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What is the exact value of cos(2 x) if tan( x) = -3/4 and x is in quadrant IV?
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-1/2
how did you come up wit that?
1+tan^2 x = sec^2 x have you seen this formula?
no I havent
K there are three trigonometric identities this is one of them. Can you solve it now?
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also I think -1/2 is not the correct answer. As it is in 4th quad cos is always +ve
there is this direct formula which you can use cos 2x = (1 - tan^2 x) / (1 + tan^2 x)
|dw:1396900036382:dw| from the triangle in 4th quadrant you can get sin and cos \[\sin x = -\frac{3}{5}\] \[\cos x = \frac{4}{5}\] the double angle identity is: \[\cos 2x = \cos^2 x - \sin^2 x\]
ok so cos is 3/2?
wait no im lost
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cos is always between +1 and -1 lol
cos2x = 16/25-9/25
7/25= cos2x?
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