Nested trig

Calculus Level pending

Let f ( x ) ≡ arcsin ⁡ ( cos ⁡ x ) f\left( x \right) \equiv \arcsin { (\cos { x } ) } where x x is measured in radians and 0 < x < π 0<x<\pi

What is the value of the following expression?

∣ f ′ ( 2 π 13 ) ∣ \left| f^{ \prime }\left( \frac { 2\pi }{ 13 } \right) \right|


The answer is 1.

Andrew Sellek by Andrew Sellek , 24, United Kingdom

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1 solution

First, let y = f ( x ) y=f(x)

Then, let u = cos ⁡ x u = \cos x

Therefore,

d y d u = 1 1 − u 2 \frac{dy}{du} = \frac {1}{\sqrt{1-u^2}}

d u d x = − sin ⁡ x \frac{du}{dx} = -\sin x

Combining these gives:

d y d x = 1 1 − cos ⁡ 2 x × − sin ⁡ x \frac{dy}{dx} = \frac {1} {\sqrt{1-\cos^2 x}} \times -\sin x

d y d x = − sin ⁡ x ∣ sin ⁡ x ∣ \frac{dy}{dx} = \frac {-\sin x} {|\sin x|}

In the given domain, sin ⁡ x \sin x is always positive. Therefore,

d y d x = − 1 = f ′ ( x ) \frac{dy}{dx} = -1 = f'(x)

So

∣ f ′ ( x ) ∣ = 1 |f'(x)| = 1

In the domain then, the differential is always equal to 1 \boxed{1}

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