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If $\sin ^{-1} 1+\sin ^{-1} \frac{4}{5}=\sin ^{-1} x$, then what is $x$ equal to?
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The correct answer is:
$3 / 5$
Let $\sin ^{-1}(1)+\sin ^{-1}\left(\frac{4}{5}\right)=\sin ^{-1} \mathrm{x}$
$\operatorname{Let} \sin ^{-1}(1)=\theta \Rightarrow \sin \theta=1 \Rightarrow \cos \theta=0$
and $\sin ^{-1}\left(\frac{4}{5}\right)=\phi \Rightarrow \sin \phi=\left(\frac{4}{5}\right) \Rightarrow \cos \phi=\sqrt{1-\frac{16}{25}}$
$\begin{aligned} &=\sqrt{\frac{9}{25}}=\frac{3}{5} \\ \therefore & \sin ^{-1} \mathrm{x}=\theta+\phi \\ \Rightarrow & \mathrm{x}=\sin (\theta+\phi)=\sin \theta \cos \phi+\cos \theta \sin \phi \\ &=1 \times \frac{3}{5}+0 \times \frac{4}{5} \\ \Rightarrow & \mathrm{x}=\frac{3}{5} \end{aligned}$
$\operatorname{Let} \sin ^{-1}(1)=\theta \Rightarrow \sin \theta=1 \Rightarrow \cos \theta=0$
and $\sin ^{-1}\left(\frac{4}{5}\right)=\phi \Rightarrow \sin \phi=\left(\frac{4}{5}\right) \Rightarrow \cos \phi=\sqrt{1-\frac{16}{25}}$
$\begin{aligned} &=\sqrt{\frac{9}{25}}=\frac{3}{5} \\ \therefore & \sin ^{-1} \mathrm{x}=\theta+\phi \\ \Rightarrow & \mathrm{x}=\sin (\theta+\phi)=\sin \theta \cos \phi+\cos \theta \sin \phi \\ &=1 \times \frac{3}{5}+0 \times \frac{4}{5} \\ \Rightarrow & \mathrm{x}=\frac{3}{5} \end{aligned}$
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