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Question: Prove that \[{\sin ^4}\theta + {\cos ^4}\theta = 1 - 2{\sin ^2}\theta {\cos ^2}\theta \]...

Prove that sin⁡4θ+cos⁡4θ=1−2sin⁡2θcos⁡2θ{\sin ^4}\theta + {\cos ^4}\theta = 1 - 2{\sin ^2}\theta {\cos ^2}\theta

Explanation

Solution

In this question, we need to take anyone's side of the equation and then put identities of trigonometry and squares as per our requirement of the question and simplify the equation till that point where our L.H.S is equal to R.H.S.

Formula used:
(a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1

Complete step-by-step answer:
Below is the equation we need to equate
sin⁡4θ+cos⁡4θ=1−2sin⁡2θcos⁡2θ{\sin ^4}\theta + {\cos ^4}\theta = 1 - 2{\sin ^2}\theta {\cos ^2}\theta
First of all we will simplify L.H.S. of the above equation.
L.H.S. sin⁡4θ+cos⁡4θ......(1){\sin ^4}\theta + {\cos ^4}\theta ......(1)
Here, we let sin⁡2θ=x{\sin ^2}\theta = x
And, cos⁡2θ=y{\cos ^2}\theta = y
Now, we will put the values which we have let above in equation (1)
=x2+y2......(2)= {x^2} + {y^2}......(2)
Here, we will use an identity (a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab which we can write as per our requirement as: a2+b2=(a+b)2−2ab{a^2} + {b^2} = {(a + b)^2} - 2aband put in equation (2)
=(x+y)2−2xy......(3)= {(x + y)^2} - 2xy......(3)
We will put original values of xx and yyin the equation (3)
=(sin⁡2θ+cos⁡2θ)2−2sin⁡2θcos⁡2θ......(4)= {({\sin ^2}\theta + {\cos ^2}\theta )^2} - 2{\sin ^2}\theta {\cos ^2}\theta ......(4)
We know a trigonometry identity where sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 we will use this value in the equation (4)
=1−2sin⁡2θcos⁡2θ= 1 - 2{\sin ^2}\theta {\cos ^2}\theta = R.H.S. of our equation
Hence proved L.H.S. = R.H.S

Note: Second method to solve the above question:
sin⁡4θ+cos⁡4θ=1−2sin⁡2θcos⁡2θ{\sin ^4}\theta + {\cos ^4}\theta = 1 - 2{\sin ^2}\theta {\cos ^2}\theta
First of all we will take the negative term left side from right side of equation.
sin⁡4θ+cos⁡4θ+2sin⁡2θcos⁡2θ=1......(1){\sin ^4}\theta + {\cos ^4}\theta + 2{\sin ^2}\theta {\cos ^2}\theta = 1......(1)
We will simplify only L.H.S. of the equation
We let sin⁡2θ=x{\sin ^2}\theta = x
And, cos⁡2θ=y{\cos ^2}\theta = y
Now, put these values in the equation (1)
=x2+y2+2xy......(2)= {x^2} + {y^2} + 2xy......(2)
As we know, an identity is suitable for above equation, which is (a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab we will put this identity in the equation (2)
And we will get =(x+y)2......(3) = {(x + y)^2}......(3)
We will take square root of the equation (3)
=(x+y)2......(4)= \sqrt {{{(x + y)}^2}} ......(4)
We know that square root of a pair will be individual value, and put original value ofxx and yyin the equation (4)
And, we know a trigonometry identity where sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
=x+y=sin⁡2θ+cos⁡2θ=1= x + y = {\sin ^2}\theta + {\cos ^2}\theta = 1 = R.H.S.
Hence proved, L.H.S = R.H.S