Question

If $sin(si{n}^{-1}\frac{1}{5}+co{s}^{-1}x)=1$, then find the value of x.

Answer 1

$Given,sin(si{n}^{-1}\frac{1}{5}+co{s}^{-1}x)=1\Rightarrow si{n}^{-1}\frac{1}{5}+co{s}^{-1}x=si{n}^{-1}1[\because sin\theta =x\Rightarrow \theta =si{n}^{-1}x]\Rightarrow si{n}^{-1}\frac{1}{5}+co{s}^{-1}x=si{n}^{-1}\left(sin\frac{\pi }{2}\right)[\because sin\left(\frac{\pi }{2}\right)=1]\Rightarrow si{n}^{-1}\frac{1}{5}+co{s}^{-1}x=\frac{\pi }{2}\Rightarrow si{n}^{-1}\frac{1}{5}=\frac{\pi }{2}-co{s}^{-1}x\Rightarrow si{n}^{-1}\frac{1}{5}=si{n}^{-1}x[\because si{n}^{-1}x+co{s}^{-1}x=\frac{\pi }{2}]$

$\Rightarrow x=\frac{1}{5}$