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Question: If \(y = {\sin ^{ - 1}}(\cos x) + {\cos ^{ - 1}}(\sin x)\), prove that \(\dfrac{{dy}}{{dx}} = - 2\)....

If y=sin⁡−1(cos⁡x)+cos⁡−1(sin⁡x)y = {\sin ^{ - 1}}(\cos x) + {\cos ^{ - 1}}(\sin x), prove that dydx=−2\dfrac{{dy}}{{dx}} = - 2.

Explanation

Solution

In this question we are given an equation and we have to prove that the differentiation of the given equation is -2. We will start by differentiating the given equation and move forward by simplifying it using some simple trigonometric identities. Then, we can cancel the like terms and it will give us the desired result.

Formula used: 1) sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1
2) d(cos⁡−1x)dx=−11−x2\dfrac{{d({{\cos }^{ - 1}}x)}}{{dx}} = \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}
3) d(sin⁡−1x)dx=11−x2\dfrac{{d({{\sin }^{ - 1}}x)}}{{dx}} = \dfrac{1}{{\sqrt {1 - {x^2}} }}
4) cos⁡−1x+sin⁡−1x=π2{\cos ^{ - 1}}x + {\sin ^{ - 1}}x = \dfrac{\pi }{2}

Complete step-by-step answer:
We are given that y=sin⁡−1(cos⁡x)+cos⁡−1(sin⁡x)y = {\sin ^{ - 1}}(\cos x) + {\cos ^{ - 1}}(\sin x). We will start by differentiating the given equation.
y=sin⁡−1(cos⁡x)+cos⁡−1(sin⁡x)y = {\sin ^{ - 1}}(\cos x) + {\cos ^{ - 1}}(\sin x)
Differentiate with respect to x,
⇒dydx=d[sin⁡−1(cos⁡x)]dx+d[cos⁡−1(sin⁡x)]dx\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d[{{\sin }^{ - 1}}(\cos x)]}}{{dx}} + \dfrac{{d[{{\cos }^{ - 1}}(\sin x)]}}{{dx}}
Using formula (2) and (3), and chain rule-
⇒dydx=−sin⁡x1−cos⁡2x+(−cos⁡x1−sin⁡2x)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \sin x}}{{\sqrt {1 - {{\cos }^2}x} }} + \left( {\dfrac{{ - \cos x}}{{\sqrt {1 - {{\sin }^2}x} }}} \right)
We know that sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1
Hence, 1−cos⁡2x=sin⁡2x1 - {\cos ^2}x = {\sin ^2}x and 1−sin⁡2x=cos⁡2x1 - {\sin ^2}x = {\cos ^2}x.
Using these trigonometric identities,
⇒dydx=−sin⁡xsin⁡2x−cos⁡xcos⁡2x\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \sin x}}{{\sqrt {{{\sin }^2}x} }} - \dfrac{{\cos x}}{{\sqrt {{{\cos }^2}x} }}
Solving further,
⇒dydx=−sin⁡xsin⁡x−cos⁡xcos⁡x\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \sin x}}{{\sin x}} - \dfrac{{\cos x}}{{\cos x}}
On simplifying,
⇒dydx=−1−1\Rightarrow \dfrac{{dy}}{{dx}} = - 1 - 1
⇒dydx=−2\Rightarrow \dfrac{{dy}}{{dx}} = - 2
Hence, proved.

Note: The given question can also be solved by the following method. In this method, we use inverse trigonometry identity to solve the question. In the previous method, we started by differentiating the given equation. However, in this method we will start by substituting the identities and will differentiate on a later stage.
We are given y=sin⁡−1(cos⁡x)+cos⁡−1(sin⁡x)y = {\sin ^{ - 1}}(\cos x) + {\cos ^{ - 1}}(\sin x)
We know that sin⁡−1x+cos⁡−1x=π2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}
Hence, sin⁡−1x=π2−cos⁡−1x{\sin ^{ - 1}}x = \dfrac{\pi }{2} - {\cos ^{ - 1}}x , and cos⁡−1x=π2−sin⁡−1x{\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}x.
Substituting this inverse trigonometric identity in the given equation,
⇒y=π2−cos⁡−1(cos⁡x)+π2−sin⁡−1(sin⁡x)\Rightarrow y = \dfrac{\pi }{2} - {\cos ^{ - 1}}(\cos x) + \dfrac{\pi }{2} - {\sin ^{ - 1}}(\sin x)
Simplifying the above equation,
⇒y=π2+π2−x−x\Rightarrow y = \dfrac{\pi }{2} + \dfrac{\pi }{2} - x - x ….. (cos⁡−1(cos⁡x)=x{\cos ^{ - 1}}(\cos x) = x)
Adding and subtracting,
⇒y=π−2x\Rightarrow y = \pi - 2x
Differentiate with respect to x,
⇒dydx=d(π−2x)dx\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d(\pi - 2x)}}{{dx}}
⇒dydx=−2\Rightarrow \dfrac{{dy}}{{dx}} = - 2
Hence, proved.