using the exact values of the sine and cosine of both 1/6∏ and 3/4∏ and one of the sum and difference formulas, show that the exact value of cos(7/12∏) is 1/4(√2-√6)

any help would be greatly welcomed...thanks

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- Apr 6th 2009, 06:55 AMjordan1trigonometry problem????
using the exact values of the sine and cosine of both 1/6∏ and 3/4∏ and one of the sum and difference formulas, show that the exact value of cos(7/12∏) is 1/4(√2-√6)

any help would be greatly welcomed...thanks

- Apr 6th 2009, 07:08 AMstapel
Follow the instructions:

You know that 7/12 = 3/4 - 1/6 = 9/12 - 2/12.

You have memorized the exact values for the sine and cosine of the (1/4)(pi) and (1/6)(pi).

You have been given some sum and difference identities.

So restate the (7/12)(pi) in terms of (3/4)(pi) and (1/6)(pi), apply the appropriate identities, and then simplify the result, using what you know about the periodicity of the sine and cosine waves.

If you get stuck, please reply showing how far you have gotten in working through the steps. Thank you! :D - Apr 6th 2009, 07:10 AMe^(i*pi)
$\displaystyle sin(\frac{\pi}{6}) = \frac{1}{2}$

$\displaystyle cos(\frac{\pi}{6}) = \frac{\sqrt3}{2}$

$\displaystyle sin(\frac{3\pi}{4}) = \frac{\sqrt2}{2}$

$\displaystyle cos(\frac{3\pi}{4}) = -\frac{\sqrt2}{2}$

$\displaystyle \frac{7\pi}{12} = \frac{3\pi}{4} - \frac{\pi}{6}$

$\displaystyle cos(A-B) = cosAcosB + sinAsinB$

$\displaystyle A = \frac{3\pi}{4}$ and $\displaystyle B = \frac{\pi}{6}$