Question

How do you verify ${{\cos }^{4}}x-{{\sin }^{4}}x=1-2{{\sin }^{2}}x$ ?

Answer 1

Hint: In this question we have been asked to verify the given expression ${{\cos }^{4}}x-{{\sin }^{4}}x=1-2{{\sin }^{2}}x$ . For doing that we will simplify the left hand side of the expression using a trigonometric identity given as ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and try to reduce it as right hand side of the expression.

Complete step by step solution:
Now consider from the question we have been asked to verify the given expression ${{\cos }^{4}}x-{{\sin }^{4}}x=1-2{{\sin }^{2}}x$ .
From the basics of concepts of algebra we are aware of an identity given as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.Now we will use this identity and simplify the left hand side of the given trigonometric expression as follows $\Rightarrow {{\cos }^{4}}x-{{\sin }^{4}}x=\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)$ .
Now we will use the basic trigonometric identity given as $\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)=1$ for further simplify the left hand side of the expression and reducing it into the right hand side of the given expression which is $1-2{{\sin }^{2}}x$ . If the finalized reduced left hand side of the expression is equal to the given right hand side of the expression then the given trigonometric expression is verified.
Now by using the identity we will have
$\begin{align}
  & \Rightarrow 1\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)=\left( 1-{{\sin }^{2}}x-{{\sin }^{2}}x \right) \\
 & =1-2{{\sin }^{2}}x \\
\end{align}$.
Now we can conclude that the reduced left hand side of the given trigonometric expression is equal to the given right hand side of the expression. Therefore the expression is verified.

Note: While answering questions of this type we should be sure with our concepts that we apply and simplifications that we perform. Similarly we have two more trigonometric identities mathematically given as
 $\begin{align}
  & 1+{{\tan }^{2}}x={{\sec }^{2}}x \\
 & 1+{{\cot }^{2}}x={{\csc }^{2}}x \\
\end{align}$