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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use the following trigonometric identity, such that:

`tan x = (sin x)/(cos x)`

The problem provides the information that `tan x = 4/5` , hence, you need to substitute `4/5` in identity above, such that:

`4/5 = (sin x )/(cos x)`

`5 sin x = 4 cos x => sin x = (4 cos x)/5`

You need to use the following basic identity, such that:

`sin^2 x + cos^2 x = 1 => ((4cos x )^2)/25 + cos^2 x = 1`

`16cos^2 x + 25 cos^2 x = 25 => 41cos^2 x = 25`

`cos^2 x = 25/41 => cos x = +-sqrt(25/41 )`

The problem provides the information that x is an acute angle, hence `cos (x) = 5/sqrt41.`

`tan x = sin x/(5/sqrt41) => 4/5 = sin x*sqrt 41/5`

`sin x = 4/sqrt 41`

Hence, evaluating sin x and cos x, under the given conditions, yields `cos x = 5/sqrt41` , `sin x = 4/sqrt 41.`

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The value of sin x and cos x has to be determined such that tan x=4/5. Also, x lies between 0 and 90 degrees.

tan x = 4/5

Use the relation `1 + tan^2x = sec^2x`

`sec^2x = (4/5)^2 + 1`

`sec^2x = (16/25) + 1`

`sec^2x = 41/25`

`sec x = sqrt 41/5`

`cos x = 5/sqrt 41`

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

= `1 - 25/41`

= `16/41`

`sin x = 4/sqrt 41`

The required value of sin x is `4/sqrt 41` and cos x = `5/sqrt 41`

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

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