Question

Find the value of ${\sin ^{ - 1}}(\cos \dfrac{{33\pi }}{5})$

Answer 1

Hint: First, analyze the given information which is in the trigonometric form.
The trigonometric functions are useful whenever trigonometric functions are involved in a given expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
Change the given cosine value to the sine value, so that we can simply apply the inverse trigonometric property
Formula used:
The given trigonometric function and its own inverse will cancel each other, like ${\sin ^{ - 1}}(\sin ) = 1$
The cosine function can be rewritten in the sine function as $\cos \theta = \sin (\dfrac{\pi }{2} - \theta )$

Complete step by step answer:
Given that ${\sin ^{ - 1}}(\cos \dfrac{{33\pi }}{5})$ and we need to find its value. Now using the addition operation, we can rewrite the given values as $33 = 3 + 30$
Applying these values in the above we get
${\sin ^{ - 1}}(\cos \dfrac{{33\pi }}{5}) = {\sin ^{ - 1}}(\cos \dfrac{{(30 + 3)\pi }}{5})$
Further solving we get
$ \Rightarrow {\sin ^{ - 1}}(\cos \dfrac{{30\pi + 3\pi }}{5}) = {\sin ^{ - 1}}(\cos (6\pi + \dfrac{{3\pi }}{5}))$ (canceling the common terms)
Since the value $6\pi $ is an even integer so the quadrant will not be changed and thus we have $ \Rightarrow {\sin ^{ - 1}}(\cos (6\pi + \dfrac{{3\pi }}{5})) = {\sin ^{ - 1}}(\cos \dfrac{{3\pi }}{5})$ where ${\sin ^{ - 1}}(\cos (n\pi + \theta )) = {\sin ^{ - 1}}(\cos \theta )$
for any $n$ as even integers.
Thus, applying the formula that
$\cos \theta = \sin (\dfrac{\pi }{2} - \theta )$ and we get $ \Rightarrow {\sin ^{ - 1}}(\cos \dfrac{{3\pi }}{5}) = {\sin ^{ - 1}}(\sin (\dfrac{\pi }{2} - \dfrac{{3\pi }}{5}))$
Now by the cross multiplication, we have $ \Rightarrow {\sin ^{ - 1}}(\sin (\dfrac{\pi }{2} - \dfrac{{3\pi }}{5})) = {\sin ^{ - 1}}(\sin \dfrac{{5\pi - 6\pi }}{{10}})$
Further solving we get $ \Rightarrow {\sin ^{ - 1}}(\sin \dfrac{{5\pi - 6\pi }}{{10}}) = {\sin ^{ - 1}}(\sin (\dfrac{{ - \pi }}{{10}}))$
Since the trigonometric function and its own inverse will cancel each other, like ${\sin ^{ - 1}}(\sin ) = 1$
Hence, we get $ \Rightarrow {\sin ^{ - 1}}(\sin (\dfrac{{ - \pi }}{{10}})) = \dfrac{{ - \pi }}{{10}}$
Therefore, we get ${\sin ^{ - 1}}(\cos \dfrac{{33\pi }}{5}) = \dfrac{{ - \pi }}{{10}}$

Note:
In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like $\dfrac{{\sin }}{{\cos }} = \tan $and $\tan = \dfrac{1}{{\cot }}$
the cosine function can be rewritten in the sine function as $\cos \theta = \sin (\dfrac{\pi }{2} - \theta )$
we try to convert the given into some form of the sine value so that we easily cancel using the inverse property.