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

Find the value of sin⁡−1(cos⁡33π5){\sin ^{ - 1}}(\cos \dfrac{{33\pi }}{5})

Explanation

Solution

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{\sin ^{ - 1}}(\sin ) = 1
The cosine function can be rewritten in the sine function as cos⁡θ=sin⁡(π2−θ)\cos \theta = \sin (\dfrac{\pi }{2} - \theta )

Complete step by step answer:
Given that sin⁡−1(cos⁡33π5){\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+3033 = 3 + 30
Applying these values in the above we get
sin⁡−1(cos⁡33π5)=sin⁡−1(cos⁡(30+3)π5){\sin ^{ - 1}}(\cos \dfrac{{33\pi }}{5}) = {\sin ^{ - 1}}(\cos \dfrac{{(30 + 3)\pi }}{5})
Further solving we get
⇒sin⁡−1(cos⁡30π+3π5)=sin⁡−1(cos⁡(6π+3π5))\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 6π6\pi is an even integer so the quadrant will not be changed and thus we have ⇒sin⁡−1(cos⁡(6π+3π5))=sin⁡−1(cos⁡3π5) \Rightarrow {\sin ^{ - 1}}(\cos (6\pi + \dfrac{{3\pi }}{5})) = {\sin ^{ - 1}}(\cos \dfrac{{3\pi }}{5}) where sin⁡−1(cos⁡(nπ+θ))=sin⁡−1(cos⁡θ){\sin ^{ - 1}}(\cos (n\pi + \theta )) = {\sin ^{ - 1}}(\cos \theta )
for any nn as even integers.
Thus, applying the formula that
cos⁡θ=sin⁡(π2−θ)\cos \theta = \sin (\dfrac{\pi }{2} - \theta ) and we get ⇒sin⁡−1(cos⁡3π5)=sin⁡−1(sin⁡(π2−3π5)) \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 ⇒sin⁡−1(sin⁡(π2−3π5))=sin⁡−1(sin⁡5π−6π10) \Rightarrow {\sin ^{ - 1}}(\sin (\dfrac{\pi }{2} - \dfrac{{3\pi }}{5})) = {\sin ^{ - 1}}(\sin \dfrac{{5\pi - 6\pi }}{{10}})
Further solving we get ⇒sin⁡−1(sin⁡5π−6π10)=sin⁡−1(sin⁡(−π10)) \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{\sin ^{ - 1}}(\sin ) = 1
Hence, we get ⇒sin⁡−1(sin⁡(−π10))=−π10 \Rightarrow {\sin ^{ - 1}}(\sin (\dfrac{{ - \pi }}{{10}})) = \dfrac{{ - \pi }}{{10}}
Therefore, we get sin⁡−1(cos⁡33π5)=−π10{\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 sin⁡cos⁡=tan⁡\dfrac{{\sin }}{{\cos }} = \tan and tan⁡=1cot⁡\tan = \dfrac{1}{{\cot }}
the cosine function can be rewritten in the sine function as cos⁡θ=sin⁡(π2−θ)\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.