Deriving trig functions where the angle is an inverse trig function. Example: sin(acos x).
Inverse Functions
sin(asin t) = t
cos(acos t) = t
tan(atan t) = t
Other such calculations can be derived from the three Pythagorean trig identities:
sin^2 x + cos^2 x = 1
sec^2 = 1 + tan^2 x
1 + cot^2 x = csc^2 x
sin^2 x + cos^2 x = 1
Also known as (sin x)^2 + (cos x)^2 = 1
Let x = asin t where -1 ≤ t ≤ 1
(sin(asin t))^2 + (cos(asin t))^2 = 1
t^2 + (cos(asin t))^2 = 1
(cos(asin t))^2 = 1 - t^2
cos(asin t) = √(1 - t^2)
Let x = acos t, -1 ≤ t ≤ 1
(sin(acos t))^2 + (cos(acos t))^2 = 1
(sin(acos t))^2 = 1 - (cos(acos t))^2
(sin(acos t))^2 = 1 - t^2
sin(acos t) = √(1 - t^2)
sec^2 x = 1 + tan^2 x
sec^2 x = 1 + tan^2 x
1/cos^2 x = 1 + tan^2 x
Let x = atan t, -1 ≤ t ≤ 1
1/(cos(atan t))^2 = 1 + (tan(atan t))^2
1/((cos(atan t))^2 = 1 + t^2
(cos(atan t))^2 = 1/(1 + t^2)
cos(atan t) = √(1/(1 + t^2))
Let x = acos t, -1 ≤ t ≤ 1
1/(cos(acos t))^2 = 1 + (tan(acos t))^2
1/t^2 = 1 + (tan(acos t))^2
(tan(acos t))^2 = 1/t^2 - 1
(tan(acos t))^2 = (1 - t^2)/t^2
(tan(acos t)) = √(1 - t^2)/t
1 + cot^2 x = csc^2 x
1 + cot^2 x = csc^2 x
1 + 1/tan^2 x = 1/sin^2 x
Let x = asin t, -1 ≤ t ≤ 1
1 + 1/(tan(asin t))^2 = 1/(sin(asin t))^2
1/(tan(asin t))^2 = 1/t^2 - 1
1/(tan(asin t))^2 = (1 - t^2)/t^2
(tan(asin t))^2 = t^2/(1 - t^2)
tan(asin t) = t/√(1 - t^2)
Let x = atan t, -1 ≤ t ≤ 1
1 + 1/(tan(atan t))^2 = 1/(sin(atan t))^2
1 + 1/t^2 = 1/(sin(atan t))^2
(t^2 + 1)/t^2 = 1/(sin(atan t))^2
(sin(atan t))^2 = t^2/(t^2 + 1)
sin(atan t) = t/√(t^2 +1)
Summary:
cos(asin t) = √(1 - t^2)
sin(acos t) = √(1 - t^2)
cos(atan t) = √(1/(1 + t^2))
tan(acos t) = √(1 - t^2)/t
tan(asin t) = t/√(1 - t^2)
sin(atan t) = t/√(t^2 +1)
This blog is property of Edward Shore. 2014
A blog is that is all about mathematics and calculators, two of my passions in life.
Friday, October 17, 2014
sin(atan x), sin(acos x), cos(asin x), cos(atan x), tan(asin x), tan(acos x)
BA-54: Real Estate Programs
BA-54: Real Estate Programs BA-54: Chris won this calculator at HHC 2024 and donated it to me. Much appreciation as always. ...