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.
$\displaystyle \cos(7\pi/12)$
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} $
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
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!
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