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November 27, 2014

November 27, 2014

Posted by **Jiskhaa** on Monday, April 8, 2013 at 2:31am.

sin theta = 3/4 , pi/2 < theta < pi

- Trig -
**Reiny**, Monday, April 8, 2013 at 7:59amgiven: sinØ = 3/4, and Ø is in quadrant II

then by Pythagoras, y= 3, x = ? , r = 4

x^2 + 9 = 16

x = ±√5 , but in II x = -√7

so cosØ = -√7/4

sin 2Ø = 2sinØcosØ

= 2(3/4)(-√7/4) = -6√7/16

= -3√7/8

cos 2Ø = cos^2 Ø - sin^2 Ø = 7/16 - 9/16

= -2/16

= - 1/8

tan 2Ø = sin 2Ø / cos 2Ø

= (-3√7/8) / (-1/8)

= 3√7

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