aussy123
Given: costheta= 4/5, sin x =12/13 ,theta is in the third quadrant, and x is in the fourth quadrant; evaluate cos 2x.
119/169
119/169
11/13



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anonymous
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you don't need the sine part

aussy123
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then how do I start

terenzreignz
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Two variables?

anonymous
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\[\cos(2x)=2\cos^2(x)1\]

tomhoffhim
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dw:1364529999084:dw

anonymous
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oh i see there is a \(\theta\) and an \(x\)

aussy123
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costheta= 4/5
I need cosx

anonymous
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you are asked for \(\cos(2x)\) where \(\sin(x)=\frac{12}{13}\)

anonymous
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\[\cos(2x)=12\sin^2(x)\]

aussy123
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When I do it I get 169

aussy123
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a denominator of 169

anonymous
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there is no \(\theta \) in your question

tomhoffhim
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ok, I'll do it for you, give me a second, this is an easy question

aussy123
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yea I had to put theta for that

tomhoffhim
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satellite, you don't know what you're doing..

anonymous
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\[\cos(x)=12\sin^2(x)\]
\[\cos(x)=1(\frac{12}{13})^2\]\

anonymous
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there is your 169

anonymous
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\[\cos(2x)=1(\frac{12}{13})^2\]
\[\cos(2x)=1\frac{144}{169}\]

anonymous
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\[\cos(2x)=\frac{25}{169}\]

aussy123
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I know, thats what I get, but it isnt a choice

anonymous
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your question asks for \(\cos(2x)\) and doesn't mention any \(\theta\)

aussy123
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I know but that is what is already given

anonymous
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oh damn i made a mistake!! sorry

tomhoffhim
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I'm too lazy to do it
but you need to draw the triangles and either use theta or x for the angle

aussy123
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lol its okay, Ive made plenty

anonymous
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\[\cos(2x)=12\sin^2(x)\]
\[\cos(2x)=12\times (\frac{12}{13})^2\]
\[cos(2x)=12\times \frac{144}{169}\]

anonymous
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\[\cos(2x)=1\frac{288}{169}\]\[\cos(2x)=\frac{119}{169}\]

aussy123
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Lol this is a lot of work, can you use this same formula for tan2Theta

aussy123
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nvm. I found the formula, Thank you!