Trigonometry & Inverse Trigonometry
Properties of Inverse Trigonometric Functions
Grade 12
Question:
<p>If <i>a</i> = 3sin<sup>−1</sup>(23/25) and <i>β</i> = 3cos<sup>−1</sup>(23/25), where the inverse trigonometric functions take only the principal values, then the correct option(s) is/are</p>
<p>(a) cos <i>β</i> > 0</p>
<p>(b) sin <i>β</i> < 0</p>
<p>(c) cos (<i>a</i> + <i>β</i>) > 0</p>
<p>(d) cos <i>a</i> < 0</p>
Step-by-Step Solution
Key Concept: We need to find the ranges of α and β using properties of inverse trigonometric functions, then evaluate trigonometric values and inequalities. The key is recognizing that sin⁻¹(23/25) and cos⁻¹(23/25) are complementary angles in the principal range.
Step 1: Establish relationships between $\alpha$, $\beta$, and $\theta$
Let $\theta = \sin^{-1}\left(\frac{23}{25}\right)$.
Since $\frac{23}{25} > 0$, the principal value of $\theta$ lies in the interval $\left(0, \frac{\pi}{2}\right)$.
Using the identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$, we can write $\cos^{-1}\left(\frac{23}{25}\right) = \frac{\pi}{2} - \sin^{-1}\left(\frac{23}{25}\right) = \frac{\pi}{2} - \theta$.
Given the definitions of $\alpha$ and $\beta$:
$\alpha = 3\sin^{-1}\left(\frac{23}{25}\right) = 3\theta$
$\beta = 3\cos^{-1}\left(\frac{23}{25}\right) = 3\left(\frac{\pi}{2} - \theta\right) = \frac{3\pi}{2} - 3\theta$
Step 2: Determine the range of $\theta$
We have $\sin \theta = \frac{23}{25} = 0.92$.
We know that $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \approx 0.866$ and $\sin\left(\frac{\pi}{2}\right) = 1$.
Since $0.866 < 0.92 < 1$, it follows that $\frac{\pi}{3} < \theta < \frac{\pi}{2}$.
Step 3: Determine the ranges of $\alpha$ and $\beta$
Using the range of $\theta$ from Step 2:
For $\alpha = 3\theta$:
$3 \cdot \frac{\pi}{3} < 3\theta < 3 \cdot \frac{\pi}{2}$
$\pi < \alpha < \frac{3\pi}{2}$
For $\beta = \frac{3\pi}{2} - 3\theta$:
First, multiply the inequality for $\theta$ by $-3$:
$-3 \cdot \frac{\pi}{2} < -3\theta < -3 \cdot \frac{\pi}{3}$
$-\frac{3\pi}{2} < -3\theta < -\pi$
Now, add $\frac{3\pi}{2}$ to all parts of the inequality:
$\frac{3\pi}{2} - \frac{3\pi}{2} < \frac{3\pi}{2} - 3\theta < \frac{3\pi}{2} - \pi$
$0 < \beta < \frac{\pi}{2}$
Step 4: Evaluate $\cos \beta$
From Step 3, $\beta \in \left(0, \frac{\pi}{2}\right)$. This means $\beta$ lies in the first quadrant.
In the first quadrant, the cosine function is positive.
Therefore, $\cos \beta > 0$.
Step 5: Evaluate $\cos(\alpha + \beta)$
Summing the expressions for $\alpha$ and $\beta$:
$\alpha + \beta = 3\theta + \left(\frac{3\pi}{2} - 3\theta\right) = \frac{3\pi}{2}$
Now, calculate $\cos(\alpha + \beta)$:
$\cos(\alpha + \beta) = \cos\left(\frac{3\pi}{2}\right) = 0$
Since $0$ is not strictly greater than $0$, the statement $\cos(\alpha + \beta) > 0$ is false.
Correct Answer: a, c