Evaulate 2x exactly... help me please?

Question

hi, my final is tonight and I'm trying to study up.

if cot x 3/7
and sin x < 0
evaulate cos 2x exactly

thanks a lot for any help!

cot x = 3/7!

sorry................

Answer 1

cot(x) = adjacent side /opposite side

since cot(x) = 3/7

so adj. side = 3 and

opposite side = 7

hypotenuse = sqrt(opp.side^2 + adj.side^2)

hypotenuse = sqrt(49 + 9) = sqrt(58)

so sin(x) = opp. side/hypotenuse = - 7/sqrt(58) (since sin(x)<0)

now cos(2x) = 1 - 2sin^2(x)

cos(2x) = 1 - 2(-7/sqrt(58))^2

cos(2x) = 1 - 2(49/58)

cos(2x) = 1- 49/29

cos(2x) = -20/29

Answer 2

cotx = 3/7 which is positive
sinx < 0 and since cotx = cos/sin then cos x must be negative also. Quadrant III

cotx = adj/opp and by pythagorean theorem, this makes the hypotenuse = sqrt(3^2 + 7^2) = sqrt(58)

then sinx = opp/hyp = -7/sqrt58
cos 2x = 1 - 2sin^2x
cos2x = 1 - 2(-7/sqrt58)^2
cos2x = 1 - 2(49/58 )
cos 2x = 1 - 49/29
cos 2x = -20/29

Answer 3

If cotx > 0 and sinx<0, then x terminates in the third quadrant.
A quick sketch and the Pythagorean Theorem reveals that
cos(x) = -3/sqrt(58). But cos(2x) = 2cos^2(x) - 1, so substituting, cos(2x) = -20/29, upon reducing the resulting fraction.

Answer 4

help us help you
what do you mean by
cot x 3/7
if it was a simple fraction, it would come before the x, yes?
is the 3/7 an exponent?
use parentheses and brackets like cot[(3/7)x]
or the carat sign, "^" (shift 6), to denote an exponent and use parentheses there as well like cot[x^(3/7)]
anybody who tries to answer without knowing specifically will be working an answer and most likely in the wrong direction.
standing by!