Inverse Trigonometric Functions Question 3
Question 3 - 2024 (31 Jan Shift 1)
For $\alpha, \beta, \gamma \neq 0$. If $\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pi$ and $(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta$, then $\gamma$ equal to
(1) $\frac{\sqrt{3}}{2}$
(2) $\frac{1}{\sqrt{2}}$
(3) $\frac{\sqrt{3}-1}{2 \sqrt{2}}$
(4) $\sqrt{3}$
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Answer (1)
Solution
Let $\sin ^{-1} \alpha=\mathrm{A}, \sin ^{-1} \beta=\mathrm{B}, \sin ^{-1} \gamma=\mathrm{C}$
$\mathrm{A}+\mathrm{B}+\mathrm{C}=\pi$
$(\alpha+\beta)^{2}-\gamma^{2}=3 \alpha \beta$
$\alpha^{2}+\beta^{2}-\gamma^{2}=\alpha \beta$
$\frac{\alpha^{2}+\beta^{2}-\gamma^{2}}{2 \alpha \beta}=\frac{1}{2}$
$\Rightarrow \cos \mathrm{C}=\frac{1}{2}$
$\sin \mathrm{C}=\gamma$
$\cos \mathrm{C}=\sqrt{1-\gamma^{2}}=\frac{1}{2}$
$\gamma=\frac{\sqrt{3}}{2}$
