Question

Simplify the expression ${{\cos }^{4}}x-{{\sin }^{4}}x$

Answer 1

Hint: Use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Convert ${{\cos }^{4}}x-{{\sin }^{4}}x$ in the form of ${{a}^{2}}-{{b}^{2}}$ and use the previously mentioned identity. Use the fact that ${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$. Hence simplify the expression. Alternatively use Euler's identity to solve the expression. Use the fact that $\cos x=\dfrac{{{e}^{ix}}+{{e}^{-ix}}}{2}$ and $\sin x=\dfrac{{{e}^{ix}}-{{e}^{-ix}}}{2i}$. Use the binomial theorem to expand the terms ${{\left( {{e}^{ix}}+{{e}^{-ix}} \right)}^{4}}$ and ${{\left( {{e}^{ix}}-{{e}^{-ix}} \right)}^{4}}$ . Finally, simplify and use the identity ${{e}^{ix}}+{{e}^{-ix}}=2\cos x$ and hence find the simplified expression.

Complete step-by-step answer:
Let $S={{\cos }^{4}}x-{{\sin }^{4}}x$
We have S $={{\cos }^{4}}x-{{\sin }^{4}}x$
We know that ${{\cos }^{4}}x={{\left( {{\cos }^{2}}x \right)}^{2}}$ and ${{\sin }^{4}}x={{\left( {{\sin }^{2}}x \right)}^{2}}$
Hence, we have
${{\cos }^{4}}x-{{\sin }^{4}}x={{\left( {{\cos }^{2}}x \right)}^{2}}-{{\left( {{\sin }^{2}}x \right)}^{2}}$
Hence, we have
S $={{\left( {{\cos }^{2}}x \right)}^{2}}-{{\left( {{\sin }^{2}}x \right)}^{2}}$
We know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
Using the above identity, we get
S $=\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)$
We know that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$
Hence, we have
S $=\left( 1 \right)\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)$
We know that ${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$
Using the above identity, we get
S $=\cos 2x$
Hence S= cos2x
Hence, we have
${{\cos }^{4}}x-{{\sin }^{4}}x=\cos 2x$ which is the required simplified form of the expression.

Note: Alternative solution:
We know from Euler’s identity $\cos x+i\sin x={{e}^{ix}}$
Hence, we have $\cos x+i\sin x={{e}^{ix}}\text{ (i)}$
Replace x by -x, we get
$\cos x-i\sin x={{e}^{-ix}}\text{ (ii)}$
Adding equation (i) and equation (ii), we get
$2\cos x={{e}^{ix}}+{{e}^{-ix}}\Rightarrow \cos x=\dfrac{{{e}^{ix}}+{{e}^{-ix}}}{2}$
Subtracting equation (i) and equation (ii), we get
$\sin x=\dfrac{{{e}^{ix}}-{{e}^{-ix}}}{2i}$
Hence, we have
S $={{\left( \dfrac{{{e}^{ix}}+{{e}^{-ix}}}{2} \right)}^{4}}-{{\left( \dfrac{{{e}^{ix}}-{{e}^{-ix}}}{2i} \right)}^{4}}$
We know that ${{\left( a+b \right)}^{4}}={{a}^{4}}+4{{a}^{3}}b+6{{a}^{2}}{{b}^{2}}+4a{{b}^{3}}+{{b}^{4}}$
Using the above identity, we get
 $S=\dfrac{{{e}^{i4x}}+4{{e}^{i2x}}+6+4{{e}^{-i2x}}+{{e}^{-i4x}}}{16}-\dfrac{{{e}^{i4x}}-4{{e}^{i2x}}+6-4{{e}^{-i2x}}+{{e}^{-i4x}}}{16}$
Hence, we have
$S=\dfrac{8\left( {{e}^{i2x}}+{{e}^{-i2x}} \right)}{16}$
We know that ${{e}^{ix}}+{{e}^{-ix}}=2\cos x$
Replacing x by 2x, we get
${{e}^{i2x}}+{{e}^{-i2x}}=2\cos 2x$
Hence, we have $S=\dfrac{16}{16}\cos 2x$
Hence, we have S =cos2x, which is the required simplified expression.