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)
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