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anonymous
 one year ago
Given sec(theta) = 4/3 and 90 degrees < (theta) <180 degrees ; find sin2(theta)
anonymous
 one year ago
Given sec(theta) = 4/3 and 90 degrees < (theta) <180 degrees ; find sin2(theta)

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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.290 degrees < (theta) <180 degrees tells us that we are restricted in the 2nd quadrant... so all sine values are going to be positive...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.2\[\sin^2x\] or \[\sin2x\] for the last part of the question. They have different meanings.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\sin^2\theta............ \sin2\theta\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2alright... sin2theta has an identity \[2\sin\theta \cos\theta\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2given that sec(theta) = 4/3 sec(theta) =\[\frac{1}{\frac{4}{3}} \rightarrow \frac{3}{4}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2it'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

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1434340094426:dw so we need to use the pythagorean theorem a^2+b^2=c^2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2in this case we solve for a. a^2+3^2=4^2 a^2+9=16 a^2 = 169 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
 one year ago
Best ResponseYou've already chosen the best response.2sec(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

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0It's perfectly possible to have a side length that isn't a whole number example: dw:1434340852199:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0and I think you're thinking of this triangle dw:1434340917667:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2O_O! really? so that sine is \[\frac{\sqrt{7}}{4} \] ??

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0yeah sin(theta) = sqrt(7)/4

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0use that to find sin(2*theta)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2ok so all we need to do is solve \[\sin(2 \theta) = 2 \sin\theta \cos\theta\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2OP said sin 2 theta! No exponents.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2sine is positive due to that 2nd quadrant restriction so \[2\times \frac{\sqrt{7}}{4} \times \frac{3}{4}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[2 \times \frac{3\sqrt{7}}{16}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{6\sqrt{7}}{16}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{3\sqrt{7}}{8}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you. I was confused a bit but now I see where I messed up.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2can you give me medal ^_^
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