Question

Quick trig question:

For a question, I have :

Find the exact value

cos(-7pi/12); what do I do?

Should I do this

cos{pi/6-3pi/4]

cosxcosy+sinxsiny

√3/2 * -√2/2 = √3-√2/4 + 1/2*√2/2

√3
---
4

Can anyone explain what I did wrong?

Answer 1

The answers I have is √6 + 2

√6/4 - √2/4

√2 - √6

√2/4 - √6/4

Answer 2

well, 1/6 - 3/4 = -1/12; \(\bf \ne \) -7/12

Answer 3

Hmmm?


1/6 = 2/12

3/4 = 9/12

2 - 9 -7
---- = --
12 12

Answer 4

I can see -4/12 - 3/12 = -7/12

Answer 5

ohhh yea... hmmm, you're right, I spoke too soon, I forgot a small piece on the numerator

Answer 6

cosx cosy+sinx siny
= (√3/2) * (-√2/2) + (1/2) (√2/2)
= -√6/4 + √2/4

Answer 7

Ahhhhhh; I forgot to multiply the square roots :P.

Answer 8

yeah .. :)