Finding exact values for trigonometric functions

Hello,

I am given a righ-triangle, where the h = sqroot(17), a =1, and b = 4. I am suppose to find the exact value of the trigonometric sin 4x. So, I use the double angle formula, and what results is 2 sin 2x cos 2x. Then, I use the the double angle formulas again, and get

2[(2 sin x cos x)(1 - sin^2 x)]. After inputing the ratios for the angles, provided from the dimensions of the right-triangle, I get 256/289 as my exact answer; yet, the true

answer--according to the book--is 240/289. Does anyone see where I went wrong?

Re: Finding exact values for trigonometric functions

Hello, Bashyboy!

Quote:

I am given a righ-triangle, where: h = sqrt(17), a = 1, and b = 4.

I am suppose to find the exact value of the trigonometric sin 4x.

So, I use the double angle formula, and what results is 2 sin 2x cos 2x.

Then, I use the the double angle formulas again,

. . and get: 2(2 sin x cos x)(1 - sin^2 x) .Here!

$\displaystyle \cos2x \:=\:2\cos^2\!x\!-\!1\,\text{ or }\,1\!-\!2\sin^2x\,\text{ or }\,\cos^2\!x\!-\!\sin^2\!x$

You were probably thinking of: .$\displaystyle \cos^2\!x \:=\:1-\sin^2\!x$

Re: Finding exact values for trigonometric functions

Oh, blimey. Thank you very much!