$\cos (7 π / 12)$ in surd form

By Anonymous on Monday, March 12, 2001 - 04:14 pm :

How do you evaluate something like this in surd form?

\cos (7 π / 12) = ?

By Michael Doré (Md285) on Monday, March 12, 2001 - 04:29 pm :

You can use the double angle formula. We have:

$\cos (7 π / 6) = 2 \cos^{2} (7 π / 12) - 1$

(using $\cos 2 A = 2 \cos^{2} A - 1$ )

The left hand side is $- \cos (π / 6) = - \sqrt{3} / 2$ .

$\cos^{2} (7 π / 12) = 1 / 2 - \sqrt{3} / 4$

Therefore (as $\cos (7 π / 12 < 0$ ) we see the answer is:

$- \sqrt{1 / 2 - \sqrt{3} / 4}$

By Kerwin Hui (Kwkh2) on Monday, March 12, 2001 - 04:33 pm :

Another way is to note that

7 π / 12 = π / 3 + π / 4

. So

$\cos (7 π / 12) = \cos (π / 4) \cos (π / 3) - \sin (π / 4) \sin (π / 3) = 1 / \sqrt{2} \times 1 / 2 - 1 / \sqrt{2} \times \sqrt{3} / 2 = 2^{- 1.5} (1 - 3^{0.5})$

Kerwin

By Anonymous on Monday, March 12, 2001 - 04:55 pm :

Thank you Michael, Kerwin.

Michael, how did you figure out that

\cos (7 π / 12) < 0

By Barkley Bellinger (Bb246) on Monday, March 19, 2001 - 04:45 pm :

A previous anonymous questioner asked how Michael knew that

\cos (7 π / 12) < 0

. If you know your cosine graph then you will know that

\cos (x) < 0

for

π / 2 < x < π

, which

7 π / 12

satisfies.