Question

Find the exact value of cos (19pi/12)

Answer 1

I keep getting this question marked wrong
I use cos(5pi/6+3pi/4)=cos5pi/6cos3pi/4+sin5pi/6sin3pi/4
this gets me -sqr3/2*-sqr2/2+1/2*sqr2/2
when I simplify this i get sqr6/4+sqr2/2 or (sqr6 + sqr 2)/4 , but it still gets marked wrong.
> Could you show me where the mistake in my work is? Thank you c:

Answer 2

\[\frac{ \sqrt{3}-1 }{ 2\sqrt{2}}\]

Answer 3

the mistake is in your expansion

cos(A + B) = cos(A)cos(B) - sin(A)sin(B)

Answer 4

@campbell_st the sign switches?

Answer 5

yep... cos(A + B) = cos(A)cos(B) - sin(A)sin(B)
cos(A - B)= cos(A)cos(B) + sin(A)sin(B)

Answer 6

Thank you! No wonder I kept messing cos up!

Answer 7

glad to help