Question

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

Answer 1

Hint: The functions sine, cosine, and tangent of an angle are sometimes referred to as the primary or basic trigonometric functions. Trigonometric identities are the equations involving the trigonometric functions that are true for every value of the variables involved. These identities are true for right-angled triangles. The Pythagorean identity of sine function is \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\].

Complete step by step answer:
As per the given question, we need to prove the given trigonometric expression using trigonometric identities and algebraic formulae. Here, we are given the expression \[{{\sin }^{6}}x+{{\cos }^{6}}x=1-3{{\sin }^{2}}x{{\cos }^{2}}x\] which we need to verify whether it is correct or not by solving one side of the equation.
Let us consider the equation \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\].
In the given expression, if we assume \[\sin \theta \] to be a and \[\cos \theta \] to be b, then we get the expression as
\[\Rightarrow \]\[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]
\[\Rightarrow {{a}^{2}}+{{b}^{2}}=1\]
Now cubing the above equation on both sides, we get
\[\Rightarrow {{({{a}^{2}}+{{b}^{2}})}^{3}}={{1}^{3}}\]
As we know the algebraic formulae \[{{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}\], we can write the above equation as
\[\Rightarrow ({{a}^{6}}+{{b}^{6}}+3{{a}^{4}}{{b}^{2}}+3{{a}^{2}}{{b}^{4}})=1\]
On substituting \[\sin \theta \] and \[\cos \theta \] in a and b, then we get the expression as
\[\Rightarrow \]\[{{\sin }^{6}}x+{{\cos }^{6}}x+3{{\sin }^{4}}x{{\cos }^{2}}x+3{{\sin }^{2}}x{{\cos }^{4}}x=1\]
According to the problem, the left-hand side of the equation is \[{{\sin }^{6}}x+{{\cos }^{6}}x\]. So, we shift the other two terms to the right-hand side of the equation. Then, we get
\[\Rightarrow \]\[{{\sin }^{6}}x+{{\cos }^{6}}x=1-3{{\sin }^{4}}x{{\cos }^{2}}x-3{{\sin }^{2}}x{{\cos }^{4}}x\]
Since in right-hand side \[3{{\sin }^{2}}x{{\cos }^{2}}x\] is common in both the terms, we can group them
\[\Rightarrow {{\sin }^{6}}x+{{\cos }^{6}}x=1-3{{\sin }^{2}}x{{\cos }^{2}}x({{\sin }^{2}}x+{{\cos }^{2}}x)\]
We know that \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\].
On substituting this value in the above equation, we get
\[\Rightarrow {{\sin }^{6}}x+{{\cos }^{6}}x=1-3{{\sin }^{2}}x{{\cos }^{2}}x(1)\]
\[\Rightarrow \]\[{{\sin }^{6}}x+{{\cos }^{6}}x=1-3{{\sin }^{2}}x{{\cos }^{2}}x\]
Hence proved.

Note:
In order to solve such types of questions, we need to have enough knowledge of trigonometric functions and identities. We also need to know the algebraic formulae to simplify the expressions. We must avoid calculation mistakes to get the expected answers.