Question

Prove that \[{\cos ^6}\left( x \right) + {\sin ^6}\left( x \right) = \dfrac{1}{8}\left( {5 + 3\cos 4x} \right)?\]

Answer 1

Hint: In this question, we need to prove that \[{\cos ^6}\left( x \right) + {\sin ^6}\left( x \right) = \dfrac{1}{8}\left( {5 + 3\cos 4x} \right)\] . Here, we will consider the left-hand side and use the formula from exponents to rewrite it. After that we will use algebraic and trigonometric identities to simplify and determine the proof.

Formula used:
Algebraic identities:
\[{a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)\]
\[{a^2} + {b^2} = {\left( {a - b} \right)^2} + 2ab\]
Trigonometric identities:
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
\[{\cos ^2}\theta - {\sin ^2}\theta = \cos 2\theta \]
\[2{\cos ^2}\theta = 1 + \cos 2\theta \]
\[\sin 2\theta = 2\sin \theta \cos \theta \]
\[2{\sin ^2}\theta = 1 - \cos 2\theta \]

Complete step by step answer:
We need to prove
\[{\cos ^6}\left( x \right) + {\sin ^6}\left( x \right) = \dfrac{1}{8}\left( {5 + 3\cos 4x} \right)\]
Now let us consider the LHS to prove the condition,
So, \[LHS = {\cos ^6}\left( x \right) + {\sin ^6}\left( x \right)\]
From the formula of exponents, we know that
\[{\left( {{x^p}} \right)^q} = {x^{pq}}\]
Therefore, we can write \[{\cos ^6}\left( x \right) + {\sin ^6}\left( x \right)\] as \[{\left( {{{\cos }^2}x} \right)^3} + {\left( {{{\sin }^2}x} \right)^3}\]
So, left-hand side becomes,
\[LHS = {\left( {{{\cos }^2}x} \right)^3} + {\left( {{{\sin }^2}x} \right)^3}\]
Now we know that
\[{a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)\]
Here, \[a = {\cos ^2}x\] and \[b = {\sin ^2}x\]
On substituting the values in the formula, we get
\[ = \left( {{{\cos }^2}x + {{\sin }^2}x} \right)\left( {{{\left( {{{\cos }^2}x} \right)}^2} + {{\left( {{{\sin }^2}x} \right)}^2} - {{\cos }^2}x \cdot {{\sin }^2}x} \right)\]
We know from the trigonometric identity that
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Thus, by substituting the value we have
\[ = \left( 1 \right)\left( {{{\left( {{{\cos }^2}x} \right)}^2} + {{\left( {{{\sin }^2}x} \right)}^2} - {{\cos }^2}x \cdot {{\sin }^2}x} \right)\]
Now we know that
\[{a^2} + {b^2} = {\left( {a - b} \right)^2} + 2ab\]
Here, \[a = {\cos ^2}x\] and \[b = {\sin ^2}x\]
On substituting the values in the formula, we have
\[ = \left( {{{\left( {{{\cos }^2}x - {{\sin }^2}x} \right)}^2} + 2{{\cos }^2}x \cdot {{\sin }^2}x - {{\cos }^2}x \cdot {{\sin }^2}x} \right)\]
On simplifying we get
\[ = \left( {{{\left( {{{\cos }^2}x - {{\sin }^2}x} \right)}^2} + {{\cos }^2}x \cdot {{\sin }^2}x} \right)\]
As we know from the trigonometric identity that
\[{\cos ^2}\theta - {\sin ^2}\theta = \cos 2\theta \]
Therefore, we get
\[ = \left( {{{\left( {\cos 2x} \right)}^2} + {{\cos }^2}x \cdot {{\sin }^2}x} \right)\]
\[ = \left( {{{\cos }^2}\left( {2x} \right) + {{\cos }^2}x \cdot {{\sin }^2}x} \right)\]
Multiply and divide by \[4\] we get
\[ = \dfrac{1}{4}\left( {4{{\cos }^2}\left( {2x} \right) + 4{{\cos }^2}x \cdot {{\sin }^2}x} \right)\]
Now using identity
\[2{\cos ^2}\theta = 1 + \cos 2\theta \]
Therefore, we get
\[ = \dfrac{1}{4}\left( {2\left( {1 + \cos 4x} \right) + 4{{\cos }^2}x \cdot {{\sin }^2}x} \right)\]
\[ = \dfrac{1}{4}\left( {2\left( {1 + \cos 4x} \right) + {{\left( {2\cos x \cdot \sin x} \right)}^2}} \right)\]
As we know that
\[\sin 2\theta = 2\sin \theta \cos \theta \]
Therefore, we have
\[ = \dfrac{1}{4}\left( {2\left( {1 + \cos 4x} \right) + {{\sin }^2}2x} \right)\]
On simplifying, we get
\[ = \dfrac{1}{4}\left( {2 + 2\cos 4x + {{\sin }^2}2x} \right)\]
Now multiply and divide by \[2\]
\[ = \dfrac{1}{{4 \cdot 2}}\left( {4 + 4\cos 4x + 2{{\sin }^2}2x} \right)\]
\[ = \dfrac{1}{8}\left( {4 + 4\cos 4x + 2{{\sin }^2}2x} \right)\]
Now using identity
\[2{\sin ^2}\theta = 1 - \cos 2\theta \]
we get
\[ = \dfrac{1}{8}\left( {4 + 4\cos 4x + 1 - \cos 4x} \right)\]
On simplifying, we get
\[ = \dfrac{1}{8}\left( {5 + 3\cos 4x} \right)\]
which is equal to the RHS
Hence proved.

Note: Whenever you come across these kinds of problems, always start from the more complex side. Also, you need to keep a check on both LHS and RHS while solving. Like in this question, when you were solving your LHS, you had to check in what terms your RHS is present, and then according to that you need to change your LHS to prove the question.