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 $\gamma$ equals
Solution
<p>Let $\sin ^{-1} \alpha=A, \sin ^{-1} \beta=B, \sin ^{-1} \gamma=C$</p>
<p>$$\begin{aligned}
& \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \\
& (\alpha+\beta)^2-\gamma^2=3 \alpha \beta \\
& \alpha^2+\beta^2-\gamma^2=\alpha \beta \\
& \frac{\alpha^2+\beta^2-\gamma^2}{2 \alpha \beta}=\frac{1}{2} \\
& \Rightarrow \cos \mathrm{C}=\frac{1}{2} \\
& \sin \mathrm{C}=\gamma \\
& \cos \mathrm{C}=\sqrt{1-\gamma^2}=\frac{1}{2} \\
& \gamma=\frac{\sqrt{3}}{2}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range
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