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 one year ago
cos theta = 4/5, 0˚< theta < 90˚ use the information given to find sin2theta, cos2theta, and tan 2theta
 one year ago
cos theta = 4/5, 0˚< theta < 90˚ use the information given to find sin2theta, cos2theta, and tan 2theta

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satellite73
 one year ago
Best ResponseYou've already chosen the best response.1you need \(\sin(\theta)\) so compute these do you know how to find it?

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1dw:1358823887705:dw

alanli123
 one year ago
Best ResponseYou've already chosen the best response.0sin 2 theta = 2 sin theta cos theta cos 2 theta = 2 cos ^2 theta  1 tan 2 theta = sin 2 theta / cos 2 theta

abbyl94
 one year ago
Best ResponseYou've already chosen the best response.0do i plug in 4/5 where theta is?

Azteck
 one year ago
Best ResponseYou've already chosen the best response.1If costheta=4/5 then sintheta= 3/5 \[\sin2\theta=2\sin\theta \cos\theta\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1there is a picture of an angle whose cosine is \(\frac{4}{5}\) you find the third side by pythagoras, or recalling the 3  4  5 right triangle this tells you \(\sin(\theta)=\frac{3}{5}\)

abbyl94
 one year ago
Best ResponseYou've already chosen the best response.0so 2 sin theta = 2 sin 3/5 cos 4/5

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1no do not make that mistake, it is a common one do not use \(\theta=\frac{4}{5}\) use \(\cos(\theta)=\frac{4}{5}\) so \[\sin(2\theta)=2\times \frac{4}{5}\times \frac{3}{5}\]

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1no not the cosine and sine of those numbers, those numbers ARE the cosine and sine

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1use the numbers themselves, they are your sines and cosines

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1your good now right?

abbyl94
 one year ago
Best ResponseYou've already chosen the best response.0im still confused a little ..
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