TrigonometryWhat are the values of sinx and cosx for the acute angle x such that tanx=4/5 ?
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.`
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`
`sin x = 4/sqrt 41`
The required value of sin x is `4/sqrt 41` and cos x = `5/sqrt 41`
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