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Question: Prove that: \({\sin ^2}(x) = \left( {\dfrac{1}{2}} \right)(1 - \cos 2x)\)...

Prove that: sin⁡2(x)=(12)(1−cos⁡2x){\sin ^2}(x) = \left( {\dfrac{1}{2}} \right)(1 - \cos 2x)

Explanation

Solution

We have to prove the given trigonometric equation here in the question. For that we take either side of the equation and using trigonometric identities we obtain the other side or we apply identities in the equation itself and equate the equation in the end.

Complete solution step by step:
Firstly we write down the trigonometric equation given in the question:
sin⁡2x=(12)(1−cos⁡2x){\sin ^2}x = \left( {\dfrac{1}{2}} \right)\left( {1 - \cos 2x} \right)
Now we can use two methods to prove the above equation true. While using the first method, we have to take one side of the equation and using trigonometric identities we get to the other side of the equation i.e.
Taking the RHS of the equation we have
(12)(1−cos⁡2x)  - - - - - - (1)\left( {\dfrac{1}{2}} \right)\left( {1 - \cos 2x} \right)\,{\text{ - - - - - - (1)}}
We know that
cos⁡2x=1−2sin⁡2x=2cos⁡2x−1\cos 2x = 1 - 2{\sin ^2}x = 2{\cos ^2}x - 1
We have to get the LHS of the equation which consist of sine function so we take the first two parts of the identity and put it in equation (1) and we have
(12)(1−(1−2sin⁡2x))=(12)(1−1+2sin⁡2x) =(12)(2sin⁡2x) =sin⁡2x  \left( {\dfrac{1}{2}} \right)\left( {1 - \left( {1 - 2{{\sin }^2}x} \right)} \right) = \left( {\dfrac{1}{2}} \right)\left( {1 - 1 + 2{{\sin }^2}x} \right) \\\ = \left( {\dfrac{1}{2}} \right)\left( {2{{\sin }^2}x} \right) \\\ = {\sin ^2}x \\\
So we have got to our LHS and hence proved the given equation true.
Now we use the other method where we apply the identities in the equation itself to change then equation in one trigonometric function and after simplifying, in the end, equate the equation like this
sin⁡2x=(12)(1−cos⁡2x){\sin ^2}x = \left( {\dfrac{1}{2}} \right)(1 - \cos 2x)
We change the whole function into sine function using the identity
cos⁡2x=1−2sin⁡2x =2cos⁡2x−1  \cos 2x = 1 - 2{\sin ^2}x \\\ = 2{\cos ^2}x - 1 \\\
So we have

⇒sin⁡2x=(12)(1−(1−2sin⁡2x)) ⇒sin⁡2x=2sin⁡2x2 ⇒sin⁡2x=sin⁡2x ∴1=1 \Rightarrow {\sin ^2}x = \left( {\dfrac{1}{2}} \right)\left( {1 - \left( {1 - 2{{\sin }^2}x} \right)} \right) \\\ \Rightarrow {\sin ^2}x = \dfrac{{2{{\sin }^2}x}}{2} \\\ \Rightarrow {\sin ^2}x = {\sin ^2}x \\\ \therefore 1 = 1 \\\

We have equated the equation and hence proved the expression true.

Note: While encountering questions like these our aim is always to convert the whole equation into sine or cosine trigonometric functions, use standard identities to simplify and prove the expression true. For a simpler equations method second is apt and for complex and large equations we should use the first one.