Question

Prove that sin^8a-cos^8a=(1-2cos^2a)(1-2sin^2acos^2a)

Answer 1

Answer:

Hence Proved

Step-by-step explanation:

$L.H.S.= sin^{8}a-cos^{8}a \\=(sin^{4}a-cos^{4}a)(sin^{4}a+cos^{4}a)\\=(sin^{2}a-cos^{2}a)(sin^{2}a+cos^{2}a)(sin^{4}a+cos^{4}a)...........(1)\\As, (sin^{2}a+cos^{2}a)=1 => sin^{2}a=1- cos^{2}a......................(2)\\and, (sin^{2}a+cos^{2}a)^{2} = (sin^{4}a+cos^{4}a +2 sin^{2}a*cos^{2}a)\\=> 1=(sin^{4}a+cos^{4}a +2 sin^{2}a*cos^{2}a)\\=>sin^{4}a+cos^{4}a=(1 -2 sin^{2}a*cos^{2}a)..............(3)\\\\\\Substituting (2) and (3) in (1)\\L.H.S. = ((1-cos^{2}a)-cos^{2}a)(1)(1 -2 sin^{2}a*cos^{2}a)$$=(1-2cos^2a)(1-2sin^2acos^2a)$

Answer 2

Here is your answer ✌!

LHS,

sin^8a - cos^8a

( sin^4a )² - ( cos^4a )²

(sin^4a - cos^4a) (sin^4a + cos^4a)

[ (sin²a)² - (cos²a)² ] [ ( sin²a + cos²a) -

2sin²a cos²a ]

(sin²a + cos²a)(sin²a - cos²a) [ 1 - 2sin²a

cos²a ]

(sin²a - cos²a) ( 1 - 2sin²a cos²a )

we know , cos 2a = cos²a - sin²a

- cos 2a ( 1 - 2sin²a cos ²a )

And also, cos 2a = 2cos²a - 1

- ( 2cos²a - 1 ) ( 1 - 2sin²a cos²a )

( 1 - 2cos²a ) ( 1 - 2sin²a cos²a )

Hence proved

Hope it's help you ✌!