anonymous
  • anonymous
Given sec(theta) = -4/3 and 90 degrees < (theta) <180 degrees ; find sin2(theta)
Mathematics
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SOLVED
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chestercat
  • chestercat
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UsukiDoll
  • UsukiDoll
90 degrees < (theta) <180 degrees tells us that we are restricted in the 2nd quadrant... so all sine values are going to be positive...
anonymous
  • anonymous
If sec(theta) = -4/3 cos(theta) = -3/4 Use sin^2(x)+cos^2(x)=1 to solve for sin^2(x). Plug in -3/4 for cos(x), square it, and subtract the result from 1.
UsukiDoll
  • UsukiDoll
\[\sin^2x\] or \[\sin2x\] for the last part of the question. They have different meanings.

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UsukiDoll
  • UsukiDoll
\[\sin^2\theta............ \sin2\theta\]
anonymous
  • anonymous
its Sin2theta
UsukiDoll
  • UsukiDoll
alright... sin2theta has an identity \[2\sin\theta \cos\theta\]
UsukiDoll
  • UsukiDoll
given that sec(theta) = -4/3 sec(theta) =\[\frac{1}{-\frac{4}{3}} \rightarrow \frac{-3}{4}\]
UsukiDoll
  • UsukiDoll
it's negative due to the fact that we have the restriction of 90 degrees
UsukiDoll
  • UsukiDoll
|dw:1434340094426:dw| so we need to use the pythagorean theorem a^2+b^2=c^2
UsukiDoll
  • UsukiDoll
in this case we solve for a. a^2+3^2=4^2 a^2+9=16 a^2 = 16-9 a^2 = 7 a = \[\sqrt{7}\] ugh one problem... I know we can't pick the negative result for a because negative sides are nonexistent and doesn't make sense... I don't think decimals can be used either... not sure.
UsukiDoll
  • UsukiDoll
sec(theta) = -4/3 cos(theta) = -3/4 in 2nd quadrant hmmm... cos(theta) 3/4 = adjacent/hypotenuse and then solve for a... that should be ...why the heck do I have square root of 7 ? sine theta is \[\frac{\sqrt{7}}{4}\] ... I thought all sides are supposed to have whole numbers
UsukiDoll
  • UsukiDoll
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
It's perfectly possible to have a side length that isn't a whole number example: |dw:1434340852199:dw|
jim_thompson5910
  • jim_thompson5910
and I think you're thinking of this triangle |dw:1434340917667:dw|
UsukiDoll
  • UsukiDoll
O_O! really? so that sine is \[\frac{\sqrt{7}}{4} \] ??
jim_thompson5910
  • jim_thompson5910
yeah sin(theta) = sqrt(7)/4
jim_thompson5910
  • jim_thompson5910
use that to find sin(2*theta)
UsukiDoll
  • UsukiDoll
ok so all we need to do is solve \[\sin(2 \theta) = 2 \sin\theta \cos\theta\]
UsukiDoll
  • UsukiDoll
OP said sin 2 theta! No exponents.
UsukiDoll
  • UsukiDoll
sine is positive due to that 2nd quadrant restriction so \[2\times \frac{\sqrt{7}}{4} \times \frac{-3}{4}\]
UsukiDoll
  • UsukiDoll
\[2 \times \frac{-3\sqrt{7}}{16}\]
UsukiDoll
  • UsukiDoll
\[\frac{-6\sqrt{7}}{16}\]
UsukiDoll
  • UsukiDoll
\[\frac{-3\sqrt{7}}{8}\]
UsukiDoll
  • UsukiDoll
that's it.
anonymous
  • anonymous
Thank you. I was confused a bit but now I see where I messed up.
UsukiDoll
  • UsukiDoll
can you give me medal ^_^
anonymous
  • anonymous
sure :)
UsukiDoll
  • UsukiDoll
yay

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