# Finding exact values for trigonometric functions

• Dec 11th 2011, 09:27 AM
Bashyboy
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?
• Dec 11th 2011, 09:49 AM
Soroban
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!

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

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

• Dec 11th 2011, 12:41 PM
Bashyboy
Re: Finding exact values for trigonometric functions
Oh, blimey. Thank you very much!