Posted by Lucy on Sunday, August 31, 2008 at 12:03am.
If cos theta = 0.8 and 270<theta<360, find the exact value of sin 2theta
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

PreCalculuscheck answers  Mohamed, Sunday, August 31, 2008 at 12:23am
1) 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

PreCalculuscheck answers  Reiny, Sunday, August 31, 2008 at 12:36am
2) 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

PreCalculuscheck answers  Mohamed, Sunday, August 31, 2008 at 12:39am
1) 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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