Find sin 2x, cos 2x, and tan 2x from the given infomation

csc x=4, tan x < 0

thanks but I'm still stumped by what they mean about how and what the double angle formula is. The double angle formula confuses me

To find sin 2x, cos 2x, and tan 2x, we can use the trigonometric identities.

Given the information that csc x = 4 and tan x < 0, we can find the values of sin x, cos x, and tan x.

First, we know that csc x is the reciprocal of sin x. Therefore, sin x = 1/csc x = 1/4.

Since tan x < 0, we know that tan x is negative. From the definition of tan x, we know that tan x = sin x / cos x. Since sin x = 1/4, cos x = cos x = √(1 - sin^2 x) = √(1 - (1/4)^2) = √(1 - 1/16) = √(15/16) = √15/4. Since tan x is negative, cos x must be positive.

Now, we can use the double angle trigonometric identities to find sin 2x, cos 2x, and tan 2x.

sin 2x = 2sin x cos x
cos 2x = cos^2 x - sin^2 x
tan 2x = (2tan x) / (1 - tan^2 x)

Substituting the values we found earlier, we have:

sin 2x = 2(1/4)(√15/4) = √15 / 8
cos 2x = (√15/4)^2 - (1/4)^2 = 15/16 - 1/16 = 14/16 = 7/8
tan 2x = (2(1/4)) / (1 - (1/4)^2) = 1/2 / (15/16) = 16/30 = 8/15

Therefore, sin 2x = √15 / 8, cos 2x = 7/8, and tan 2x = 8/15.

csc positive, tan negative means QII

since sinx = 1/cscx,

sinx = 1/4
cosx = √15/4
tanx = 1/√15

Now just apply your double-angle formulas

sin2x = 2 sinx cosx

= 2(1/4)(√15/4) = 2√15/16

cos2x = cos^2x - sin^2 x
= 15/16 - 1/16 = 7/8

tan2x = 2tanx/(1-tan^2 x)
= 2/√15 / (1 - 1/15)
= 2/√15 * 15/14
= √15/7