Question

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

Answer 1

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))

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

sin(2x) = 2sin(x)cos(x)
sin(2x) = 2(3/sqrt(13))(-2/sqrt(13))
sin(2x) = 2(-6/13)
sin(2x) = -12/13

Answer 6

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 :)

Answer 7

that is ridiculous.