Question

18) PT the following identities:
Sin^8 theta - cos^8 theta = (sin^2 theta - cos^2 theta ) (1-2 sin^2 theta cos^2 theta )​

Answer 1

Solution :

$\bf{sin^{8}\theta - cos^{8}\theta = (sin^{2}\theta - cos^{2}\theta)(1 - 2sin^{2}\theta cos^{2}\theta)}$

From the above equation , we get :

• LHS = $\bf{sin^{8}\theta - cos^{8}\theta}$

• RHS = $\bf{sin^{2}\theta - cos^{2}\theta)(1 - 2sin^{2}\theta cos^{2}\theta)}$

By solving the LHS of the equation , we get :

$:\implies \bf{sin^{8}\theta - cos^{8}\theta}$

By using the identity and substituting it in the equation, we get :

$\bf{a^{8} - b^{8} = (a + b)(a - b)(a^{2} + b^{2})(a^{4} + a^{4})}$

$:\implies \bf{(sin\theta - cos\theta)(sin\theta + cos\theta)(sin^{2}\theta + cos^{2}\theta)(sin^{4}\theta + cos^{4}\theta)}$

$:\implies \bf{(sin^{2}\theta - cos^{2}\theta)(sin^{2}\theta + cos^{2}\theta)(sin^{4}\theta + cos^{4}\theta)}$

$:\implies \bf{(sin^{2}\theta - cos^{2}\theta)(sin^{2}\theta + cos^{2}\theta)\{(sin^{2}\theta)^{2} + (cos^{2}\theta)^{2}\}}$

By using the identity and substituting it in the equation , we get :

$\bf{a^{2} + b^{2} = (a + b)^{2} - 2ab}$

$:\implies \bf{(sin^{2}\theta - cos^{2}\theta)(sin^{2}\theta + cos^{2}\theta)\{(sin^{2}\theta + cos^{2}\theta)^{2} - 2sin^{2}\theta cos^{2}\theta\}}$

By using the identity and substituting it in the equation , we get :

$\bf{sin^{2}\theta + cos^{2}\theta = 1}$

$:\implies \bf{(sin^{2}\theta - cos^{2}\theta)(1)(1^{2} - 2sin^{2}\theta cos^{2}\theta)}$

$:\implies \bf{(sin^{2}\theta - cos^{2}\theta)(1 - 2sin^{2}\theta cos^{2}\theta)}$

$\boxed{\therefore \bf{LHS = (sin^{2}\theta - cos^{2}\theta)(1 - 2sin^{2}\theta cos^{2}\theta)}}$

Now by putting the LHS and RHS together , we get :

$:\implies \bf{LHS = RHS}$

$:\implies \bf{(sin^{2}\theta - cos^{2}\theta)(1 - 2sin^{2}\theta cos^{2}\theta) = (sin^{2}\theta - cos^{2}\theta)(1 - 2sin^{2}\theta cos^{2}\theta)}$

Hence it is proved that :

$\boxed{\therefore \bf{sin^{8}\theta - cos^{8}\theta = (sin^{2}\theta - cos^{2}\theta)(1 - 2sin^{2}\theta cos^{2}\theta)}}$

Proved !!