Question

How to prove ${\sin ^6}x + {\cos ^6}x = 1 - 3{\sin ^2}x{\cos ^2}x$?

Answer 1

In this question, we need to prove ${\sin ^6}x + {\cos ^6}x = 1 - 3{\sin ^2}x{\cos ^2}x$. Here, we will consider LHS and use the formula from exponents to rewrite it. Then we will apply algebraic and trigonometric identities to evaluate and determine the proof.

Complete step-by-step solution:

We need to prove ${\sin ^6}x + {\cos ^6}x = 1 - 3{\sin ^2}x{\cos ^2}x$.

Now let us consider the LHS to prove the condition,
LHS=${\sin ^6}x + {\cos ^6}x$

From the formula of exponents, we know that ${\left( {{a^x}} \right)^m} = {a^{xm}}$.

Therefore, now we can rewrite ${\sin ^6}x + {\cos ^6}x$ as ${\left( {{{\sin }^2}x} \right)^3} + {\left( {{{\cos }^2}x} \right)^3}$.
LHS=${\left( {{{\sin }^2}x} \right)^3} + {\left( {{{\cos }^2}x} \right)^3}$

Now, we know that ${a^3} + {b^3} = {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right)$

Here, $a = {\sin ^2}x$ and $b = {\cos ^2}x$

Substituting the values in the formula, we have,
${\left( {{{\sin }^2}x} \right)^3} + {\left( {{{\cos }^2}x} \right)^3} = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3} - 3{\sin ^2}x{\cos ^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)$
We know from the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$.

Thus, by substituting the value we have,
${\left( {{{\sin }^2}x} \right)^3} + {\left( {{{\cos }^2}x} \right)^3} = {\left( 1 \right)^3} - 3{\sin ^2}x{\cos ^2}x\left( 1 \right)$
${\left( {{{\sin }^2}x} \right)^3} + {\left( {{{\cos }^2}x} \right)^3} = 1 - 3{\sin ^2}x{\cos ^2}x$
${\sin ^6}x + {\cos ^6}x = 1 - 3{\sin ^2}x{\cos ^2}x$

Hence proved, RHS=LHS.

Note: Whenever we come across these kinds of questions, we always start from the more complex side. This is because it is a lot easier to eliminate terms to make a complex function simple than to find ways to introduce terms to make a simple function complex. Take one step, watch one step. Providing
trigonometric functions are an art. There are often several ways to get to the answer. Naturally, some methods are more elegant and short while other methods are crude, massive and ugly. However, the key point to note is that whichever way we take, as long as we can get to the final destination, we will get the marks. Providing trigonometric functions becomes a piece of cake after we have conquered a massive number of questions and expose ourselves to all the different varieties of questions. So, practice well to make these problems like a piece of cake.