall the standard reciprocal and quotient relationships.
Step-by-Step Solution
$\sec\theta=\dfrac{1}{\cos\theta}$. [1.0 Mark]
$\cot\theta=\dfrac{\cos\theta}{\sin\theta}$. [1.0 Mark]
$\cot\theta=\dfrac{\cos\theta}{\sin\theta}$. [1.0 Mark]
Question 30
Hint available
If $A=30^\circ$, verify that $\tan 2A = \dfrac{2\tan A}{1-\tan^2A}$.
Key Idea
Substitute $A=30^\circ$ on both sides and check they match.
Step-by-Step Solution
LHS: $\tan2A=\tan60^\circ=\sqrt3$. [1.0 Mark]
RHS: $\dfrac{2\tan30^\circ}{1-\tan^230^\circ}=\dfrac{2/\sqrt3}{1-1/3}=\dfrac{2/\sqrt3}{2/3}=\sqrt3$. Since LHS $=$ RHS, verified. [1.0 Mark]
RHS: $\dfrac{2\tan30^\circ}{1-\tan^230^\circ}=\dfrac{2/\sqrt3}{1-1/3}=\dfrac{2/\sqrt3}{2/3}=\sqrt3$. Since LHS $=$ RHS, verified. [1.0 Mark]
Question 31
Hint available
If $\sec A = 2$, find the value of $A$ and hence evaluate $\dfrac{1}{\tan A}$.
Key Idea
Recall the standard angle for which secant equals 2.
Step-by-Step Solution
$\sec A=2\Rightarrow \cos A=\dfrac12 \Rightarrow A=60^\circ$. [1.0 Mark]
$\dfrac{1}{\tan A}=\cot60^\circ=\dfrac{1}{\sqrt3}$. [1.0 Mark]
$\dfrac{1}{\tan A}=\cot60^\circ=\dfrac{1}{\sqrt3}$. [1.0 Mark]
Question 32
Hint available
Show that $(1-\cos^2\theta)\,\text{cosec}^2\theta = 1$.
Key Idea
Use the identity $1-\cos^2\theta=\sin^2\theta$, then simplify with the reciprocal relation.
Step-by-Step Solution
$1-\cos^2\theta=\sin^2\theta$ (Pythagorean identity). [1.0 Mark]
$\sin^2\theta\times\text{cosec}^2\theta=\sin^2\theta\times\dfrac{1}{\sin^2\theta}=1$. Hence shown. [1.0 Mark]
$\sin^2\theta\times\text{cosec}^2\theta=\sin^2\theta\times\dfrac{1}{\sin^2\theta}=1$. Hence shown. [1.0 Mark]
Question 33
Hint available
In the given right triangle, find all six trigonometric ratios of $\theta$.
Key Idea
Apply each of the six ratio definitions using the given side lengths.
Step-by-Step Solution
$\sin\theta=\dfrac{7}{25}$, $\cos\theta=\dfrac{24}{25}$, $\tan\theta=\dfrac{7}{24}$. [1.0 Mark]
$\text{cosec}\,\theta=\dfrac{25}{7}$, $\sec\theta=\dfrac{25}{24}$. [1.0 Mark]
$\cot\theta=\dfrac{24}{7}$. [1.0 Mark]
$\text{cosec}\,\theta=\dfrac{25}{7}$, $\sec\theta=\dfrac{25}{24}$. [1.0 Mark]
$\cot\theta=\dfrac{24}{7}$. [1.0 Mark]
Question 34
Hint available
If $3\tan A = 4$, find the value of $\dfrac{4\cos A - \sin A}{2\cos A + \sin A}$.
Key Idea
Divide numerator and denominator by cos A to express the fraction entirely in terms of tan A, then substitute.
Step-by-Step Solution
$\tan A=\dfrac43$. Dividing numerator and denominator of the given expression by $\cos A$: $\dfrac{4-\tan A}{2+\tan A}$. [1.0 Mark]
$=\dfrac{4-\frac43}{2+\frac43}=\dfrac{\frac{8}{3}}{\frac{10}{3}}$. [1.0 Mark]
$=\dfrac{8}{10}=\dfrac45$. [1.0 Mark]
$=\dfrac{4-\frac43}{2+\frac43}=\dfrac{\frac{8}{3}}{\frac{10}{3}}$. [1.0 Mark]
$=\dfrac{8}{10}=\dfrac45$. [1.0 Mark]
Question 35
Hint available
Verify that $\dfrac{1-\tan^2A}{1+\tan^2A}=1-2\sin^2A$ for $A=30^\circ$.
Key Idea
Substitute the standard value $A=30^\circ$ into both sides and confirm they are equal.
Step-by-Step Solution
LHS: $\dfrac{1-\tan^230^\circ}{1+\tan^230^\circ}=\dfrac{1-\frac13}{1+\frac13}=\dfrac{\frac23}{\frac43}=\dfrac12$. [1.0 Mark]
RHS: $1-2\sin^230^\circ=1-2\left(\dfrac12\right)^2=1-\dfrac12=\dfrac12$. [1.0 Mark]
Since LHS $=$ RHS $=\dfrac12$, the identity is verified for $A=30^\circ$. [1.0 Mark]
RHS: $1-2\sin^230^\circ=1-2\left(\dfrac12\right)^2=1-\dfrac12=\dfrac12$. [1.0 Mark]
Since LHS $=$ RHS $=\dfrac12$, the identity is verified for $A=30^\circ$. [1.0 Mark]
Question 36
Hint available
If $\sin\theta=\dfrac{a}{b}$, find $\cos\theta$ and $\tan\theta$ in terms of $a$ and $b$.
Key Idea
Use the Pythagorean identity to express cos θ in terms of a and b, then find tan θ.
Step-by-Step Solution
$\cos^2\theta=1-\sin^2\theta=1-\dfrac{a^2}{b^2}=\dfrac{b^2-a^2}{b^2}$. [1.0 Mark]
$\cos\theta=\dfrac{\sqrt{b^2-a^2}}{b}$ (taking the positive root for an acute angle). [1.0 Mark]
$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{a/b}{\sqrt{b^2-a^2}/b}=\dfrac{a}{\sqrt{b^2-a^2}}$. [1.0 Mark]
$\cos\theta=\dfrac{\sqrt{b^2-a^2}}{b}$ (taking the positive root for an acute angle). [1.0 Mark]
$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{a/b}{\sqrt{b^2-a^2}/b}=\dfrac{a}{\sqrt{b^2-a^2}}$. [1.0 Mark]
Question 37
Hint available
Evaluate: $\dfrac{2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ}{ \tan^2 60^\circ}$.
Key Idea
Substitute all standard values carefully and simplify step by step.
Step-by-Step Solution
$\tan45^\circ=1,\ \cos30^\circ=\dfrac{\sqrt3}{2},\ \sin60^\circ=\dfrac{\sqrt3}{2},\ \tan60^\circ=\sqrt3$. [1.0 Mark]
Numerator $=2(1)^2+\left(\dfrac{\sqrt3}{2}\right)^2-\left(\dfrac{\sqrt3}{2}\right)^2=2+0=2$ (since the last two terms are equal and cancel). [1.0 Mark]
Denominator $=(\sqrt3)^2=3$. So the expression $=\dfrac23$. [1.0 Mark]
Numerator $=2(1)^2+\left(\dfrac{\sqrt3}{2}\right)^2-\left(\dfrac{\sqrt3}{2}\right)^2=2+0=2$ (since the last two terms are equal and cancel). [1.0 Mark]
Denominator $=(\sqrt3)^2=3$. So the expression $=\dfrac23$. [1.0 Mark]
Question 38
Hint available
If $\sec\theta + \tan\theta = x$, show that $\sec\theta - \tan\theta = \dfrac{1}{x}$.
Key Idea
Multiply $(\sec\theta+\tan\theta)$ by $(\sec\theta-\tan\theta)$ and use the identity $\sec^2\theta-\tan^2\theta=1$.
Step-by-Step Solution
$(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=\sec^2\theta-\tan^2\theta=1$ (standard identity). [1.5 Marks]
Given $\sec\theta+\tan\theta=x$, substituting: $x(\sec\theta-\tan\theta)=1$. [1.0 Mark]
$\sec\theta-\tan\theta=\dfrac{1}{x}$. Hence shown. [0.5 Mark]
Given $\sec\theta+\tan\theta=x$, substituting: $x(\sec\theta-\tan\theta)=1$. [1.0 Mark]
$\sec\theta-\tan\theta=\dfrac{1}{x}$. Hence shown. [0.5 Mark]
Question 39
Hint available
If $A=B=45^\circ$, verify that $\cos(A+B) = \cos A \cos B - \sin A \sin B$.
Key Idea
Evaluate both sides independently using standard values and compare.
Step-by-Step Solution
LHS: $\cos(45^\circ+45^\circ)=\cos90^\circ=0$. [1.0 Mark]
RHS: $\cos45^\circ\cos45^\circ-\sin45^\circ\sin45^\circ=\dfrac{1}{\sqrt2}\times\dfrac{1}{\sqrt2}-\dfrac{1}{\sqrt2}\times\dfrac{1}{\sqrt2}=\dfrac12-\dfrac12=0$. [1.0 Mark]
Since LHS $=$ RHS $=0$, the relation is verified for $A=B=45^\circ$. [1.0 Mark]
RHS: $\cos45^\circ\cos45^\circ-\sin45^\circ\sin45^\circ=\dfrac{1}{\sqrt2}\times\dfrac{1}{\sqrt2}-\dfrac{1}{\sqrt2}\times\dfrac{1}{\sqrt2}=\dfrac12-\dfrac12=0$. [1.0 Mark]
Since LHS $=$ RHS $=0$, the relation is verified for $A=B=45^\circ$. [1.0 Mark]
Question 40
Hint available
If $x = 2\sin^2\theta$ and $y = 2\cos^2\theta + 1$, find the value of $x+y$.
Key Idea
Add the two expressions and use the Pythagorean identity to simplify.
Step-by-Step Solution
$x+y=2\sin^2\theta+2\cos^2\theta+1$. [1.0 Mark]
$=2(\sin^2\theta+\cos^2\theta)+1=2(1)+1$. [1.0 Mark]
$=3$. [1.0 Mark]
$=2(\sin^2\theta+\cos^2\theta)+1=2(1)+1$. [1.0 Mark]
$=3$. [1.0 Mark]
Question 41
Hint available
Simplify: $\sin^4\theta - \cos^4\theta$ in terms of $\sin^2\theta$ and $\cos^2\theta$ only, and then reduce it to a single trigonometric term.
Key Idea
Factorise as a difference of squares, then apply the Pythagorean identity.
Step-by-Step Solution
$\sin^4\theta-\cos^4\theta=(\sin^2\theta-\cos^2\theta)(\sin^2\theta+\cos^2\theta)$. [1.0 Mark]
Since $\sin^2\theta+\cos^2\theta=1$, this simplifies to $\sin^2\theta-\cos^2\theta$. [1.0 Mark]
This can also be written as $1-2\cos^2\theta$ (substituting $\sin^2\theta=1-\cos^2\theta$). [1.0 Mark]
Since $\sin^2\theta+\cos^2\theta=1$, this simplifies to $\sin^2\theta-\cos^2\theta$. [1.0 Mark]
This can also be written as $1-2\cos^2\theta$ (substituting $\sin^2\theta=1-\cos^2\theta$). [1.0 Mark]
Question 42
Hint available
If $\tan A = \dfrac{1}{\sqrt3}$, find the value of $\dfrac{\sin A + \cos A}{\text{cosec}\,A}$.
Key Idea
Identify the standard angle, find all needed ratios, then substitute.
Step-by-Step Solution
$\tan A=\dfrac{1}{\sqrt3}\Rightarrow A=30^\circ$. [1.0 Mark]
$\sin30^\circ=\dfrac12,\ \cos30^\circ=\dfrac{\sqrt3}{2},\ \text{cosec}\,30^\circ=2$. [1.0 Mark]
$\dfrac{\sin A+\cos A}{\text{cosec}\,A}=\dfrac{\frac12+\frac{\sqrt3}{2}}{2}=\dfrac{1+\sqrt3}{4}$. [1.0 Mark]
$\sin30^\circ=\dfrac12,\ \cos30^\circ=\dfrac{\sqrt3}{2},\ \text{cosec}\,30^\circ=2$. [1.0 Mark]
$\dfrac{\sin A+\cos A}{\text{cosec}\,A}=\dfrac{\frac12+\frac{\sqrt3}{2}}{2}=\dfrac{1+\sqrt3}{4}$. [1.0 Mark]
Question 43
Hint available
If $\sec\theta = \dfrac{13}{5}$, find the values of all the other five trigonometric ratios of $\theta$.
Key Idea
Use the reciprocal relation to find cos θ, then the Pythagorean identity to find sin θ, and finally derive the rest.
Step-by-Step Solution
$\cos\theta=\dfrac{1}{\sec\theta}=\dfrac{5}{13}$. [1.0 Mark]
$\sin^2\theta=1-\cos^2\theta=1-\dfrac{25}{169}=\dfrac{144}{169}\Rightarrow\sin\theta=\dfrac{12}{13}$. [1.5 Marks]
$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{12/13}{5/13}=\dfrac{12}{5}$. [1.0 Mark]
$\text{cosec}\,\theta=\dfrac{1}{\sin\theta}=\dfrac{13}{12}$. [0.75 Mark]
$\cot\theta=\dfrac{1}{\tan\theta}=\dfrac{5}{12}$. [0.75 Mark]
$\sin^2\theta=1-\cos^2\theta=1-\dfrac{25}{169}=\dfrac{144}{169}\Rightarrow\sin\theta=\dfrac{12}{13}$. [1.5 Marks]
$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{12/13}{5/13}=\dfrac{12}{5}$. [1.0 Mark]
$\text{cosec}\,\theta=\dfrac{1}{\sin\theta}=\dfrac{13}{12}$. [0.75 Mark]
$\cot\theta=\dfrac{1}{\tan\theta}=\dfrac{5}{12}$. [0.75 Mark]
Question 44
Hint available
Prove that: $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$.
Key Idea
Combine the two fractions on the LHS over a common denominator, then simplify the numerator using the Pythagorean identity.
Step-by-Step Solution
LHS $=\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}$. Taking the common denominator $\cos A(1+\sin A)$: [1.0 Mark]
$=\dfrac{\cos^2A+(1+\sin A)^2}{\cos A(1+\sin A)}$. [1.0 Mark]
Expand the numerator: $\cos^2A+1+2\sin A+\sin^2A=(\sin^2A+\cos^2A)+1+2\sin A=1+1+2\sin A=2+2\sin A$. [1.5 Marks]
$=\dfrac{2(1+\sin A)}{\cos A(1+\sin A)}=\dfrac{2}{\cos A}$. [1.0 Mark]
$=2\sec A$. Hence proved. [0.5 Mark]
$=\dfrac{\cos^2A+(1+\sin A)^2}{\cos A(1+\sin A)}$. [1.0 Mark]
Expand the numerator: $\cos^2A+1+2\sin A+\sin^2A=(\sin^2A+\cos^2A)+1+2\sin A=1+1+2\sin A=2+2\sin A$. [1.5 Marks]
$=\dfrac{2(1+\sin A)}{\cos A(1+\sin A)}=\dfrac{2}{\cos A}$. [1.0 Mark]
$=2\sec A$. Hence proved. [0.5 Mark]
Question 45
Hint available
Prove that: $\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot\theta}{1-\tan\theta}=1+\sec\theta\,\text{cosec}\,\theta$.
Key Idea
Convert everything to $\sin\theta$ and $\cos\theta$, combine over a common denominator, and simplify using $\sin^2\theta+\cos^2\theta=1$.
Step-by-Step Solution
Write $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$ and $\cot\theta=\dfrac{\cos\theta}{\sin\theta}$. The first term becomes $\dfrac{\sin\theta/\cos\theta}{1-\cos\theta/\sin\theta}=\dfrac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}$. [1.5 Marks]
Similarly, the second term becomes $\dfrac{\cos\theta/\sin\theta}{1-\sin\theta/\cos\theta}=\dfrac{\cos^2\theta}{\sin\theta(\cos\theta-\sin\theta)}=-\dfrac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}$. [1.5 Marks]
Adding: $\dfrac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}-\dfrac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}=\dfrac{1}{\sin\theta-\cos\theta}\left(\dfrac{\sin^2\theta}{\cos\theta}-\dfrac{\cos^2\theta}{\sin\theta}\right)$. [1.0 Mark]
$=\dfrac{1}{\sin\theta-\cos\theta}\times\dfrac{\sin^3\theta-\cos^3\theta}{\sin\theta\cos\theta}$. Using $\sin^3\theta-\cos^3\theta=(\sin\theta-\cos\theta)(\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta)$, this becomes $\dfrac{\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta}{\sin\theta\cos\theta}=\dfrac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta}$. [0.75 Mark]
$=\dfrac{1}{\sin\theta\cos\theta}+1=\text{cosec}\,\theta\sec\theta+1=1+\sec\theta\,\text{cosec}\,\theta$. Hence proved. [0.25 Mark]
Similarly, the second term becomes $\dfrac{\cos\theta/\sin\theta}{1-\sin\theta/\cos\theta}=\dfrac{\cos^2\theta}{\sin\theta(\cos\theta-\sin\theta)}=-\dfrac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}$. [1.5 Marks]
Adding: $\dfrac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}-\dfrac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}=\dfrac{1}{\sin\theta-\cos\theta}\left(\dfrac{\sin^2\theta}{\cos\theta}-\dfrac{\cos^2\theta}{\sin\theta}\right)$. [1.0 Mark]
$=\dfrac{1}{\sin\theta-\cos\theta}\times\dfrac{\sin^3\theta-\cos^3\theta}{\sin\theta\cos\theta}$. Using $\sin^3\theta-\cos^3\theta=(\sin\theta-\cos\theta)(\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta)$, this becomes $\dfrac{\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta}{\sin\theta\cos\theta}=\dfrac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta}$. [0.75 Mark]
$=\dfrac{1}{\sin\theta\cos\theta}+1=\text{cosec}\,\theta\sec\theta+1=1+\sec\theta\,\text{cosec}\,\theta$. Hence proved. [0.25 Mark]
Question 46
Hint available
If $x=r\sin A\cos B$, $y=r\sin A\sin B$ and $z=r\cos A$, prove that $x^2+y^2+z^2=r^2$.
Key Idea
Square and add all three expressions, then factor out common terms and apply the Pythagorean identity twice.
Step-by-Step Solution
$x^2+y^2=r^2\sin^2A\cos^2B+r^2\sin^2A\sin^2B=r^2\sin^2A(\cos^2B+\sin^2B)$. [1.5 Marks]
Using $\sin^2B+\cos^2B=1$: $x^2+y^2=r^2\sin^2A(1)=r^2\sin^2A$. [1.0 Mark]
$x^2+y^2+z^2=r^2\sin^2A+r^2\cos^2A=r^2(\sin^2A+\cos^2A)$. [1.5 Marks]
Using $\sin^2A+\cos^2A=1$: $x^2+y^2+z^2=r^2(1)=r^2$. Hence proved. [1.0 Mark]
Using $\sin^2B+\cos^2B=1$: $x^2+y^2=r^2\sin^2A(1)=r^2\sin^2A$. [1.0 Mark]
$x^2+y^2+z^2=r^2\sin^2A+r^2\cos^2A=r^2(\sin^2A+\cos^2A)$. [1.5 Marks]
Using $\sin^2A+\cos^2A=1$: $x^2+y^2+z^2=r^2(1)=r^2$. Hence proved. [1.0 Mark]
Question 47
Hint available
[Case Study]
A gardener is laying out a triangular flower bed in the shape of a right triangle, as shown, with the right angle at the corner where two straight edges of the bed meet. The two straight edges measure $6$ m and $8$ m, and the sloped edge (hypotenuse) measures $10$ m. The angle marked $\theta$ is at one corner of the bed.
(a) Find $\sin\theta$. [1 Mark]
(b) Find $\cos\theta$. [1 Mark]
(c) Verify that $\sin^2\theta+\cos^2\theta=1$ for this triangle. [1 Mark]
(d) Find $\tan\theta$. [1 Mark]
A gardener is laying out a triangular flower bed in the shape of a right triangle, as shown, with the right angle at the corner where two straight edges of the bed meet. The two straight edges measure $6$ m and $8$ m, and the sloped edge (hypotenuse) measures $10$ m. The angle marked $\theta$ is at one corner of the bed.
(a) Find $\sin\theta$. [1 Mark]
(b) Find $\cos\theta$. [1 Mark]
(c) Verify that $\sin^2\theta+\cos^2\theta=1$ for this triangle. [1 Mark]
(d) Find $\tan\theta$. [1 Mark]
Key Idea
Case study on trigonometry ratios and identities.
Step-by-Step Solution
(a) Find $\sin\theta$. [1 Mark]
$\sin\theta=\dfrac{6}{10}=\dfrac35$. [1.0 Mark]
(b) Find $\cos\theta$. [1 Mark]
$\cos\theta=\dfrac{8}{10}=\dfrac45$. [1.0 Mark]
(c) Verify that $\sin^2\theta+\cos^2\theta=1$ for this triangle. [1 Mark]
$\left(\dfrac35\right)^2+\left(\dfrac45\right)^2=\dfrac{9}{25}+\dfrac{16}{25}=\dfrac{25}{25}=1$. Verified. [1.0 Mark]
(d) Find $\tan\theta$. [1 Mark]
$\tan\theta=\dfrac{6}{8}=\dfrac34$. [1.0 Mark]
$\sin\theta=\dfrac{6}{10}=\dfrac35$. [1.0 Mark]
(b) Find $\cos\theta$. [1 Mark]
$\cos\theta=\dfrac{8}{10}=\dfrac45$. [1.0 Mark]
(c) Verify that $\sin^2\theta+\cos^2\theta=1$ for this triangle. [1 Mark]
$\left(\dfrac35\right)^2+\left(\dfrac45\right)^2=\dfrac{9}{25}+\dfrac{16}{25}=\dfrac{25}{25}=1$. Verified. [1.0 Mark]
(d) Find $\tan\theta$. [1 Mark]
$\tan\theta=\dfrac{6}{8}=\dfrac34$. [1.0 Mark]
Question 48
Hint available
[Case Study]
A carpenter uses a standard $30^\circ$-$60^\circ$-$90^\circ$ set-square (a common tool in a geometry box) to mark angles while cutting wood for a triangular shelf bracket.
(a) What is the value of $\sin 30^\circ + \sin 60^\circ$? [1 Mark]
(b) What is the value of $\cos 30^\circ \times \cos 60^\circ$? [1 Mark]
(c) The carpenter needs to check that $\tan 30^\circ \times \tan 60^\circ = 1$. Verify this. [1 Mark]
(d) Which pair of standard angles gives equal sine and cosine values? [1 Mark]
A carpenter uses a standard $30^\circ$-$60^\circ$-$90^\circ$ set-square (a common tool in a geometry box) to mark angles while cutting wood for a triangular shelf bracket.
(a) What is the value of $\sin 30^\circ + \sin 60^\circ$? [1 Mark]
(b) What is the value of $\cos 30^\circ \times \cos 60^\circ$? [1 Mark]
(c) The carpenter needs to check that $\tan 30^\circ \times \tan 60^\circ = 1$. Verify this. [1 Mark]
(d) Which pair of standard angles gives equal sine and cosine values? [1 Mark]
Key Idea
Case study on trigonometry ratios and identities.
Step-by-Step Solution
(a) What is the value of $\sin 30^\circ + \sin 60^\circ$? [1 Mark]
$\sin30^\circ+\sin60^\circ=\dfrac12+\dfrac{\sqrt3}{2}=\dfrac{1+\sqrt3}{2}$. [1.0 Mark]
(b) What is the value of $\cos 30^\circ \times \cos 60^\circ$? [1 Mark]
$\cos30^\circ\times\cos60^\circ=\dfrac{\sqrt3}{2}\times\dfrac12=\dfrac{\sqrt3}{4}$. [1.0 Mark]
(c) The carpenter needs to check that $\tan 30^\circ \times \tan 60^\circ = 1$. Verify this. [1 Mark]
$\tan30^\circ\times\tan60^\circ=\dfrac{1}{\sqrt3}\times\sqrt3=1$. Verified. [1.0 Mark]
(d) Which pair of standard angles gives equal sine and cosine values? [1 Mark]
At $45^\circ$, $\sin45^\circ=\cos45^\circ=\dfrac{1}{\sqrt2}$. [1.0 Mark]
$\sin30^\circ+\sin60^\circ=\dfrac12+\dfrac{\sqrt3}{2}=\dfrac{1+\sqrt3}{2}$. [1.0 Mark]
(b) What is the value of $\cos 30^\circ \times \cos 60^\circ$? [1 Mark]
$\cos30^\circ\times\cos60^\circ=\dfrac{\sqrt3}{2}\times\dfrac12=\dfrac{\sqrt3}{4}$. [1.0 Mark]
(c) The carpenter needs to check that $\tan 30^\circ \times \tan 60^\circ = 1$. Verify this. [1 Mark]
$\tan30^\circ\times\tan60^\circ=\dfrac{1}{\sqrt3}\times\sqrt3=1$. Verified. [1.0 Mark]
(d) Which pair of standard angles gives equal sine and cosine values? [1 Mark]
At $45^\circ$, $\sin45^\circ=\cos45^\circ=\dfrac{1}{\sqrt2}$. [1.0 Mark]
Question 49
Hint available
[Case Study]
A triangular sail for a small sailing boat is shaped like a right triangle, with sides $5$ m, $12$ m, and $13$ m as shown, and the angle at one corner marked $\phi$.
(a) Find $\sin\phi$ and $\cos\phi$. [1 Mark]
(b) Find $\sec\phi$ and $\text{cosec}\,\phi$. [1 Mark]
(c) Verify that $1+\tan^2\phi=\sec^2\phi$ for this sail. [1 Mark]
(d) Find the value of $\tan\phi \times \cot\phi$. [1 Mark]
A triangular sail for a small sailing boat is shaped like a right triangle, with sides $5$ m, $12$ m, and $13$ m as shown, and the angle at one corner marked $\phi$.
(a) Find $\sin\phi$ and $\cos\phi$. [1 Mark]
(b) Find $\sec\phi$ and $\text{cosec}\,\phi$. [1 Mark]
(c) Verify that $1+\tan^2\phi=\sec^2\phi$ for this sail. [1 Mark]
(d) Find the value of $\tan\phi \times \cot\phi$. [1 Mark]
Key Idea
Case study on trigonometry ratios and identities.
Step-by-Step Solution
(a) Find $\sin\phi$ and $\cos\phi$. [1 Mark]
$\sin\phi=\dfrac{5}{13}$; $\cos\phi=\dfrac{12}{13}$. [1.0 Mark]
(b) Find $\sec\phi$ and $\text{cosec}\,\phi$. [1 Mark]
$\sec\phi=\dfrac{13}{12}$; $\text{cosec}\,\phi=\dfrac{13}{5}$. [1.0 Mark]
(c) Verify that $1+\tan^2\phi=\sec^2\phi$ for this sail. [1 Mark]
$\tan\phi=\dfrac{5}{12}$; $1+\tan^2\phi=1+\dfrac{25}{144}=\dfrac{169}{144}$; and $\sec^2\phi=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}$. Both sides match, verified. [1.0 Mark]
(d) Find the value of $\tan\phi \times \cot\phi$. [1 Mark]
$\tan\phi\times\cot\phi=1$ (since $\cot\phi$ is the reciprocal of $\tan\phi$, for any value of $\phi$ where both are defined). [1.0 Mark]
$\sin\phi=\dfrac{5}{13}$; $\cos\phi=\dfrac{12}{13}$. [1.0 Mark]
(b) Find $\sec\phi$ and $\text{cosec}\,\phi$. [1 Mark]
$\sec\phi=\dfrac{13}{12}$; $\text{cosec}\,\phi=\dfrac{13}{5}$. [1.0 Mark]
(c) Verify that $1+\tan^2\phi=\sec^2\phi$ for this sail. [1 Mark]
$\tan\phi=\dfrac{5}{12}$; $1+\tan^2\phi=1+\dfrac{25}{144}=\dfrac{169}{144}$; and $\sec^2\phi=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}$. Both sides match, verified. [1.0 Mark]
(d) Find the value of $\tan\phi \times \cot\phi$. [1 Mark]
$\tan\phi\times\cot\phi=1$ (since $\cot\phi$ is the reciprocal of $\tan\phi$, for any value of $\phi$ where both are defined). [1.0 Mark]
Question 50
Hint available
[Case Study]
A student is designing a company logo using two right triangles of different sizes but with the same shape (i.e. the same acute angle $\theta$ at one corner in both). In the smaller triangle, the side opposite $\theta$ is $3$ cm and the hypotenuse is $5$ cm. In the larger triangle, the hypotenuse is $15$ cm.
(a) Find $\sin\theta$ using the smaller triangle. [1 Mark]
(b) Since both triangles have the same angle $\theta$, what can you say about the value of $\sin\theta$ computed from the larger triangle? [1 Mark]
(c) Find the side opposite $\theta$ in the larger triangle. [1 Mark]
(d) Find $\cos\theta$ for the smaller triangle. [1 Mark]
A student is designing a company logo using two right triangles of different sizes but with the same shape (i.e. the same acute angle $\theta$ at one corner in both). In the smaller triangle, the side opposite $\theta$ is $3$ cm and the hypotenuse is $5$ cm. In the larger triangle, the hypotenuse is $15$ cm.
(a) Find $\sin\theta$ using the smaller triangle. [1 Mark]
(b) Since both triangles have the same angle $\theta$, what can you say about the value of $\sin\theta$ computed from the larger triangle? [1 Mark]
(c) Find the side opposite $\theta$ in the larger triangle. [1 Mark]
(d) Find $\cos\theta$ for the smaller triangle. [1 Mark]
Key Idea
Case study on trigonometry ratios and identities.
Step-by-Step Solution
(a) Find $\sin\theta$ using the smaller triangle. [1 Mark]
$\sin\theta=\dfrac{3}{5}$. [1.0 Mark]
(b) Since both triangles have the same angle $\theta$, what can you say about the value of $\sin\theta$ computed from the larger triangle? [1 Mark]
Since the ratios of a given angle are always the same regardless of the triangle's size, $\sin\theta$ computed from the larger triangle will also equal $\dfrac35$. [1.0 Mark]
(c) Find the side opposite $\theta$ in the larger triangle. [1 Mark]
$\sin\theta=\dfrac35=\dfrac{\text{opposite}}{15}\Rightarrow\text{opposite}=9$ cm. [1.0 Mark]
(d) Find $\cos\theta$ for the smaller triangle. [1 Mark]
$\cos^2\theta=1-\sin^2\theta=1-\dfrac{9}{25}=\dfrac{16}{25}\Rightarrow\cos\theta=\dfrac45$. [1.0 Mark]
$\sin\theta=\dfrac{3}{5}$. [1.0 Mark]
(b) Since both triangles have the same angle $\theta$, what can you say about the value of $\sin\theta$ computed from the larger triangle? [1 Mark]
Since the ratios of a given angle are always the same regardless of the triangle's size, $\sin\theta$ computed from the larger triangle will also equal $\dfrac35$. [1.0 Mark]
(c) Find the side opposite $\theta$ in the larger triangle. [1 Mark]
$\sin\theta=\dfrac35=\dfrac{\text{opposite}}{15}\Rightarrow\text{opposite}=9$ cm. [1.0 Mark]
(d) Find $\cos\theta$ for the smaller triangle. [1 Mark]
$\cos^2\theta=1-\sin^2\theta=1-\dfrac{9}{25}=\dfrac{16}{25}\Rightarrow\cos\theta=\dfrac45$. [1.0 Mark]