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Geometry Level 2

1 − sin ⁡ 2 x − sin ⁡ 2 x = cos ⁡ 2 x \sqrt{1-\sin^2x} - \sin^2x = \cos^2x

Solve for x x (in degrees) in the interval [ 0 ∘ , 36 0 ∘ ) [ 0^ \circ,360^ \circ) .

0 180 No solution 0, 180

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Peter Orton by Peter Orton , 24, Philippines

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1 solution

Peter Orton
Feb 20, 2015

1 − s i n 2 x − s i n 2 x = c o s 2 x \sqrt{1-sin^2x} - sin^2x = cos^2x c o s 2 x = 1 \sqrt{cos^2x} = 1 ∣ c o s x ∣ = 1 |cosx| = 1 c o s x = 1 cosx = 1 or c o s x = − 1 cosx = -1 x = 0 , 180 x = {0, 180}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Yes, a common mistake students make is to assume x 2 = x \sqrt{ x^2} = x only.

In case anyone didn't catch the intermediate step:

1 − sin ⁡ 2 x = cos ⁡ 2 x = ∣ cos ⁡ x ∣ \sqrt{1-\sin^{2} x} = \sqrt{\cos^{2} x} = |\cos x|

∣ cos ⁡ x ∣ = sin ⁡ 2 x + cos ⁡ 2 x = 1 |\cos x| = \sin^{2} x + \cos^{2} x = 1

Jake Lai - 6 years, 3 months ago

Surely c o s ( x ) = 1 ⇒ x = 360 ? cos(x) = 1\Rightarrow\ x= 360 \ ?

Curtis Clement - 6 years, 3 months ago

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