# What are the values of sinx and cosx for the acute angle x such that tanx = 4/5 ?

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We know that tan x = adjacent side/ opposite side. Now consider x to be angle in a right angled triangle such that tan x = 4/5.

The hypotenuse of the right angled triangle is equal to sqrt( 4^2 + 5^2) = sqrt (16 + 25) = sqrt 41.

sin x = opposite side / hypotenuse or 4 / sqrt 41.

cos x = adjacent side / hypotenuse = 5 / sqrt 41.

**The required value of sin x is 4/ sqrt 41 and cos x is 5/ sqrt 41.**

To check:

arc tan (4/5) = 38.65 degrees

arc sin (4/ sqrt 41)= 38.65 degrees

arc cos (5/ sqrt 41) = 38.65 degrees

We know that the tangent function is a ratio of the opposite cathetus and adjacent cathetus.

We'll recall that the opposite side to the acute angle, in the unit circle, is the y component. But y component, in the unit circle, is the value of the sine function.

We also know that the adjacent side to the acute angle, in the unit circle, is the x component. But x component, in the unit circle, is the value of the cosine function.

tan x = sin x/cos x

But tan x = 4/5

4/5 = sin x/cos x

We'll apply the fundamental formula of trigonometry:

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

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

cos x = 1/sqrt(16/25 + 1)

cos x = +/- sqrt (25/41)

cos x = +/- 5sqrt41/41

sin x = +/-sqrt (1 - 25/41)

sin x = +/-sqrt 16/25

sin x = +/- 4/5

tanx = 4/5.

To find sinx and cosx for the acute angle x.

We know if tanx = t, is given, then sinx = t/(sqrt(1+t^2) and cosx = 1/sqrt(1+t^2).

t = tanx = 4/5.

Therefore sinx = t/sqrt(1+t^2) .

sinx = (4/5)/sqrt{1+(4/5)^2}.

sinx = 4/sqrt{5^2+4^2}

sinx = 4/sqrt(25+16).

sinx = 4/(41)

cosx = 1/sqrt(1+t^2). = 1/sqrt{1+(4/5)^2}

cosx = 5/sqrt(5^2+4^2)

cosx = 5/sqrt(41)