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# `4sin(pi/8)sin(pi/8)`

susank7780 | Student, Grade 11 | Salutatorian

Posted May 23, 2013 at 1:25 AM via web

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`4sin(pi/8)sin(pi/8)`

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ishpiro | Teacher | (Level 1) Associate Educator

Posted May 23, 2013 at 1:45 AM (Answer #1)

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`` I was not sure if the problem `4sin(pi/8)sin(pi/8)` was reallly supposed to be `4sin(pi/8)cos(pi/8)` , so here are the solutions to both:

`4sin(pi/8)sin(pi/8) = 4sin^2(pi/8)`

Apply the one of the half-angle identities:

`sin^2(x/2) = (1 - cosx)/2`

Since `pi/8 = 1/2*(pi/4)` , `sin^2(pi/8) = (1 - cos(pi/4))/2`

Recall that `pi/4` is 45 degrees and `cos(pi/4) = sqrt(2)/2`

Then, `4sin^2(pi/8) = 4*(1 - sqrt(2)/2)/2 = 4*(2 - sqrt(2))/4=2 - sqrt(2)`

The answer is `2 - sqrt(2)`

IF this problem was supposed to read

`4sin(pi/8)cos(pi/8)` , then the solution is as follows:

Apply the double-angle identity for sine:

sin2x = 2sinxcosx

Then, `2sin(pi/8)cos(pi/8) = sin(2*(pi/8)) = sin(pi/4)`

Thus, `4sin(pi/8)cos(pi/8) = 2sin(pi/4)`

Recall than `pi/4` is 45 degrees and `sin(pi/4) = sqrt(2)/2`

So `4sin(pi/8)cos(pi/8) = 2* sqrt(2)/2 = sqrt(2)`