What is the exact value of sin(arccos 3/5) and why?

Question

What is the exact value of sin(arccos 3/5) and why?

Answer 1

Since the range of arccos is [0, π], we know that for any x, sin (arccos x) is positive, and so √sin² (arccos x) = (sin arccos x). Then by basic trigonometry we have:

sin (arccos x) = √sin² (arccos x) = √(1 - cos² (arccos x)) = √(1 - x²).

So in particular, sin (arccos (3/5)) = √(1 - (3/5)²) = √(1 - 9/25) = √(16/25) = 4/5.

Answer 2

cos = adj/hyp

so for the triangle we have the hypotenuse = 5 and one side = 3 so we can do:

5^2 = 3^2+x^2
25 = 9 +x^2
16 = x^2
x = 4

So since sin = opp/hyp

the sin (arccos (3/5)) = 4/5

Answer 3

sin(arccos(3/5)

let arccos(3/5) = x so sin(arccos(3/5)) = sinx

when arccos(3/5) = x

by definition cosx = 3/5

sin^2(x) = (1-cos^2(x)) = (1-9/25) = 16/25

sinx=4/5 (since range of arccos is from 0,pi sin(arccos) is always +)

so sin(arccos(3/5) = sinx = 4/5

Answer 4

Just complete the triangle. Arccos 3/5 is the angle whose cos is 3/5. i.e. adjacent over hypotenuse is 3/5. By Pythagoras, opposite side is 4.

The sin (opposite over hypotenuse) of the angle whose cos is 3/5 is therefore 4/5

Answer 5

The arccos of 3/5 is the angle whose cosine equals 3/5, or 53.13 degrees. The sine of this angle equals 4/5.

Answer 6

arccos 3/5 = 53.13º
sin 53.13º = 0.6 = 3/5

Answer 7

tan(cos^-1(-3/5)) = ? Let, cos^-1(-3/5) = x, ===> cosx = -3/5, SO, adjasent = -3, & hypotenus = 5, SO, opposite = sqrt[5^2 - (-3)^2] = +/- 4, Thus, tan(cos^-1(-3/5)) = 4/(-3) = -4/3 >=======================< ANSWER ALSO, tan(cos^-1(-3/5)) = -4/(-3) = 4/3 >=======================< ANSWER