Question

Find the value of x which satisfy the equation $si{n}^{-1}8/17+si{n}^{-1}3/5=co{s}^{-1}x$.

Answer 1

${\sin }^{-1}\left(\frac{8}{17}\right)+{\sin }^{-1}\left(\frac{3}{5}\right)={\cos }^{-1}\left(x\right)$

To find $\to x=2$

Tabe $LHS$

${\sin }^{-1}\left(\frac{8}{17}\right)+{\sin }^{-1}\left(\frac{3}{5}\right)$

Put ${\sin }^{-1}\left(\frac{8}{17}\right)=p\Rightarrow \frac{8}{17}=\sin P$

${\sin }^{-1}\left(\frac{3}{5}\right)=q\Rightarrow \frac{3}{5}=\sin q$

$\Rightarrow \cos p=\sqrt{1-{\sin }^{2}p}=\sqrt{1-{\left(\frac{8}{17}\right)}^2}=\sqrt{\frac{225}{289}}\frac{15}{17}$

$\cos q=\sqrt{1-{\sin }^{2}q}=\sqrt{1-{\left(\frac{3}{5}\right)}^2}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

$\Rightarrow$ we know, $\cos (p+q)={\cos }^{-1}[\cos p\cos q-\sin p\sin q]$

$\therefore p+q={\cos }^{-1}[\cos p\cos q-\sin p\sin q]$

$p+q={\cos }^{-1}[\frac{15}{17}\times \frac{4}{5}-\frac{8}{17}\times \frac{3}{5}]$

$p+q={\cos }^{-1}[\frac{60}{85}-\frac{24}{25}]$

[Put values of $p$ & $q$]

${\sin }^{-1}\left(\frac{8}{17}\right)+{\sin }^{-1}\left(\frac{3}{5}\right)={\cos }^{-1}\left(\frac{36}{85}\right)$

Hence, value of $x=\frac{35}{85}$