# if tgx=1/3, the is the value of sinx or coxI know, the cotgx is inverted value, so its 3 (i hope). I know, that i can calculete arctg 1/3 and after it calculate the sinx and cosx, but i think, that...

if tgx=1/3, the is the value of sinx or cox

I know, the cotgx is inverted value, so its 3 (i hope). I know, that i can calculete arctg 1/3 and after it calculate the sinx and cosx, but i think, that there is better solution...

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The enunciation of the problem has requested the values of sin x and cos x, not the values of the angle x!

We'll apply Pythagorean identity:

(sin x)^2 + (cos x)^2 = 1

We'll divide by (cos x)^2:

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

We'll replace tan x by 1/3

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

10/9 = 1/(cos x)^2 => cos x = +sqrt 9/10

Since it is not specified what is the range of values of x, the values of cosine function may be positive or negative.

cos x = -3sqrt10/10 or cos x = +3sqrt10/10

sin x = sqrt (1 - 9/10)

sin x = sqrt10/10 or sin x = -sqrt10/10

**The values of sin x and cos x, when tan x = 1/3, are: sin x = sqrt10/10 or sin x = -sqrt10/10 and cos x = -3sqrt10/10 or cos x = +3sqrt10/10.**