Question

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

Answer 1

Hint: We will first convert Left hand side terms i.e. LHS into cos terms using the trigonometry identity given as ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ . Then, we will use formula of ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ 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 ${{\sin }^{4}}A-{{\cos }^{4}}A$ . And the right-hand side (RHS) is $1-2{{\cos }^{2}}A$ . So, we will first try to convert LHS in cos terms as RHS is in cos term. So, we can write
$LHS={{\sin }^{4}}A-{{\cos }^{4}}A$
$LHS={{\left( {{\sin }^{2}}A \right)}^{2}}-{{\cos }^{4}}A$
Now, we will use the identity i.e. ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ . So, we can write the equation as
$LHS={{\left( 1-{{\cos }^{2}}A \right)}^{2}}-{{\cos }^{4}}A$
Now, we will use the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ where here, $a=1,b={{\cos }^{2}}A$ . So, we get as
$LHS=1-2{{\cos }^{2}}A+{{\cos }^{4}}A-{{\cos }^{4}}A$
We will cancel out positive negative terms, we will get as
$LHS=1-2{{\cos }^{2}}A=RHS$
Thus, this is equal to RHS. So, we have proved $LHS=RHS$ .
Thus, proved.

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