Questions on Double & Half Angles.? pleaaase help im so confused!?

Question

1.) If cos y = 1/2, and 3π/2 < y < 2π, find cos 2y

2.) If tan B = 4/3, and B lies in Quadrant III, find sin(B/2)

3.) The Expression 2sinx cosx is equivalent to which of the following:
a. 2sinx
b. sin 2x
c. cos 2x
d. tan 2x

4.)The expression 2cos^2(15°)-1 has the same value as:
a. sin 15°
b. cos 15°
c. sin 30°
d. cos 30°

5. If siny = -(7/25), find cos 2y

6. If sin B = 0.6, and B terminates in Quadrant II, find cos (B/2)

7. If cos A = -(5/13) and π < A < 3π/2, find cos(1/2)A

Answer 1

1.)

cos(y) = (1/2)
y = (pi/3) or (5pi/3)

to keep it (3pi/2) < y < 2pi, y = (5pi/3)

cos(2(5pi/3)) = (-1/2)

ANS : cos(2y) = (-1/2)

Another way to do it

cos(2y) = cos(y)^2 - sin(y)^2

a^2 + b^2 = c^2
1^2 + b^2 = 2^2
1 + b^2 = 4
b^2 = 3
b = sqrt(3)

sin(y) = sqrt(3)/2

cos(2y) = cos(y)^2 - sin(y)^2
cos(2y) = (1/2)^2 - (sqrt(3)/2)^2
cos(2y) = (1/4) - (3/4)
cos(2y) = (-2/4)
cos(2y) = (-1/2)

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2.)
tan(B) = (4/3)

4^2 + 3^2 = c^2
16 + 9 = c^2
c^2 = 25
c = 5

cos(B) = 3/5

sin(B/2) = sqrt((1/2)(1 - cos(B)))
sin(B/2) = sqrt((1/2)(1 - (3/5)))
sin(B/2) = sqrt((1/2)(2/5))
sin(B/2) = sqrt(1/5)
sin(B/2) = sqrt(5)/5

ANS : sin(B/2) = sqrt(5)/5

Another way to do it, and probably the easiest way in this case

tan(B) = 4/3
B = tan^-1(4/3)

sin((tan^-1(4/3))/2) = sqrt(5)/5

but then you wouldn't know how to put it as sqrt(5)/5, you would get .447213595...

3.)
2sin(x)cos(x) = sin(2x)
ANS : B.

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4.)
2cos(15)^2 - 1 = cos(2(15)) = cos(30)
ANS : D.

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5.)
sin(y) = (-7/25)

(-7)^2 + b^2 = 25^2
49 + b^2 = 625
b^2 = 576
b = 24

cos(y) = 24/25

cos(2y) = cos(y)^2 - sin(y)^2
cos(2y) = (24/25)^2 - (-7/25)^2
cos(2y) = (576/625) - (49/625)
cos(2y) = (527/625)

using sin(y) = (-7/25), y = sin^-1(-7/25), then plugging that into cos(2y), will give you the same answer.

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6.)
sin(B) = .6
sin(B) = (6/10)
sin(B) = (3/5)

3^2 + b^2 = 5^2
9 + b^2 = 25
b^2 = 16
b = 4
cos(B) = 4/5

cos(B/2) = sqrt((1/2)(1 + cos(B))
cos(B/2) = sqrt((1/2)(1 + (4/5))
cos(B/2) = sqrt((1/2)(9/5))
cos(B/2) = sqrt(9/10)
cos(B/2) = 3/sqrt(10)
cos(B/2) = 3sqrt(10)/10

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7.)
by this do you mean cos((1/2)A) or sqrt(cos(A))
Whatever it is, maybe you can see what to do with the sites i provided you or the work i have done above.

On the #2 and #6, i'm not 100% certain if i have them right because of the quadrant that they need to be in.

Answer 2

You can do these with sin2A = 2sinAcosA and
cos2A = 2(cosA)^2 - 1
#2 also uses the fact that tanA = sinA/cosA
You may also need (sinA)^2 + (cosA)^2 = 1