proving trigonometry question

khoo88 SG

sin^8(a)+cos^8(a)=1-4sin^2(a)cos^2(a)+2sin^4(a)cos^4(a)

show me the full step on how to prove this question

tq in advance

wee_ws SG

Hi,

The approach is to start with the side with more terms, because it is easier to simplify to less. Then we just express everything in terms of sine and see what happens in the course of the proof. Here goes:

RHS = 1 - 4 {sin^2 (a)} {cos^2 (a)} + 2 {sin^4 (a)} {cos^4 (a)}

= 1 - 4 {sin^2 (a)} {1 - sin^2 (a)} + 2 {sin^4 (a)} {1 - sin^2 (a)}^2

= 1 - 4 sin^2 (a) + 6 sin^4 (a) - 4 sin^6 (a) + 2 sin^8 (a)

= {1 - sin^2 (a)}^4 + sin^8 (a)

I find the question challenging for students as it requires knowledge of Pascal's triangle to know that (1 - x^2)^4 = 1 - 4x^2 + 6x^4 - 4x^6 + x^8.

Thanks!

Cheers,
Wen Shih

eagle SG

Alternatively,

sin^2 x + cos^2 x = 1

(sin^2 x + cos^2 x)^4 = 1

sin^8 x + 4 sin^6 x cos^2 x + 6 sin^4 x cos^4 x + 4 sin^2 x cos^6 x + cos^8 x = 1

sin^8 x + cos^8 x + 4 sin^4 x (1 - cos^2 x) cos^2 x + 4 sin^2 x cos^4 x (1 - sin^2 x) + 6 sin^4 x cos^4 x = 1

sin^8 x + cos^8 x + 4 sin^4 x cos^2 x - 4 sin^4 x cos^4 x + 4 sin^2 x cos^4 x + 4 sin^4 x cos^4 x + 6 sin^4 x cos^4 x = 1

sin^8 x + cos^8 x - 2sin^4 x cos^4 x + 4 sin^2 x (1 - cos^2 x) cos^2 x + + 4 sin^2 x cos^2 x (1 - sin^2 x) = 1

rearranging gives u the answer.

Tedious, but no necessity to work backwards from pascal triangle.

wee_ws SG

Hi,

In either method, binomial theorem is used. So, it will be good to take note of the close relationship between trigonometric proving and binomial theorem when powers are involved. Thanks!

Cheers,
Wen Shih