# Some Trig Help: Find the exact value of cos(−5pi/12)

• Jul 3rd 2009, 03:31 PM
mvho
Some Trig Help: Find the exact value of cos(−5pi/12)
I've used the sum and difference identity for cos and here is my answer:

$
\frac{\sqrt{3}+\sqrt{2}}{4}
$

Cos -5pi/12 I translate as -75 degrees, so I used my known angles which were:

pi/-6 and -pi/4.

I also know that -5pi/12 is in the 4th region so the value should be positive (All Students Take Calculus).

From the Cosine Sum formula we have:

Cos(a+b)=Cos(a)Cos(b)-Sin(a)Sin(b)
Cos(-pi/4 + -pi/6)= Cos(-pi/4)Cos(-pi/6)-Sin(-pi/4)Sin(-pi/6)
Cos(-pi/4 + -pi/6)=(sqrt2/2)(sqrt3/2)-(sqrt2/2)(1/2)

.....

So what Did I do wrong guys?
• Jul 3rd 2009, 04:02 PM
yeongil
$\cos \left(-\frac{5\pi}{12}\right)$
\begin{aligned}
&= \cos \left(-\frac{\pi}{4} - \frac{\pi}{6}\right) \\
&= \cos \left(-\frac{\pi}{4}\right) \cos \left(\frac{\pi}{6}\right) + \sin \left(-\frac{\pi}{4}\right) \sin \left(\frac{\pi}{6}\right) \\
&= \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right) \\
&= \frac{\sqrt{6} - \sqrt{2}}{4}
\end{aligned}

01
• Jul 3rd 2009, 04:11 PM
mvho
Thanks for the explanation, looks like I rushed through it.