Trigonometry! #117

Geometry Level 2

Find the value of 3 csc ⁡ 2 0 ∘ − sec ⁡ 2 0 ∘ \sqrt{3}\csc20^{\circ}-\sec20^{\circ}

This problem is part of the set Trigonometry .


The answer is 4.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

This expression can be written as

3 sin ⁡ ( 2 0 ∘ ) − 1 cos ⁡ ( 2 0 ∘ ) = 3 cos ⁡ ( 2 0 ∘ ) − sin ⁡ ( 2 0 ∘ ) sin ⁡ ( 2 0 ∘ ) cos ⁡ ( 2 0 ∘ ) = \dfrac{\sqrt{3}}{\sin(20^{\circ})} - \dfrac{1}{\cos(20^{\circ})} = \dfrac{\sqrt{3}\cos(20^{\circ}) - \sin(20^{\circ})}{\sin(20^{\circ})\cos(20^{\circ})} =

2 ∗ ( 3 2 cos ⁡ ( 2 0 ∘ ) − 1 2 sin ⁡ ( 2 0 ∘ ) ) 1 2 sin ⁡ ( 4 0 ∘ ) = \dfrac{2*(\frac{\sqrt{3}}{2}\cos(20^{\circ}) - \frac{1}{2}\sin(20^{\circ}))}{\frac{1}{2}\sin(40^{\circ})} =

4 ∗ ( cos ⁡ ( 3 0 ∘ ) cos ⁡ ( 2 0 ∘ ) − sin ⁡ ( 3 0 ∘ ) sin ⁡ ( 2 0 ∘ ) ) sin ⁡ ( 4 0 ∘ ) = 4 cos ⁡ ( 5 0 ∘ ) sin ⁡ ( 4 0 ∘ ) = 4 . \dfrac{4*(\cos(30^{\circ})\cos(20^{\circ}) - \sin(30^{\circ})\sin(20^{\circ}))}{\sin(40^{\circ})} = \dfrac{4\cos(50^{\circ})}{\sin(40^{\circ})} = \boxed{4}.

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@brian charlesworth By the way, could you please post a solution for this problem? I can't seem to figure it out.

Omkar Kulkarni - 6 years, 3 months ago

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I've just posted a solution to the problem you have referred to. It might not be what you had in mind, but it does the job. :)

Brian Charlesworth - 6 years, 3 months ago

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Yeah thanks!

Omkar Kulkarni - 6 years, 3 months ago

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