how do i find cos(arcsin 7/25)?

Question

Thanks!

thank you guys soo much!!

Answer 1

let θ = sin^-1(7/25) then sinθ = 7/25

then cos(arcsin7/25) = cos θ cuz we let arcsin7/25 = θ


from sinθ = 7/25 you can draw a right triangle
sinθ = opp/hyp
therefore opp = 7, hyp = 25
adjacent = √(25^2 -7^2) = √(576) = 24

cos θ = adj/ hyp = 24/25

final answer 24/25

Answer 2

sqrt [ 1 - (7/25)^2 ]
sqrt [ 1 - 49/625 ]
sqrt [ 576/625 ]
24 / 25

Here's the other way to think of it.

Given a right triangle with a leg of length 7 and a hypotenuse of length 25. The other leg, using Pythagorean Theorem, has a length of 24.

So cos (arcsin 7/25) = 24/25

Answer 3

Draw a right triangle and label the hypotenuse 25 and one of the sides 7. Find the 3rd side using the Pythagorean Thm. Now the cos(arcsine 7/25) will be that 3rd side over 25.

that's it! :)

Answer 4

This means that the opposite/hyp= 7/25 (this is a right triangle)
Use the Pythagorean Theorem to find the adjacent side
(which is 24)
cosine is opposite/hyp
so the answer is 24/25

Answer 5

let arcsin(7/25) = x

by definition

sinx = 7/25

so cosx = sqrt(1-sin^2(x))

cosx = sqrt(1 - 49/625)

cosx = sqrt(576/625) = 24/25

so cos(arcsin7/25) = cos(x) = 24/25

Answer 6

Let θ = arc (sin 7/25)
x ² = 25 ² - 7 ²
x ² = 576
x = 24
cos θ = 24/25

Answer 7

set sinu = 7/25
to get cos u , its equal to = root(1-sin^2u)
cos(u) = root(1-0.28^2)