SOLUTION: please help me find the exact values of: tan2u given sinu= -4/5 and pi< u < 3pi/2

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Question 622815: please help me find the exact values of:
tan2u
given sinu= -4/5 and pi< u < 3pi/2

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
There is a formula that expresses tan(2u) in terms of tan(u):
tan%282u%29+=+%282%2Atan%28u%29%29%2F%281-tan%5E2%28u%29%29
So if we can find tan(u) we can use this formula to find tan(2u).

We are given that sin(u) = -4/5. We can use one of the Pythagorean identities, cos%5E2%28a%29+=+1-+sin%5E2%28a%29, to use sin(u) to find cos(u). And once we have cos(u) we can use the fact that tan(u) = sin(u)/cos(u) to find tan(u).

cos%5E2%28u%29+=+1+-+sin%5E2%28u%29
cos%5E2%28u%29+=+1+-+%28-4%2F5%29%5E2
cos%5E2%28u%29+=+1+-+16%2F25
cos%5E2%28u%29+=+25%2F25+-+16%2F25
cos%5E2%28u%29+=+9%2F25
Now we find the square root of each side. But which square root should we use? The positive one or the negative one? Well, since we are told that pi%3C+u+%3C+3pi%2F2 we know that u terminates in the 3rd quadrant. And in the 3rd quadrant cos is negative. So we should use the negative square root:
cos%28u%29+=+-sqrt%289%2F25%29
cos%28u%29+=+-3%2F5

Now we can find tan(u):
tan%28u%29+=+sin%28u%29%2Fcos%28u%29
tan%28u%29+=+%28-4%2F5%29%2F%28-3%2F5%29
tan%28u%29+=+4%2F3

Now we can find tan(2u):
tan%282u%29+=+%282%2Atan%28u%29%29%2F%281-tan%5E2%28u%29%29
tan%282u%29+=+%282%2A%284%2F3%29%29%2F%281-%284%2F3%29%5E2%29
tan%282u%29+=+%288%2F3%29%2F%281-%2816%2F9%29%29
tan%282u%29+=+%288%2F3%29%2F%289%2F9-%2816%2F9%29%29
tan%282u%29+=+%288%2F3%29%2F%28-7%2F9%29
When we divide fractions we change it to multiplying by the reciprocal:
tan%282u%29+=+%288%2F3%29%2A%28-9%2F7%29
tan%282u%29+=+-24%2F63
which reduces to:
tan%282u%29+=+-8%2F21