Posted by **AL** on Wednesday, July 11, 2012 at 10:17am.

Given that tan¤=3/4 and ¤ is in the second quadrant find sin2¤ (¤=theta)

- TRIG -
**bobpursley**, Wednesday, July 11, 2012 at 10:22am
see other post.

- TRIG -
**Bosnian**, Wednesday, July 11, 2012 at 10:48am
cos theta = + OR - 1 / sqrt ( 1 + tan ^ 2 theta )

In Quadrant II, cosine are negative so :

cos theta = - 1 / sqrt ( 1 + tan ^ 2 theta )

cos theta = - 1 / sqrt [ 1 + ( 3 / 4 ) ^ 2 theta ]

cos theta = - 1 / sqrt ( 1 + 9 / 16 )

cos theta = - 1 / sqrt ( 16 / 16 + 9 / 16 )

cos theta = - 1 / sqrt ( 25 / 16 )

cos theta = - 1 / ( 5 / 4 )

cos theta = - 4 / 5

sin ^ 2 theta + cos ^ 2 theta = 1

sin ^ 2 theta = 1 - cos ^ 2 theta

sin theta = + OR - sqrt ( 1 - cos ^ 2 theta )

In Quadrant II, sine are positive so :

sin theta = sqrt ( 1 - cos ^ 2 theta )

sin theta = sqrt [ 1 - ( - 4 / 5 ) ^ 2 ]

sin theta = sqrt ( 1 - 16 / 25 )

sin theta = sqrt ( 25 / 25 - 16 / 25 )

sin theta = sqrt ( 9 / 25 )

sin theta = 3 / 5

sin 2 theta = 2 sin theta cos theta

sin 2 theta = 2 * 3 / 5 * - 4 / 5

sin 2 theta = - 24 / 25

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