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abbyl94
 3 years ago
cos theta = 4/5, 0˚< theta < 90˚ use the information given to find sin2theta, cos2theta, and tan 2theta
abbyl94
 3 years ago
cos theta = 4/5, 0˚< theta < 90˚ use the information given to find sin2theta, cos2theta, and tan 2theta

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

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

alanli123
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0do i plug in 4/5 where theta is?

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

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

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

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