Question

Prove the following: $\left( {{\cos }^{4}}x+{{\sin }^{4}}x \right)=\dfrac{1}{2}\left( 2-{{\sin }^{2}}2x \right)$.

Answer 1

Hint: In order to solve this question, we need to know a few trigonometric identities like ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and $2\sin x\cos x=\sin 2x$. Also, we should know some algebraic identities like ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. By using these identities, we can prove the given expression.

Complete step-by-step answer:

In this question, we have been asked to prove that $\left( {{\cos }^{4}}x+{{\sin }^{4}}x \right)=\dfrac{1}{2}\left( 2-{{\sin }^{2}}2x \right)$. To prove this, we will first consider the left hand side (LHS), so we will write it as,

$LHS={{\cos }^{4}}x+{{\sin }^{4}}x$

We can also express the terms as squares, so we get the LHS as,

$LHS={{\left( {{\cos }^{2}}x \right)}^{2}}+{{\left( {{\sin }^{2}}x \right)}^{2}}$

We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. From this we can find the value of ${{a}^{2}}+{{b}^{2}}$ as ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$.Now, we can apply this formula to the terms in the LHS where we can consider $a={{\cos }^{2}}x$ and $b={{\sin }^{2}}x$. So we can write the LHS as,

$LHS={{\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)}^{2}}-2{{\cos }^{2}}x{{\sin }^{2}}x$

And we know that, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. So, we will get,

$LHS={{\left( 1 \right)}^{2}}-2{{\cos }^{2}}x{{\sin }^{2}}x$

Now, we will multiply and divide the LHS by 2. So we will get,

$LHS=\dfrac{1}{2}\left[ 2-4{{\cos }^{2}}x{{\sin }^{2}}x \right]$

We can express the term $4{{\cos }^{2}}x{{\sin }^{2}}x$ as a square that is ${{\left( 2{{\sin }^{2}}x{{\cos }^{2}}x \right)}^{2}}$.

Hence we will get the LHS as, $LHS=\dfrac{1}{2}\left[ 2-{{\left( 2\sin x\cos x \right)}^{2}} \right]$

We know that, $2\sin x\cos x=\sin 2x$. So, we will get the LHS as,

$LHS=\dfrac{1}{2}\left[ 2-{{\sin }^{2}}x \right]$

We know that $\dfrac{1}{2}\left[ 2-{{\sin }^{2}}x \right]$ is the RHS of the expression given in the question. Thus we get that $LHS=RHS$. Hence we have proved the expression, that is, $\left( {{\cos }^{4}}x+{{\sin }^{4}}x \right)=\dfrac{1}{2}\left( 2-{{\sin }^{2}}2x \right)$.

Note: In this question, we can easily start from the right hand side (RHS) by writing $2=2{{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}$ and ${{\sin }^{2}}2x={{\left( 2\sin x\cos x \right)}^{2}}$. And then on simplification, we will get the answer and we can prove the expression.