Trigonometry! #121

Geometry Level 2

Given that A A and B B are complementary acute angles, then

cos ⁡ A sin ⁡ B − cos ⁡ A sin ⁡ B = ? \sqrt{\frac{\cos A}{\sin B}-\cos A \sin B} = \, \color{#D61F06}{?}

Assume A A and B B are positive.

cos ⁡ A \cos A sin ⁡ A \sin A tan ⁡ A \sqrt{\tan A} tan ⁡ A \tan A

Omkar Kulkarni by Omkar Kulkarni , 22, India

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3 solutions

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Eli Ross Staff
May 16, 2016

Relevant wiki: Trigonometric Co-function Identities

Note that sin ⁡ B = sin ⁡ ( 9 0 ∘ − A ) = cos ⁡ ( A ) , \sin B = \sin(90^\circ - A) = \cos(A), so the expression becomes cos ⁡ A cos ⁡ A − cos ⁡ A cos ⁡ A = 1 − cos ⁡ 2 A = sin ⁡ 2 A = ∣ sin ⁡ A ∣ = sin ⁡ A , \sqrt{\frac{\cos A}{\cos A} - \cos A \cos A} = \sqrt{1-\cos^2A} = \sqrt{\sin^2 A} = \left|\sin A\right| =\sin A, where we have made use of the fact that sin ⁡ A > 0 \sin A > 0 and the pythagorean identity sin ⁡ 2 A + cos ⁡ 2 A = 1. \sin^2 A + \cos^2 A = 1.

3 Helpful 0 Interesting 1 Brilliant 1 Confused

Is this expression really equivalent to the sine? Shouldn't a restricted domain be stated? (i.e., 0 + 2 π n ≤ x ≤ π + 2 π n 0+2\pi n \leq x \leq \pi+2\pi n )

Edwin Hughes - 5 years ago

Why it can't be sin A

Rohit Pati - 2 years, 4 months ago
Mevlut Esen
May 18, 2016

An accute angle means an angle that accumulate 90 degrees toplam your angle. From that; cosA/sinB=1 and cosA×sinB=cos^2 (A) We all know 1-cos^2 (A)=sin^2 (A)

Now from here sqrt (sin^2 (A))=sinA

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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