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Question: Prove that \({{\sin }^{4}}A-{{\cos }^{4}}A=1-2{{\cos }^{2}}A\) using trigonometric identities....

Prove that sin⁡4A−cos⁡4A=1−2cos⁡2A{{\sin }^{4}}A-{{\cos }^{4}}A=1-2{{\cos }^{2}}A using trigonometric identities.

Explanation

Solution

We will first convert Left hand side terms i.e. LHS into cos terms using the trigonometry identity given as sin⁡2A+cos⁡2A=1{{\sin }^{2}}A+{{\cos }^{2}}A=1 . Then, we will use formula of (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} to expand the equation. Thus, on further solving we will get answer as that of right-hand side i.e. (RHS).

Complete step-by-step answer:
Here, we are given the left hand side (LHS) as sin⁡4A−cos⁡4A{{\sin }^{4}}A-{{\cos }^{4}}A . And the right-hand side (RHS) is 1−2cos⁡2A1-2{{\cos }^{2}}A . So, we will first try to convert LHS in cos terms as RHS is in cos term. So, we can write
LHS=sin⁡4A−cos⁡4ALHS={{\sin }^{4}}A-{{\cos }^{4}}A
LHS=(sin⁡2A)2−cos⁡4ALHS={{\left( {{\sin }^{2}}A \right)}^{2}}-{{\cos }^{4}}A
Now, we will use the identity i.e. sin⁡2A+cos⁡2A=1{{\sin }^{2}}A+{{\cos }^{2}}A=1 . So, we can write the equation as
LHS=(1−cos⁡2A)2−cos⁡4ALHS={{\left( 1-{{\cos }^{2}}A \right)}^{2}}-{{\cos }^{4}}A
Now, we will use the formula (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} where here, a=1,b=cos⁡2Aa=1,b={{\cos }^{2}}A . So, we get as
LHS=1−2cos⁡2A+cos⁡4A−cos⁡4ALHS=1-2{{\cos }^{2}}A+{{\cos }^{4}}A-{{\cos }^{4}}A
We will cancel out positive negative terms, we will get as
LHS=1−2cos⁡2A=RHSLHS=1-2{{\cos }^{2}}A=RHS
Thus, this is equal to RHS. So, we have proved LHS=RHSLHS=RHS .
Thus, proved.

Note: One other approach of proving is by converting LHS in sin function using the identity sin⁡2A+cos⁡2A=1{{\sin }^{2}}A+{{\cos }^{2}}A=1 . So, from this we will get equation as LHS=sin⁡4A−(cos⁡2A)2LHS={{\sin }^{4}}A-{{\left( {{\cos }^{2}}A \right)}^{2}} . On further using the identity we get equation as
LHS=sin⁡4A−(1−sin⁡2A)2LHS={{\sin }^{4}}A-{{\left( 1-{{\sin }^{2}}A \right)}^{2}}
On expanding the equation, we get all the terms in sine function which will be
LHS=sin⁡4A−1−sin⁡4A+2sin⁡2ALHS={{\sin }^{4}}A-1-{{\sin }^{4}}A+2{{\sin }^{2}}A
On cancelling the terms, we get LHS as
LHS=−1+2sin⁡2ALHS=-1+2{{\sin }^{2}}A .
Now, we will also convert RHS in sin terms and then compare LHS and RHS. So, RHS will be 1−2(1−sin⁡2A)1-2\left( 1-{{\sin }^{2}}A \right) . On further expanding, we get as
⇒1−2+2sin⁡2A=−1+2sin⁡2A\Rightarrow 1-2+2{{\sin }^{2}}A=-1+2{{\sin }^{2}}A
Thus, we can see the LHS=RHSLHS=RHS . Hence, proved.