find sin 2x, cos 2x, and tan 2x, if cosx= -2/sqrt13 and x terminates in quadrant 2

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find sin 2x, cos 2x, and tan 2x, if cosx= -2/sqrt13 and x terminates in quadrant 2

Mathematics
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sin(2x) has the formula: sin(2x) = 2sin(x)cos(x) And cos(2x): cos(2x) = cos^2(x) - sin^2(x) = 1 - 2sin^2(x) = 2cos^2(x) -1 And for tan(2x) we get: tan(2x) = 2tan(x)/(1-tan^2(x))
if cos(x) = -2/sqrt(13) we know how to construct a right triangle to find all the other ....stuff....we need. cos = x/r and the pythagorean theorum states: r^2 = x^2 + y^2.
y = sqrt(sqrt(13)^2 - (-2)^2) y = sqrt(13 - 4) = sqrt(9) = 3 y = 3; x=-2 ; and r = sqrt(13) sin(x) = y/r = 3/sqrt(13) tan(x) = y/x = -3/2

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Other answers:

now its all plug and play :) cos(2x) = cos^2 - sin^2 cos(2x) = (-2/sqrt(13))^2 - (3/sqrt(13))^2 cos(2x) = 4/13 - 9/13 cos(2x) = -5/13
sin(2x) = 2sin(x)cos(x) sin(2x) = 2(3/sqrt(13))(-2/sqrt(13)) sin(2x) = 2(-6/13) sin(2x) = -12/13
tan(2x) = 2tan(x) / (1-tan^2(x)) tan(2x) = 2(-3/2) / (1 - (-3/2)^2) tan(2x) = -3/ (1 - 9/4) tan(2x) = -3 / (-5/4) tan(2x) = -3(-4/5) tan(2x) = 12/5 if I didnt mess it up :)
that is ridiculous.

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