Inverse Trigonometry
System of inverse trig; find q
MMTS_Full_Test_16
Grade 12
Question:
If $\cos^{-1}\!\sqrt{p}+\cos^{-1}\!\sqrt{1-p}+\cos^{-1}\!\sqrt{1-q}=\dfrac{3\pi}{4}$, then $q$ is
Step-by-Step Solution
Key Concept: $\cos^{-1}\!\sqrt{p}+\cos^{-1}\!\sqrt{1-p}=\pi/2$ (since $p+(1-p)=1$ and both in $[0,1]$: $\sin^{-1}\!\sqrt{p}+\cos^{-1}\!\sqrt{p}=\pi/2$). So $\pi/2+\cos^{-1}\!\sqrt{1-q}=3\pi/4\Rightarrow\cos^{-1}\!\sqrt{1-q}=\pi/4\Rightarrow\sqrt{1-q}=1/\sqrt2\Rightarrow q=1/2$.
$q=1/2$. As integer: $0$.
Correct Answer: 0