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August 22, 2014

August 22, 2014

Posted by **Lucy** on Sunday, August 31, 2008 at 12:03am.

a) -0.96

b) -0.6

c) 0.96

d) 0.28

Answer: b

2) If csc theta = -5/3, and theta has its terminal side in Quadrant III, find the exact value of tan 2theta

a)24/25

b) 7/25

c) 24/7

d) -7/25

Thanks

- Pre-Calculus-check answers -
**Mohamed**, Sunday, August 31, 2008 at 12:23am1) cosθ=0.8 , θ is in fourth quadrant

θ= cos^(-1)0.8=36.86989765 + 270=306.87

so , sin 2θ= sin2(306.87)= -0.96

answer : a

- Pre-Calculus-check answers -
**Reiny**, Sunday, August 31, 2008 at 12:36am2) csc Ø = -5/3, then sin Ø = -3/5 and is in III

the cos Ø = -4/5

so tan 2Ø = sin 2Ø/cos 2Ø

= 2sinØcosØ/(cos^2 Ø - sin^2 Ø)

=2(-3/5)(-4/5)/(16/25 - 9/25)

= 24/7

- Pre-Calculus-check answers -
**Mohamed**, Sunday, August 31, 2008 at 12:39am1) cscθ = (-5/3) , θ is in third quadrant

1/sinθ = (-5/3) => sinθ= -3/5

θ= -36.8699

θ is in third quadrant

so θ = 36.8669+180 = 216.8669

tan2θ = tan2(216.8669) = 3.42724 = 24/7

answer : c

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