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Question

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

Prove the following: (cos4x+sin4x)=12(2sin22x)\left( {{\cos }^{4}}x+{{\sin }^{4}}x \right)=\dfrac{1}{2}\left( 2-{{\sin }^{2}}2x \right).

Explanation

Solution

Hint: In order to solve this question, we need to know a few trigonometric identities like sin2x+cos2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 and 2sinxcosx=sin2x2\sin x\cos x=\sin 2x. Also, we should know some algebraic identities like (a+b)2=a2+b2+2ab{{\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 (cos4x+sin4x)=12(2sin22x)\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=cos4x+sin4xLHS={{\cos }^{4}}x+{{\sin }^{4}}x

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

LHS=(cos2x)2+(sin2x)2LHS={{\left( {{\cos }^{2}}x \right)}^{2}}+{{\left( {{\sin }^{2}}x \right)}^{2}}

We know that (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab. From this we can find the value of a2+b2{{a}^{2}}+{{b}^{2}} as a2+b2=(a+b)22ab{{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=cos2xa={{\cos }^{2}}x and b=sin2xb={{\sin }^{2}}x. So we can write the LHS as,

LHS=(cos2x+sin2x)22cos2xsin2xLHS={{\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)}^{2}}-2{{\cos }^{2}}x{{\sin }^{2}}x

And we know that, sin2x+cos2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1. So, we will get,

LHS=(1)22cos2xsin2xLHS={{\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=12[24cos2xsin2x]LHS=\dfrac{1}{2}\left[ 2-4{{\cos }^{2}}x{{\sin }^{2}}x \right]

We can express the term 4cos2xsin2x4{{\cos }^{2}}x{{\sin }^{2}}x as a square that is (2sin2xcos2x)2{{\left( 2{{\sin }^{2}}x{{\cos }^{2}}x \right)}^{2}}.

Hence we will get the LHS as, LHS=12[2(2sinxcosx)2]LHS=\dfrac{1}{2}\left[ 2-{{\left( 2\sin x\cos x \right)}^{2}} \right]

We know that, 2sinxcosx=sin2x2\sin x\cos x=\sin 2x. So, we will get the LHS as,

LHS=12[2sin2x]LHS=\dfrac{1}{2}\left[ 2-{{\sin }^{2}}x \right]

We know that 12[2sin2x]\dfrac{1}{2}\left[ 2-{{\sin }^{2}}x \right] is the RHS of the expression given in the question. Thus we get that LHS=RHSLHS=RHS. Hence we have proved the expression, that is, (cos4x+sin4x)=12(2sin22x)\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(sin2x+cos2x)22=2{{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}} and sin22x=(2sinxcosx)2{{\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.