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Question: How do you simplify \({\left( {\sin x - \cos x} \right)^2}\) ?...

How do you simplify (sin⁡x−cos⁡x)2{\left( {\sin x - \cos x} \right)^2} ?

Explanation

Solution

Look at the given expression or the function (sin⁡x−cos⁡x)2{\left( {\sin x - \cos x} \right)^2} which is in the form of (a−b)2{(a - b)^2}. So we can make use of the formula (a−b)2=a2−2ab+b2{(a - b)^2} = {a^2} - 2ab + {b^2} to simplify the given expression and then by making use of required trigonometric identities we can arrive at the required solution.

Complete step-by-step answer:
Whenever they give some expression to simplify, first think of the formulas which make the simplification easier.
So now by looking at the given expression (sin⁡x−cos⁡x)2{\left( {\sin x - \cos x} \right)^2} we can relate this to (a−b)2{(a - b)^2} form. If we use (a−b)2{(a - b)^2} form we can easily simplify the given expression, which is given by: (a−b)2=a2−2ab+b2{(a - b)^2} = {a^2} - 2ab + {b^2}.
Therefore, we can write the given expression (sin⁡x−cos⁡x)2{\left( {\sin x - \cos x} \right)^2} as,
(sin⁡x−cos⁡x)2=(sin⁡x)2−2sin⁡xcos⁡x+(cos⁡x)2{\left( {\sin x - \cos x} \right)^2} = {(\sin x)^2} - 2\sin x\cos x + {(\cos x)^2}
The above equation can be rewrite as below,
⇒(sin⁡x−cos⁡x)2=sin⁡2x−2sin⁡xcos⁡x+cos⁡2x\Rightarrow {\left( {\sin x - \cos x} \right)^2} = {\sin ^2}x - 2\sin x\cos x + {\cos ^2}x or we can also write as below,
⇒(sin⁡x−cos⁡x)2=sin⁡2x+cos⁡2x−2sin⁡xcos⁡x\Rightarrow {\left( {\sin x - \cos x} \right)^2} = {\sin ^2}x + {\cos ^2}x - 2\sin x\cos x
From the trigonometric identities we know that sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 . So now by substituting sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 in the above expression we can even make the expression simple, so we get
⇒(sin⁡x−cos⁡x)2=1−2sin⁡xcos⁡x\Rightarrow {\left( {\sin x - \cos x} \right)^2} = 1 - 2\sin x\cos x
From the above expression we can notice that we have 2sin⁡xcos⁡x2\sin x\cos x which belongs to trigonometric double angle identities. Therefore from the trigonometric double angle identities we have sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x . Hence we can replace 2sin⁡xcos⁡x2\sin x\cos x by sin⁡2x\sin 2x in the above expression, we get
⇒(sin⁡x−cos⁡x)2=1−sin⁡2x\Rightarrow {\left( {\sin x - \cos x} \right)^2} = 1 - \sin 2x .

Therefore the required answer or the simplified form of (sin⁡x−cos⁡x)2{\left( {\sin x - \cos x} \right)^2} is 1−sin⁡2x1 - \sin 2x . Hence we can write (sin⁡x−cos⁡x)2=1−sin⁡2x{\left( {\sin x - \cos x} \right)^2} = 1 - \sin 2x.

Note: Whenever they give any trigonometric functions for simplification purposes then there will be the usage of trigonometric identities. So make sure to remember all the trigonometric identities formula otherwise we will feel difficulty while solving the problem. If we know the formulas then we can solve the problem very quickly.