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January 30, 2015

January 30, 2015

Posted by **anon** on Thursday, March 3, 2011 at 6:32am.

1. sin2 = (sqrt 3)/2

2. sin^2 = cos^2 + 1/2

3. sin 2x - cosx = 0

4. cos 4x = sin 2x

(((* stands for degrees)))

q1

sin2 = (sqrt 3)/2

2 = sin^-1 (sqrt3/2)

= 60*, 120*, 420*, 480*

= 30*, 60*, 210* and 240*

(not sure of the rest)

q2

sin^2 = cos^2 + 1/2

sin^2 - cos^2 = 1/2 ...(1)

sin^2 + cos^2 = 1 ...(2)

(1) + (2),

2sin^2 = 3/2

sin^2 = 3/4

sin = +-3/2

= 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*

- trigonometry (please double check this) -
**anon**, Thursday, March 3, 2011 at 6:34am"" "" is supposed to be theta.

- trigonometry (please double check this) -
**Reiny**, Thursday, March 3, 2011 at 8:21amq1 - correct

q2 is correct but here is a simpler way.

Let =

sin^2 = cos^2 + 1/2

cos^2 - sin^2 = -1/2

cos 2 = -1/2

2 = 120 , 240 ...

= 60 , 120 , 240 , 300

q3 correct

q4 also correct

good job!

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