Question

Proving Identities

Verify the identity${\sin }^{4}x-{\cos }^{4}x=\left({\sin }^{2}x-{\cos }^{2}x\right)$

Short answer

The expression${\sin }^{4}x-{\cos }^{4}x=\left({\sin }^{2}x-{\cos }^{2}x\right)$ is an identity.

Step-by-step solution

Step 1. Given information.

An expression${\sin }^{4}x-{\cos }^{4}x=\left({\sin }^{2}x-{\cos }^{2}x\right)$

Step 2. Concept used.

To answer this sort of question, we first separate the expression into two parts: LHS and RHS. Then simplify both portions. If the answers are equivalent after simplification, the supplied expression is an identity.

Step 3. Calculation.

Now, simplify LHS of ${\sin }^{4}x-{\cos }^{4}x=\left({\sin }^{2}x-{\cos }^{2}x\right)$:

Use the identity as-

$A^2-B^2=(A+B)(A-B)$

$\begin{array}{l}={\sin }^{4}x-{\cos }^{4}x\\ ={\left({\sin }^2\theta \right)}^2-{\left({\cos }^2\theta \right)}^2\\ =\left({\sin }^{2}x-{\cos }^{2}x\right)\left({\sin }^{2}x+{\cos }^{2}x\right)\\ =\left({\sin }^{2}x-{\cos }^{2}x\right)\end{array}$

RHS of the equation is$\left({\sin }^{2}x-{\cos }^{2}x\right)$

Both RHS and LHS are equal. Hence it is an identity