The problem is:

cos(-7pi/12)

I used the half angle formula to get: -(sqrt(3)-1)*sqrt(2)/(4)

Somehow or another the computer system isn't taking my answer.

Thanks for the help!

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- Nov 20th 2006, 08:46 AMqbkr21Trig Problem
The problem is:

cos(-7pi/12)

I used the half angle formula to get: -(sqrt(3)-1)*sqrt(2)/(4)

Somehow or another the computer system isn't taking my answer.

Thanks for the help! - Nov 20th 2006, 08:53 AMThePerfectHacker
First drop the negative, cosine is an even function.

$\displaystyle \cos(7\pi/12)$

Next convert to degree if that is easier for thee,

$\displaystyle \cos(7\cdot 180^o/12)=\cos(7\cdot 15^o)=\cos(105^o)$

Next you can express this angle as,

$\displaystyle \cos(180^o-75^o)=-\cos 75^o=-\sin 15^o$

Now since,

$\displaystyle \cos 30^o=\frac{\sqrt{3}}{2}$

That means,

$\displaystyle \sin 15^o=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2} $ - Nov 20th 2006, 09:08 AMqbkr21Re:
This seems to be the same answer I got, and it still doesn't want to fit into the system...Is there another way you can express this?

Thanks! - Nov 20th 2006, 10:13 AMtopsquark
Unless you changed your original answer you had:

$\displaystyle -\frac{\sqrt{3}-1}{2} \cdot \frac{\sqrt{2}}{4} \approx -0.12941$

ThePerfectHacker had:

$\displaystyle \frac{\sqrt{2 - \sqrt{3}}}{2} \approx 0.258819$

So the answers are not the same.

-Dan - Nov 20th 2006, 03:07 PMqbkr21Re:
Here are the directions the computer system is still not taking it:

Use a sum or difference formula or a half angle formula to determine the value of the trigonometric functions. Give exact answers. Do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as sqrt(2);

Thanks! - Nov 20th 2006, 03:16 PMtopsquark
- Nov 20th 2006, 08:48 PMCaptainBlack
Read what ImPerfectHacker wrote carefully.

He did not say that:

$\displaystyle \cos(105^{\circ})=\sin(15^{\circ})$

what he did say was:

$\displaystyle \cos(105^{\circ})=-\sin(15^{\circ})$

and that:

$\displaystyle

\sin(15^{\circ})=\frac{\sqrt{2-\sqrt{3}}}{2}

$

so:

$\displaystyle \cos(105^{\circ})=-\frac{\sqrt{2-\sqrt{3}}}{2}$

RonL - Nov 20th 2006, 09:32 PMqbkr21Re:
Ok thanks Captain Black