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Question: If \(\cos \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{5}} \right) + {{\cos }^{ - 1}}x} \right) = 0\) ,...

If cos⁡(sin⁡−1(25)+cos⁡−1x)=0\cos \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{5}} \right) + {{\cos }^{ - 1}}x} \right) = 0 , find x

Explanation

Solution

We can take cos inverse on both sides. Then we can rearrange and take cosine functions of both sides. Then we can simplify the equation using trigonometric identities. Then after further simplification, we will get the required value of x.

Complete step by step solution:
We have the equation cos⁡(sin⁡−1(25)+cos⁡−1x)=0\cos \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{5}} \right) + {{\cos }^{ - 1}}x} \right) = 0
We can take cos inverse on both sides. So, we will get,
⇒cos⁡−1[cos⁡(sin⁡−1(25)+cos⁡−1x)]=cos⁡−10\Rightarrow {\cos ^{ - 1}}\left[ {\cos \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{5}} \right) + {{\cos }^{ - 1}}x} \right)} \right] = {\cos ^{ - 1}}0
We know that cos and con inverse are inverse operations. So, they will cancel out. Thus, we can write
⇒(sin⁡−1(25)+cos⁡−1x)=cos⁡−10\Rightarrow \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{5}} \right) + {{\cos }^{ - 1}}x} \right) = {\cos ^{ - 1}}0
We know that cos⁡π2=0\cos \dfrac{\pi }{2} = 0 . So cos⁡−10=π2{\cos ^{ - 1}}0 = \dfrac{\pi }{2} . Thus, the equation will become,
⇒sin⁡−1(25)+cos⁡−1x=π2\Rightarrow {\sin ^{ - 1}}\left( {\dfrac{2}{5}} \right) + {\cos ^{ - 1}}x = \dfrac{\pi }{2}
Now we can take the term with x on one side. For that we can subtract both sides with sin⁡−1(25){\sin ^{ - 1}}\left( {\dfrac{2}{5}} \right) .
⇒cos⁡−1x=π2−sin⁡−1(25)\Rightarrow {\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{2}{5}} \right)
Now we can again take cosine on both sides. So, we will get,
⇒cos⁡(cos⁡−1x)=cos⁡(π2−sin⁡−1(25))\Rightarrow \cos \left( {{{\cos }^{ - 1}}x} \right) = \cos \left( {\dfrac{\pi }{2} - {{\sin }^{ - 1}}\left( {\dfrac{2}{5}} \right)} \right)
We know that cos and con inverse are inverse operations. So, they will cancel out in the LHS. Thus, we can write,
⇒x=cos⁡(π2−sin⁡−1(25))\Rightarrow x = \cos \left( {\dfrac{\pi }{2} - {{\sin }^{ - 1}}\left( {\dfrac{2}{5}} \right)} \right)
We know that cos⁡(π2−θ)=sin⁡θ\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta . So, the equation will become,
⇒x=sin⁡(sin⁡−1(25))\Rightarrow x = \sin \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{5}} \right)} \right)
As sine and sine inverse are inverse operations, they will cancel each other. So, the equation will become,
⇒x=25\Rightarrow x = \dfrac{2}{5}

Thus, the required value of x is 25\dfrac{2}{5}.

Note:
The properties of trigonometry used in this problem are,
cos⁡(π2−θ)=sin⁡θ\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta
cos⁡π2=0\cos \dfrac{\pi }{2} = 0
cos⁡(cos⁡−1x)=x\cos \left( {{{\cos }^{ - 1}}x} \right) = x
cos⁡−1(cos⁡θ)=θ{\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta
We must not expand the bracket and separate the term inside the function. We must be familiar with the values of trigonometric functions at common angles. If we know the value of the function at different angles, we can find the inverse when the value is given. The domain of sine inverse and cos inverse is from negative one to one.