Canceling Out Sins (and cosins)

Geometry Level 2

For x x such that 0 < x < π 2 0<x<\dfrac{\pi}{2} , the expression 1 − cos ⁡ 2 x sin ⁡ x + 1 − sin ⁡ 2 x cos ⁡ x = y \dfrac{\sqrt{1-\cos^2 x}}{\sin x}+\dfrac{\sqrt{1-\sin^2 x}}{\cos x}=y . Enter the value of y y .

Note that sin ⁡ 2 x = ∣ sin ⁡ x ∣ = sin ⁡ x \displaystyle \sqrt{\sin^2 x} = |\sin x| = \sin x and cos ⁡ 2 x = ∣ cos ⁡ x ∣ = cos ⁡ x ∀ 0 < x < π 2 \displaystyle \sqrt{\cos^2 x} = |\cos x| = \cos x \ \ \forall \ \ 0 < x < \frac{\pi}{2}


The answer is 2.

View wiki
Terry Yu by Terry Yu , 17, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Terry Yu
May 16, 2017

First, you have to know that s i n 2 x = 1 − c o s 2 x sin^2x=1-cos^2x and c o s 2 x = 1 − s i n 2 x cos^2x=1-sin^2x . Therefore, we can replace the equation with s i n 2 x s i n x + c o s 2 x c o s x = y \dfrac{\sqrt{sin^2x}}{sin x}+\dfrac{\sqrt{cos^2x}}{cos x}=y . This is equal to

s i n x s i n x + c o s x c o s x = 1 + 1 = 2 \dfrac{sinx}{sinx}+\dfrac{cosx}{cosx}=1+1=\large\color{#EC7300}\boxed{2}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Note that sin ⁡ 2 x = ∣ sin ⁡ x ∣ = sin ⁡ x \displaystyle \sqrt{\sin^2 x} = |\sin x| = \sin x and cos ⁡ 2 x = ∣ cos ⁡ x ∣ = cos ⁡ x ∀ 0 < x < π 2 \displaystyle \sqrt{\cos^2 x} = |\cos x| = \cos x \ \ \forall \ \ 0 < x < \frac{\pi}{2}

Zach Abueg - 4 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...