Question

How do you verify $ {\left( {\sin x + \cos x} \right)^2} = 1 + \sin 2x $ ?

Answer 1

Hint: When we are given an equation and we have to verify it; we will take one of the sides of the equals sign and will try to manipulate it in such a way that it becomes the same as the other side. Here, we have $ {\left( {\sin x + \cos x} \right)^2} $ as our LHS. As we can clearly see that it is in the form of $ {\left( {a + b} \right)^2} $ , we will apply the identity $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ to expand it and then some trigonometric formulae such as $ {\sin ^2}\theta + {\cos ^2}\theta = 1 $ and $ 2\sin \theta \cos \theta = 1 $ will help us to verify that LHS is equal to RHS.

Complete step-by-step answer:
(i)
We are given:
 $ {\left( {\sin x + \cos x} \right)^2} = 1 + \sin 2x $
In order to verify the above equation, we will take the LHS i.e.,
 $ {\left( {\sin x + \cos x} \right)^2} $
Now, we will try to manipulate it to make it the same as the RHS. Since, it is in the form of $ {\left( {a + b} \right)^2} $ , we will expand it using the identity:
 $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $
Therefore, we will get:
 $ {\left( {\sin x + \cos x} \right)^2} = {\sin ^2}x + 2\sin x\cos x + {\cos ^2}x $
(ii)
Now if we rearrange our result, it will look like:
 $ {\left( {\sin x + \cos x} \right)^2} = {\sin ^2}x + {\cos ^2}x + 2\sin x\cos x $
And as we know through the Pythagorean identity that,
 $ {\sin ^2}\theta + {\cos ^2}\theta = 1 $
We will substitute $ {\sin ^2}x + {\cos ^2}x $ in the expression we derived, with $ 1 $ . Therefore, it will become:
 $ {\left( {\sin x + \cos x} \right)^2} = 1 + 2\sin x\cos x $
(iii)
Now, as we know the double angle identity of $ \sin 2\theta $ i.e.,
 $ \sin 2\theta = 2\sin \theta \cos \theta $
We will substitute $ 2\sin x\cos x $ in the obtained expression with $ \sin 2x $ . Therefore, we will get:
 $ {\left( {\sin x + \cos x} \right)^2} = 1 + \sin 2x $
Hence, verified.
So, the correct answer is “ $ {\left( {\sin x + \cos x} \right)^2} = 1 + \sin 2x $ ”.

Note: For proving these types of equations, we should learn all the identities and formulae carefully and should use the apt identity at the correct place in order to obtain the desired results. This question could also be solved by taking RHS and manipulating it to LHS by writing $ 2\sin x\cos x $ in the place of $ \sin 2x $ and thinking of writing $ 1 $ as $ {\sin ^2}x + {\cos ^2}x $ . Once, we are able to reach till here, we can clearly see the expanded form of $ {\left( {a + b} \right)^2} $ i.e., \[{a^2} + 2ab + {b^2}\], so we will apply the same identity to obtain LHS. But as we can see, predicting LHS from RHS was more difficult than predicting RHS from LHS, therefore, we should make the correct choice on how to start the question in the first step.