find the value of

Question

find the value of

anonymous · Answer

what are the answers they give you to choose from? @msingh

anonymous · Answer

there are no answers given

anonymous · Answer

Ok well I tried my best to type it into my calculator and I got .9993910357 
I don't know if that is the right answer or not. You could try to google it.

anonymous · Answer

okay

anonymous · Answer

@amistre64

amistre64 · Answer

im thinking some trig identity might help:  cos(a+b) = cos(a)cos(b)-sin(a)sin(b)

anonymous · Answer

okay

amistre64 · Answer

cos^2(a)cos^2(b)+sin^2(a)sin^2(b) - 2k
cos^2(a)cos^2(b)+sin^2(a)sin^2(b)+ 2k

$$2cos^2(60)cos^2(b)+2sin^2(60)sin^2(b)+cos^2(b)$$

$$cos^2(b)(1+2cos^2(60))+2sin^2(60)sin^2(b)$$
$$cos^2(b)(1+2/4)+6/4~sin^2(b)$$
$$cos^2(b)(6/4)+6/4~sin^2(b)=3/2$$

anonymous · Answer

@amistre64 
thank u

amistre64 · Answer

yw

anonymous · Answer

but @amistre64 
i getting the answer 
after solving, that is 
cos^2 theta(3/2)

amistre64 · Answer

i got it down to:

$$\frac64(cos^2(\theta)+sin^2(\theta))$$
$$\frac64(1)$$

anonymous · Answer

how u got 
2sin^2(60)sin^2(b) part

anonymous · Answer

this part will be gonna cancel

amistre64 · Answer

i took and squared out the cos(a+-b) parts

amistre64 · Answer

cos(a+b) = cos(a)cos(b)-sin(a)sin(b)
cos(a-b) = cos(a)cos(b)+sin(a)sin(b)

square those and you end up with

2cos^2(a)cos^2(b)+2sin^2(a)sin^2(b)

anonymous · Answer

okay

amistre64 · Answer

with that in hand, and an extra cos^2(b) to factor with

itll simplify