Question

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

Answer 1

Hint: 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:
${(a + b)^2} = {a^2} + {b^2} + 2ab$
${\sin ^2}\theta + {\cos ^2}\theta = 1$

Complete step-by-step answer:
Below is the equation we need to equate
\[{\sin ^4}\theta + {\cos ^4}\theta = 1 - 2{\sin ^2}\theta {\cos ^2}\theta \]
First of all we will simplify L.H.S. of the above equation.
L.H.S. \[{\sin ^4}\theta + {\cos ^4}\theta ......(1)\]
Here, we let ${\sin ^2}\theta = x$
And, ${\cos ^2}\theta = y$
Now, we will put the values which we have let above in equation (1)
$ = {x^2} + {y^2}......(2)$
Here, we will use an identity ${(a + b)^2} = {a^2} + {b^2} + 2ab$ which we can write as per our requirement as: ${a^2} + {b^2} = {(a + b)^2} - 2ab$and put in equation (2)
$ = {(x + y)^2} - 2xy......(3)$
We will put original values of $x$ and $y$in the equation (3)
$ = {({\sin ^2}\theta + {\cos ^2}\theta )^2} - 2{\sin ^2}\theta {\cos ^2}\theta ......(4)$
We know a trigonometry identity where ${\sin ^2}\theta + {\cos ^2}\theta = 1$ we will use this value in the equation (4)
$ = 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:
\[{\sin ^4}\theta + {\cos ^4}\theta = 1 - 2{\sin ^2}\theta {\cos ^2}\theta \]
First of all we will take the negative term left side from right side of equation.
\[{\sin ^4}\theta + {\cos ^4}\theta + 2{\sin ^2}\theta {\cos ^2}\theta = 1......(1)\]
We will simplify only L.H.S. of the equation
We let ${\sin ^2}\theta = x$
And, ${\cos ^2}\theta = y$
Now, put these values in the equation (1)
$ = {x^2} + {y^2} + 2xy......(2)$
As we know, an identity is suitable for above equation, which is ${(a + b)^2} = {a^2} + {b^2} + 2ab$ we will put this identity in the equation (2)
And we will get $ = {(x + y)^2}......(3)$
We will take square root of the equation (3)
\[ = \sqrt {{{(x + y)}^2}} ......(4)\]
We know that square root of a pair will be individual value, and put original value of$x$ and $y$in the equation (4)
And, we know a trigonometry identity where ${\sin ^2}\theta + {\cos ^2}\theta = 1$
$ = x + y = {\sin ^2}\theta + {\cos ^2}\theta = 1$ = R.H.S.
Hence proved, L.H.S = R.H.S