Question

sin 4x = 4 sin x cos³x-4sin³x cos x. Show as an identity.

Answer 1

2sin2xcos2x = 4sinxcosx(cos^2 x - sin^2 x)

Answer 2

thats sin4x

Answer 3

skipped a few steps there , I see what you have done, but to someone asking the question it is unlikely they will be able to follow

Answer 4

i assume they know basic trig ratios

Answer 5

Have difficulty doing these, could you help with the steps so I can try and understand

Answer 6

sin(2a) = 2sin(a)cos(x)

Answer 7

cos(2a) = cos^2 (a) - sin^2 (a)

Answer 8

remember them

Answer 9

now sin(4x) = sin( 2 * 2x) = 2sin(2x)cos(2x) by applying the first formula once

Answer 10

hard to memorize, I see those in my notes. Thanks for helping

Answer 11

now we can apply the first formula again to sin(2x) , to get 2sin(x)cos(x) and we apply the second formula to the cos(2x) term to get cos^2 (x) - sin^2 (x)

Answer 12

so sin(4x) = 2 * ( 2sin(x)cos(x) ) ( cos^2 (x) - sin^2(x) )

Answer 13

= 4sin(x)cos(x) ( cos^2(x) - sin^2(x)) , then expand

Answer 14

thank you!