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Question: Prove the given trigonometric expression: \[\dfrac{{9\pi }}{8} - \dfrac{9}{4}{\sin ^{ - 1}}\dfrac{1}...

Prove the given trigonometric expression: 9π8−94sin⁡−113=94sin⁡−1223\dfrac{{9\pi }}{8} - \dfrac{9}{4}{\sin ^{ - 1}}\dfrac{1}{3} = \dfrac{9}{4}{\sin ^{ - 1}}\dfrac{{2\sqrt 2 }}{3}

Explanation

Solution

Here we are asked to prove the given trigonometric expression. For that, we will consider the left hand side expression and then we will simplify it. We will then use different inverse trigonometric identities to further simplify it in such a way that we will get the right hand side expression.

Complete step-by-step answer:
Here we are asked to prove the given trigonometric expression.
We will first consider the left hand side expression 9π8−94sin⁡−113\dfrac{{9\pi }}{8} - \dfrac{9}{4}{\sin ^{ - 1}}\dfrac{1}{3}.
Now, we will take the term 94\dfrac{9}{4} from both the terms.
9π8−94sin⁡−113=94(π2−sin⁡−113)\dfrac{{9\pi }}{8} - \dfrac{9}{4}{\sin ^{ - 1}}\dfrac{1}{3} = \dfrac{9}{4}\left( {\dfrac{\pi }{2} - {{\sin }^{ - 1}}\dfrac{1}{3}} \right)
We know from inverse trigonometric identities that π2−sin⁡−1x=cos⁡−1x\dfrac{\pi }{2} - {\sin ^{ - 1}}x = {\cos ^{ - 1}}x
Therefore, using this inverse trigonometric identity here, we get
⇒9π8−94sin⁡−113=94cos⁡−113\Rightarrow \dfrac{{9\pi }}{8} - \dfrac{9}{4}{\sin ^{ - 1}}\dfrac{1}{3} = \dfrac{9}{4}{\cos ^{ - 1}}\dfrac{1}{3}
Let cos⁡−113=θ{\cos ^{ - 1}}\dfrac{1}{3} = \theta .
So we get,
⇒9π8−94sin⁡−113=94θ\Rightarrow \dfrac{{9\pi }}{8} - \dfrac{9}{4}{\sin ^{ - 1}}\dfrac{1}{3} = \dfrac{9}{4}\theta ………… (1)\left( 1 \right)
As we have assume cos⁡−113=θ{\cos ^{ - 1}}\dfrac{1}{3} = \theta , we can write it as cos⁡θ=13\cos \theta = \dfrac{1}{3}
We know the basic trigonometric formula that sin⁡θ=1−cos⁡2θ\sin \theta = \sqrt {1 - {{\cos }^2}\theta }
Using this, we get
sin⁡θ=1−(13)2\sin \theta = \sqrt {1 - {{\left( {\dfrac{1}{3}} \right)}^2}}
On further simplifying the terms, we get
⇒sin⁡θ=1−19\Rightarrow \sin \theta = \sqrt {1 - \dfrac{1}{9}}
On subtracting the numbers, we get
⇒sin⁡θ=89\Rightarrow \sin \theta = \sqrt {\dfrac{8}{9}}
On further simplification, we get
⇒sin⁡θ=223\Rightarrow \sin \theta = \dfrac{{2\sqrt 2 }}{3}
We can write it as
⇒θ=sin⁡−1223\Rightarrow \theta = {\sin ^{ - 1}}\dfrac{{2\sqrt 2 }}{3}
Now, we will substitute the value of θ\theta in equation (1)\left( 1 \right).
9π8−94sin⁡−113=94sin⁡−1223\dfrac{{9\pi }}{8} - \dfrac{9}{4}{\sin ^{ - 1}}\dfrac{1}{3} = \dfrac{9}{4}{\sin ^{ - 1}}\dfrac{{2\sqrt 2 }}{3}
We can see that this is equal to right hand side expression. Hence, proved.

Note: Here we have used inverse trigonometric identity and basic trigonometric formulas to solve the question. Trigonometric identities are defined as the equality which contains the trigonometric functions and it is true for all values of the variable. We need to consider the complex part of the equation to prove the equation and we simplify it and make it equal to the expression of the other side of the expression.