# trigonometry (please double check this)

Solve the following trig equations. give all the positive values of the angle between 0 degrees and 360 degrees that will satisfy each. give any approximate value to the nearest minute only.

1. sin2ƒÆ = (sqrt 3)/2
2. sin^2ƒÆ = cos^2ƒÆ + 1/2
3. sin 2x - cosx = 0
4. cos 4x = sin 2x

(((* stands for degrees)))

q1
sin2ƒÆ = (sqrt 3)/2
2ƒÆ = sin^-1 (sqrt3/2)
= 60*, 120*, 420*, 480*
ƒÆ = 30*, 60*, 210* and 240*

(not sure of the rest)

q2
sin^2ƒÆ = cos^2ƒÆ + 1/2
sin^2 ƒÆ - cos^2 ƒÆ = 1/2 ...(1)

sin^2 ƒÆ + cos^2 ƒÆ = 1 ...(2)

(1) + (2),
2sin^2 ƒÆ = 3/2
sin^2 ƒÆ = 3/4
sin ƒÆ = +-�ã3/2
ƒÆ = 60*, 120*, 240* and 300*

q3
sin 2x - cosx = 0
2sinxcosx - cosx = 0
cos x (2sin x - 1) = 0
cos x = 0 or sin x = 1/2
x = 90*, 270* or
x = 30* 150*

answer, x = 30*, 90*, 150* and 270*

q4
cos 4x = sin 2x
2(sin 2x)^2 + sin 2x - 1 = 0
(2sin 2x - 1)(sin 2x + 1) = 0
sin 2x = -1, 1/2
2x = 30*, 150*, 270*, 390*, 510*, 630*
x = 15*, 75*, 135*, 195*, 255* and 315*

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1. ""ƒÆ "" is supposed to be theta.

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2. q1 - correct

q2 is correct but here is a simpler way.
Let Ø = ƒÆ
sin^2 Ø = cos^2 Ø + 1/2
cos^2 Ø - sin^2 Ø = -1/2
cos 2Ø = -1/2
2Ø = 120° , 240° ...
Ø = 60° , 120° , 240° , 300°

q3 correct

q4 also correct

good job!

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