Question

Find the value of ${\text{sin}}^{-1}\left[\text{cos}\left(\frac{53\mathrm{\pi }}{5}\right)\right]$ is :

Answer 1

Solution of trigonometric functions : We will solve the given question with in some steps

Step 1: First we write $\text{cos}\left(\frac{53\mathrm{\pi }}{5}\right)\text{= cos}\left(\frac{50\mathrm{\pi }+3\mathrm{\pi }}{5}\right)$

$=\text{cos(}\frac{50\mathrm{\pi }}{5}\text{+}\frac{3\mathrm{\pi }}{5}\text{)=cos(10π+}\frac{3\mathrm{\pi }}{5}\text{)}$

Step 2: Now we know that $\text{cos(2nπ+θ)=cosθ}$ where $\text{n∈Z}$

So $\text{cos(10π+}\frac{3\mathrm{\pi }}{5}\text{)=cos(}\frac{3\mathrm{\pi }}{5}\text{)}$

Step 3: Now we write ${\text{sin}}^{-1}\left[\cos \text{(}\frac{53\mathrm{\pi }}{5}\text{)}\right]{\text{=sin}}^{-1}\text{(cos(}\frac{3\mathrm{\pi }}{5}\text{))}$

Step 4: We know that $\text{sin(}\frac{\mathrm{\pi }}{2}\text{-θ)=cosθ}$

Hence, $\text{cos(}\frac{3\mathrm{\pi }}{5}\text{)=sin(}\frac{\mathrm{\pi }}{2}\text{-}\frac{3\mathrm{\pi }}{5}\text{)=sin(}\frac{5\mathrm{\pi }-6\mathrm{\pi }}{10}\text{)=sin(}\frac{-\mathrm{\pi }}{10}\text{)}$

Step 5: We have the formula that $\text{sin(-θ)=-sinθ}$ so

$\text{sin(}\frac{-\mathrm{\pi }}{10}\text{)=-sin(}\frac{\mathrm{\pi }}{10}\text{)}$

Step 6: ${\text{sin}}^{-1}\left[\text{cos(}\frac{53\mathrm{\pi }}{5}\text{)}\right]{\text{=sin}}^{-1}\text{(-sin(}\frac{\mathrm{\pi }}{10}{\text{))=-sin}}^{-1}\text{(sin(}\frac{\mathrm{\pi }}{10}\text{))=-}\frac{\mathrm{\pi }}{10}$ (from step 5 we put the value )

Hence, the value of ${\text{sin}}^{-1}\left[\text{cos}\left(\frac{53\mathrm{\pi }}{5}\right)\right]$ is $-\frac{\mathrm{\pi }}{10}$.