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anonymous

  • one year ago

Given sec(theta) = -4/3 and 90 degrees < (theta) <180 degrees ; find sin2(theta)

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  1. UsukiDoll
    • one year ago
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    90 degrees < (theta) <180 degrees tells us that we are restricted in the 2nd quadrant... so all sine values are going to be positive...

  2. anonymous
    • one year ago
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    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.

  3. UsukiDoll
    • one year ago
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    \[\sin^2x\] or \[\sin2x\] for the last part of the question. They have different meanings.

  4. UsukiDoll
    • one year ago
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    \[\sin^2\theta............ \sin2\theta\]

  5. anonymous
    • one year ago
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    its Sin2theta

  6. UsukiDoll
    • one year ago
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    alright... sin2theta has an identity \[2\sin\theta \cos\theta\]

  7. UsukiDoll
    • one year ago
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    given that sec(theta) = -4/3 sec(theta) =\[\frac{1}{-\frac{4}{3}} \rightarrow \frac{-3}{4}\]

  8. UsukiDoll
    • one year ago
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    it's negative due to the fact that we have the restriction of 90 degrees <theta<180 degrees which means 2nd quadrant and only sine values are positive now let's ignore the - for a bit and find out what our sine really is so cosine is adjacent/hypotenuse so that means cosine 3/4

  9. UsukiDoll
    • one year ago
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    |dw:1434340094426:dw| so we need to use the pythagorean theorem a^2+b^2=c^2

  10. UsukiDoll
    • one year ago
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    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.

  11. UsukiDoll
    • one year ago
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    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

  12. UsukiDoll
    • one year ago
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    @jim_thompson5910

  13. jim_thompson5910
    • one year ago
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    It's perfectly possible to have a side length that isn't a whole number example: |dw:1434340852199:dw|

  14. jim_thompson5910
    • one year ago
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    and I think you're thinking of this triangle |dw:1434340917667:dw|

  15. UsukiDoll
    • one year ago
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    O_O! really? so that sine is \[\frac{\sqrt{7}}{4} \] ??

  16. jim_thompson5910
    • one year ago
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    yeah sin(theta) = sqrt(7)/4

  17. jim_thompson5910
    • one year ago
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    use that to find sin(2*theta)

  18. UsukiDoll
    • one year ago
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    ok so all we need to do is solve \[\sin(2 \theta) = 2 \sin\theta \cos\theta\]

  19. UsukiDoll
    • one year ago
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    OP said sin 2 theta! No exponents.

  20. UsukiDoll
    • one year ago
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    sine is positive due to that 2nd quadrant restriction so \[2\times \frac{\sqrt{7}}{4} \times \frac{-3}{4}\]

  21. UsukiDoll
    • one year ago
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    \[2 \times \frac{-3\sqrt{7}}{16}\]

  22. UsukiDoll
    • one year ago
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    \[\frac{-6\sqrt{7}}{16}\]

  23. UsukiDoll
    • one year ago
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    \[\frac{-3\sqrt{7}}{8}\]

  24. UsukiDoll
    • one year ago
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    that's it.

  25. anonymous
    • one year ago
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    Thank you. I was confused a bit but now I see where I messed up.

  26. UsukiDoll
    • one year ago
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    can you give me medal ^_^

  27. anonymous
    • one year ago
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    sure :)

  28. UsukiDoll
    • one year ago
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    yay

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