SOLUTION: Find the exact values of sin 2θ, cos 2θ, and tan 2θ for the given value of θ. cos θ = 3/5; 0° < θ < 90° a).sin 2θ b).cos 2θ c).tan 2θ Please help! I want to

Algebra ->  Equations -> SOLUTION: Find the exact values of sin 2θ, cos 2θ, and tan 2θ for the given value of θ. cos θ = 3/5; 0° < θ < 90° a).sin 2θ b).cos 2θ c).tan 2θ Please help! I want to       Log On


   



Question 1158677: Find the exact values of sin 2θ, cos 2θ, and tan 2θ for the given value of θ.
cos θ = 3/5; 0° < θ < 90°
a).sin 2θ
b).cos 2θ
c).tan 2θ
Please help! I want to know how to solve these kind of problems so please don't just show me the answer. Thank you :)

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
If you google "double angle formulas", you can find
the cos, sin, and tan of +2theta+
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You are given that +cos%28+theta+%29+=+3%2F5+
and +0+%3C+theta+%3C+90+
This means the angle is in the 1st quadrant where the
cos, sin, and tan are all positive (+) / (+)
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(a)
+sin%28+2%2Atheta+%29+=+2%2Asin%28+theta+%29%2Acos%28+theta+%29+ ( look it up online )
Note that if +cos%28+theta+%29+=+3%2F5+, you have a 3-4-5 triangle, so
+sin%28+theta+%29+=+4%2F5+ and +tan%28+theta+%29+=+4%2F3+
+sin%28+2%2Atheta+%29+=+2%2Asin%28+theta+%29%2Acos%28+theta+%29+
+sin%28+2%2Atheta+%29+=+2%2A%28+4%2F5+%29%2A%28+3%2F5+%29+ ( all positive in the 1st quadrant )
+sin%28+2%2Atheta+%29+=+24%2F25+
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(b)
+cos%28+2%2Atheta+%29+=+2%2Acos%5E2%28+theta+%29+-+1+ ( online )
+cos%28+2%2Atheta+%29+=+2%2A%28+3%2F5+%29%5E2+-+1+
+cos%28+2%2Atheta+%29+=+18%2F25+-+1+
+cos%28+2%2Atheta+%29+=+-7%2F25+
The negative sign tells me that +2%2Atheta+ ends up in the
2nd quadrant where the cosine is negative.
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(c)
+tan%28+2%2Atheta+%29+=+%28+2%2Atan%28+theta+%29%29+%2F+%28+1+-+tan%5E2%28+theta+%29+%29+ ( online )
+tan%28+2%2Atheta+%29+=+%28+2%2A%28+4%2F3+%29%29+%2F+%28+1+-+%28+16%2F9+%29%29+
+tan%28+2%2Atheta+%29+=+%28+8%2F3+%29+%2F+%28+-7%2F9+%29+
+tan%28+2%2Atheta+%29+=+%28+8%2F3+%29%2A%28+-9%2F7+%29+
+tan%28+2%2Atheta+%29+=+-24%2F7+
It's negative because the tan is negative in the 2nd quadrant
Notice also that:
+tan%28+2%2Atheta+%29+=+sin%28+2%2Atheta+%29+%2F+cos%28+2%2Atheta+%29+
+-24%2F7+=+%28+24%2F25+%29+%2F+%28+-7%2F25+%29+
+-24%2F27+=+-24%2F7+
Also notice that I never had to find out what +theta+ was to
solve these. But from the results, I know that +theta+ MUST be
greater than 45 degrees, in order for +2%2Atheta+ to end up
in the 2nd quadrant. I'll show this using my calculator:
+cos%28+theta+%29+=+3%2F5+
+theta+=+arc+cos%28+3%2F5+%29+
+theta+=+53.13+ degrees
and +2%2Atheta++=+106.26+ which is in the 2nd quadrant
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Hope all this helps