Question

Prove that ${{\cos }^{4}}\theta -{{\sin }^{4}}\theta =\cos 2\theta $

Answer 1

Hint: Use $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$ to factorise LHS. Use trigonometric identities ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ and ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta $.

Complete step-by-step answer:
Trigonometric ratios:
There are six trigonometric ratios defined on an angle of a right-angled triangle, viz sine, cosine,
tangent, cotangent, secant and cosecant.
The sine of an angle is defined as the ratio of the opposite side to the hypotenuse.
The cosine of an angle is defined as the ratio of the adjacent side to the hypotenuse.
The tangent of an angle is defined as the ratio of the opposite side to the adjacent side.
The cotangent of an angle is defined as the ratio of the adjacent side to the opposite side.
The secant of an angle is defined as the ratio of the hypotenuse to the adjacent side.
The cosecant of an angle is defined as the ratio of the hypotenuse to the adjacent side.
Observe that sine and cosecant are multiplicative inverses of each other, cosine and secant are
multiplicative inverses of each other, and tangent and cotangent are multiplicative inverses of each
other.
Some of the identities followed by these ratios are:
[1]${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1,{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ and ${{\csc }^{2}}\theta =1+{{\cot }^{2}}\theta $
[2] $\sin 2\theta =2\sin \theta \cos \theta ,\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta $ and $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$
We have LHS $={{\cos }^{4}}\theta -{{\sin }^{4}}\theta $
Using $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$ , with $a={{\cos }^{2}}\theta $ and $b={{\sin }^{2}}\theta $, we get
LHS$=\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)$
Using ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ and ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta $, we get
LHS $=1\left( \cos 2\theta \right)=\cos 2\theta $
Hence LHS = RHS
Hence proved.

Note: [1] Pythagorean identities: The identities ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1,{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ and ${{\csc }^{2}}\theta =1+{{\cot }^{2}}\theta $ are known as the Pythagorean identities as they are a direct consequence of Pythagoras theorem.
[2] Double angle formulae: The identities $\sin 2\theta =2\sin \theta \cos \theta ,\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta $ and $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ are known as double angle formulae.