Pre-cal Help and explain please!
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Pre-cal Help and explain please!

[From: ] [author: ] [Date: 11-10-31] [Hit: ]
2.Find sin x/2, given cos x=-5/8,{for this I got 2cos 13x/2 cos -3x/2 .........
I took a test and failed it... -_-
so now i have to make corrections and anyone can help me :D
So far I'm almost done except for three last questions which I can't seem to find the answers:

1. Use the cosine of a sum identities to find cos(x+y).
sin x= 2/3 and sin y= -1/3, x in quadrant II & y in quadrant IV

{all i know is the identity: cos(A+B)=cosAcosB - sinAsinB }

2.Find sin x/2, given cos x=-5/8, with π/2
{for this I got 2cos 13x/2 cos -3x/2 .... but it's wrong and I don't even know how I got it o_o }

3.Find cosθ, given cos2θ= 1/2 and θ terminates in quadrant II

Thank you!!

-
1.

We have,

sinx = 2/3
=> sin^2x = 4/9
=> 1-cos^2x = 4/9
=> cos^2x = 5/9
=> cosx = -sqrt(5)/3 ... eq.1 [cosx is negative in quadrant II]

siny = -1/3
=> sin^2y = 1/9
=> 1-cos^2y = 1/9
=> cos^2y = 8/9
=> cosy = 2sqrt(2)/3 ... eq.2 [cosx is positive in quadrant IV]

We know,
cos(x+y) = cosxcosy - sinxsiny
= [-{2sqrt(10)}/9] +(2/9)]
= (2/9)*[1- sqrt(10)]


2.

We know,

sin(x/2) = sqrt[(1-cosx)/2]
= sqrt[{1+(5/8)}/2]
= sqrt{13/16}
= sqrt(13)/4

3.

We have,

cos2θ = 1/2
=> 2cos^2θ - 1 = 1/2
=> 2cos^2θ = 3/2
=> cos^2θ = 3/4
=> cosθ = -sqrt(3)/2 ... [Since cosθ is negative in Quadrant II]


Hope you do better on your next test!
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