# Trig Problem

• November 20th 2006, 08:46 AM
qbkr21
Trig 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!
• November 20th 2006, 08:53 AM
ThePerfectHacker
Quote:

Originally Posted by qbkr21
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!

First drop the negative, cosine is an even function.

$\cos(7\pi/12)$
Next convert to degree if that is easier for thee,
$\cos(7\cdot 180^o/12)=\cos(7\cdot 15^o)=\cos(105^o)$
Next you can express this angle as,
$\cos(180^o-75^o)=-\cos 75^o=-\sin 15^o$
Now since,
$\cos 30^o=\frac{\sqrt{3}}{2}$
That means,
$\sin 15^o=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}$
• November 20th 2006, 09:08 AM
qbkr21
Re:
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!
• November 20th 2006, 10:13 AM
topsquark
Quote:

Originally Posted by qbkr21
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!

Quote:

Originally Posted by ThePerfectHacker
First drop the negative, cosine is an even function.

$\cos(7\pi/12)$
Next convert to degree if that is easier for thee,
$\cos(7\cdot 180^o/12)=\cos(7\cdot 15^o)=\cos(105^o)$
Next you can express this angle as,
$\cos(180^o-75^o)=-\cos 75^o=-\sin 15^o$
Now since,
$\cos 30^o=\frac{\sqrt{3}}{2}$
That means,
$\sin 15^o=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}$

Quote:

Originally Posted by qbkr21
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!

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

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

So the answers are not the same.

-Dan
• November 20th 2006, 03:07 PM
qbkr21
Re:
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!
• November 20th 2006, 03:16 PM
topsquark
Quote:

Originally Posted by qbkr21
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!

You should enter:
sqrt(2 - sqrt(3))/2

or

(sqrt(2 - sqrt(3)))/2

The first one should do the trick, the second is being extra careful.

-Dan
• November 20th 2006, 08:48 PM
CaptainBlack
Quote:

Originally Posted by qbkr21
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!

He did not say that:

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

what he did say was:

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

and that:

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

so:

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

RonL
• November 20th 2006, 09:32 PM
qbkr21
Re:
Ok thanks Captain Black