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)=21(2−sin22x).
Solution
Hint: In order to solve this question, we need to know a few trigonometric identities like sin2x+cos2x=1 and 2sinxcosx=sin2x. Also, we should know some algebraic identities like (a+b)2=a2+b2+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)=21(2−sin22x). To prove this, we will first consider the left hand side (LHS), so we will write it as,
LHS=cos4x+sin4x
We can also express the terms as squares, so we get the LHS as,
LHS=(cos2x)2+(sin2x)2
We know that (a+b)2=a2+b2+2ab. From this we can find the value of a2+b2 as a2+b2=(a+b)2−2ab.Now, we can apply this formula to the terms in the LHS where we can consider a=cos2x and b=sin2x. So we can write the LHS as,
LHS=(cos2x+sin2x)2−2cos2xsin2x
And we know that, sin2x+cos2x=1. So, we will get,
LHS=(1)2−2cos2xsin2x
Now, we will multiply and divide the LHS by 2. So we will get,
LHS=21[2−4cos2xsin2x]
We can express the term 4cos2xsin2x as a square that is (2sin2xcos2x)2.
Hence we will get the LHS as, LHS=21[2−(2sinxcosx)2]
We know that, 2sinxcosx=sin2x. So, we will get the LHS as,
LHS=21[2−sin2x]
We know that 21[2−sin2x] is the RHS of the expression given in the question. Thus we get that LHS=RHS. Hence we have proved the expression, that is, (cos4x+sin4x)=21(2−sin22x).
Note: In this question, we can easily start from the right hand side (RHS) by writing 2=2(sin2x+cos2x)2 and sin22x=(2sinxcosx)2. And then on simplification, we will get the answer and we can prove the expression.