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Question: How do you verify \( \sin x + \cos x = \dfrac{{2{{\sin }^2}x - 1}}{{\sin x - \cos x}} \) ?...

How do you verify sin⁡x+cos⁡x=2sin⁡2x−1sin⁡x−cos⁡x\sin x + \cos x = \dfrac{{2{{\sin }^2}x - 1}}{{\sin x - \cos x}} ?

Explanation

Solution

Hint : Here, you are given an equality which consists of the trigonometric ratio sine and cosine and you are asked to verify the identity. What you need to do here is first consider the left-hand side of the equation and multiply and divide it by a certain term which will help you to simplify the equation. By simplifying what is meant is that bring down the left-hand side equal to the right-hand side. Also, you need to recall the trigonometric identity sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 in order to solve the question.

Complete step by step solution:
The left hand side of the equation is sin⁡x+cos⁡x\sin x + \cos x and the right hand side of the equation is 2sin⁡2x−1sin⁡x−cos⁡x\dfrac{{2{{\sin }^2}x - 1}}{{\sin x - \cos x}} . You can see that the denominator of the right hand side is sin⁡x−cos⁡x\sin x - \cos x . From this, we take an idea of multiplying and dividing the left hand side by sin⁡x−cos⁡x\sin x - \cos x . So, we get,
sin⁡x+cos⁡x=(sin⁡x−cos⁡xsin⁡x−cos⁡x)(sin⁡x+cos⁡x)\sin x + \cos x = \left( {\dfrac{{\sin x - \cos x}}{{\sin x - \cos x}}} \right)\left( {\sin x + \cos x} \right) .
Now, let us multiply the numerator of sin⁡x−cos⁡xsin⁡x−cos⁡x\dfrac{{\sin x - \cos x}}{{\sin x - \cos x}} with sin⁡x+cos⁡x\sin x + \cos x .
(sin⁡x+cos⁡x)(sin⁡x−cos⁡x) =sin⁡xsin⁡x−sin⁡xcos⁡x+cos⁡xsin⁡x−cos⁡xcos⁡x =sin⁡2x−cos⁡2x  \left( {\sin x + \cos x} \right)\left( {\sin x - \cos x} \right) \\\ = \sin x\sin x - \sin x\cos x + \cos x\sin x - \cos x\cos x \\\ = {\sin ^2}x - {\cos ^2}x \\\
So, we have sin⁡x+cos⁡x=sin⁡2x−cos⁡2xsin⁡x−cos⁡x\sin x + \cos x = \dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x - \cos x}} . Now, we will make use of the trigonometric equation which gives you the relation between the sum of squares of sine and cosine of an angle. It is sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 . From here, we can write cos⁡2x=1−sin⁡2x{\cos ^2}x = 1 - {\sin ^2}x . Let us substitute this value of cos⁡2x{\cos ^2}x in the above obtained expression.
⇒sin⁡x+cos⁡x=sin⁡2x−cos⁡2xsin⁡x−cos⁡x ⇒sin⁡x+cos⁡x=sin⁡2x−(1−sin⁡2x)sin⁡x−cos⁡x ⇒sin⁡x+cos⁡x=2sin⁡2x−1sin⁡x−cos⁡x  \Rightarrow \sin x + \cos x = \dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x - \cos x}} \\\ \Rightarrow \sin x + \cos x = \dfrac{{{{\sin }^2}x - \left( {1 - {{\sin }^2}x} \right)}}{{\sin x - \cos x}} \\\ \Rightarrow \sin x + \cos x = \dfrac{{2{{\sin }^2}x - 1}}{{\sin x - \cos x}} \\\
Hence proved. Now, you could have gone the other way round, that is, you could have first considered the right hand side of the equation and then could have followed the reverse process. But here, you are expected to get the idea of writing that 11 as 1=sin⁡2x+cos⁡2x1 = {\sin ^2}x + {\cos ^2}x .
So, we have verified that sin⁡x+cos⁡x=2sin⁡2x−1sin⁡x−cos⁡x\sin x + \cos x = \dfrac{{2{{\sin }^2}x - 1}}{{\sin x - \cos x}} .

Note : Here, we used the basic and fundamental property of trigonometry that the sum of square of sine and cosine of an angle is equal to one. You need to remember the trick we used in the second method where we expressed 11 as 1=sin⁡2x+cos⁡2x1 = {\sin ^2}x + {\cos ^2}x , this idea is crucial in solving many questions. Also, you need to memorize trigonometric properties such as 1=sin⁡2x+cos⁡2x1 = {\sin ^2}x + {\cos ^2}x .