Question

How do you prove ${(\sin x + \cos x)^4} = {(1 + 2\sin x\cos x)^2}$ ?

Answer 1

Hint: To prove the given trigonometric identity we can start with the LHS given in exponential form and expand the expression. Further we would transform the expression such that it resembles the expression given in the RHS to prove that LHS = RHS.

Formula used:
${a^m} \times {a^n} = {a^{m + n}}$
${(a + b)^2} = {a^2} + 2ab + {b^2}$
${\sin ^2}x + {\cos ^2}x = 1$

Complete step by step solution:
We have to prove that ${(\sin x + \cos x)^4} = {(1 + 2\sin x\cos x)^2}$.
We will start with the LHS, i.e. ${(\sin x + \cos x)^4}$, and try to transform the expression in such a way that it resembles the expression given in the RHS.
The expression given in the LHS is a trigonometric expression raised to exponent $4$. We can expand this expression in various ways. Here we will try to expand this expression first by writing it as (using the basic exponential property: ${a^m} \times {a^n} = {a^{m + n}}$),
${(\sin x + \cos x)^4} = {(\sin x + \cos x)^2} \times {(\sin x + \cos x)^2}$
We could have expanded it by writing $(\sin x + \cos x)$ multiplied by itself $4$times. But here we are expanding the expression as suited for our solution.
Now we can expand it further by using ${(a + b)^2} = {a^2} + 2ab + {b^2}$ ,
${(\sin x + \cos x)^2} = {\sin ^2}x + 2\sin x\cos x + {\cos ^2}x$
We know the basic trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$. Using this in the above expression we get,
${(\sin x + \cos x)^2} = {\sin ^2}x + 2\sin x\cos x + {\cos ^2}x = 1 + 2\sin x\cos x$
Thus we can write the LHS in the form of,
$
  {(\sin x + \cos x)^4} \\
   = {(\sin x + \cos x)^2} \times {(\sin x + \cos x)^2} \\
   = (1 + 2\sin x\cos x) \times (1 + 2\sin x\cos x) \\
   = {(1 + 2\sin x\cos x)^2} \\
$
Thus, we get LHS = ${(1 + 2\sin x\cos x)^2}$ = RHS.
Hence we proved the trigonometric identity ${(\sin x + \cos x)^4} = {(1 + 2\sin x\cos x)^2}$.

Note: For any value of the angle $x$ this trigonometric identity will hold true, i.e. LHS will be equal to RHS. We can also prove this identity by transforming both LHS and RHS into simpler terms. We should not get intimidated by the higher exponents in the question. We can solve them easily if our basic approach to solve the problem is clear.