You need to find sin 2x and cos 2x, hence you need to remember the formulas for din 2x and cos 2x such that:

`sin 2x = 2 sin x*cos x`

Notice that if you don't remember this formula you may always find it using `sin 2x = sin (x+x).`...

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You need to find sin 2x and cos 2x, hence you need to remember the formulas for din 2x and cos 2x such that:

`sin 2x = 2 sin x*cos x`

Notice that if you don't remember this formula you may always find it using `sin 2x = sin (x+x).` Expanding `sin(x+x)` yields:

`sin(x+x)= sin x*cos x + sin x*cos x = 2 sin x*cos x`

`cos 2x =cos^2 x- sin^2 x`

The problem provides an information that is not correct. Since`tan x = 5/3,` then sin x need to be larger than 0 and not smaller than zero, because the information `tan x = 5/3` and sin x<0 is a contradiction.

Considering `tan x = 5/3` and sin x>0 yields:

`1 + tan^2 x = 1/(cos^2 x)`

`` `1 + 25/9 = 1/(cos^2 x)`

`34/9 = 1/(cos^2 x) =gt cos^2 x = 9/34 =gt cos x = 3sqrt34/34`

Notice that cos x need to be positive to keep the positive value of `tan x = 5/3` .

You may find `sin^2 x` using the basic trigonometric formula such that:

`sin^2 x + cos^2 x = 1 =gt sin^2 x = 1 - cos^2 x`

`sin^2 x = 1 - 9/34`

`` `sin^2 x = 25/34 =gt sin x = 5sqrt34/34`

Hence, `sin 2x = 2 sin x*cos x = 30*34/34^2 = 30/34 = 15/17`

`` `cos 2x = 9/34 - 25/34 = -16/34 = -8/17`

**Hence, evaluating the values for sin 2x and cos 2x yields: `sin 2x = 15/17 ` and `cos 2x = -8/17` .**