Another Trigo Problem

Geometry Level 2

csc ⁡ 1 0 ∘ + csc ⁡ 5 0 ∘ − csc ⁡ 7 0 ∘ = ? \large \csc10^\circ + \csc 50^\circ - \csc 70^\circ = \ ?


The answer is 6.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Chew-Seong Cheong
May 24, 2017

x = csc ⁡ 1 0 ∘ + csc ⁡ 5 0 ∘ − csc ⁡ 7 0 ∘ = 1 sin ⁡ 1 0 ∘ + 1 sin ⁡ 5 0 ∘ − 1 sin ⁡ 7 0 ∘ = 1 sin ⁡ 1 0 ∘ + 1 sin ⁡ ( 6 0 ∘ − 1 0 ∘ ) − 1 sin ⁡ ( 6 0 ∘ + 1 0 ∘ ) = 1 sin ⁡ 1 0 ∘ + 1 3 2 cos ⁡ 1 0 ∘ − 1 2 sin ⁡ 1 0 ∘ − 1 3 2 cos ⁡ 1 0 ∘ + 1 2 sin ⁡ 1 0 ∘ = 1 sin ⁡ 1 0 ∘ + 4 sin ⁡ 1 0 ∘ 3 cos ⁡ 2 1 0 ∘ − sin ⁡ 2 1 0 ∘ = 1 sin ⁡ 1 0 ∘ + 4 sin ⁡ 1 0 ∘ 3 ( 1 − sin ⁡ 2 1 0 ∘ ) − sin ⁡ 2 1 0 ∘ = 1 sin ⁡ 1 0 ∘ + 4 sin ⁡ 1 0 ∘ 3 − 4 sin ⁡ 2 1 0 ∘ = 3 − 4 sin ⁡ 2 1 0 ∘ + 4 sin ⁡ 2 1 0 ∘ 3 sin ⁡ 1 0 ∘ − 4 sin ⁡ 3 1 0 ∘ = 3 sin ⁡ 3 0 ∘ = 6 \begin{aligned} x & = \csc 10^\circ + \csc 50^\circ - \csc 70^\circ \\ & = \frac 1{\sin 10^\circ} + \frac 1{\sin 50^\circ} - \frac 1{\sin 70^\circ} \\ & = \frac 1{\sin 10^\circ} + \frac 1{\sin (60^\circ-10^\circ)} - \frac 1{\sin (60^\circ+10^\circ)} \\ & = \frac 1{\sin 10^\circ} + \frac 1{\frac {\sqrt 3}2 \cos 10^\circ - \frac 12 \sin 10^\circ} - \frac 1{\frac {\sqrt 3}2 \cos 10^\circ + \frac 12 \sin 10^\circ} \\ & = \frac 1{\sin 10^\circ} + \frac {4\sin 10^\circ}{3\cos^2 10^\circ - \sin^2 10^\circ} \\ & = \frac 1{\sin 10^\circ} + \frac {4\sin 10^\circ}{3(1-\sin^2 10^\circ) - \sin^2 10^\circ} \\ & = \frac 1{\sin 10^\circ} + \frac {4\sin 10^\circ}{3 - 4\sin^2 10^\circ} \\ & = \frac {3 - 4\sin^2 10^\circ + 4\sin^2 10^\circ}{3\sin 10^\circ - 4\sin^3 10^\circ} \\ & = \frac 3{\sin 30^\circ} \\ & = \boxed{6} \end{aligned}

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cosec 10 + cosec 50 - cosec 70 ..

sec 80 + sec 40 - sec 20 . .. (1/cos 80) + (1/cos 40) - (1/cos 20) ...

(cos 40 + cos 80) / (cos 80 cos 40) - (1/cos 20) ...

Apply cos A + cos B = 2 cos ((A + B)/2) cos ((A - B)/2) ..

(2 cos 60 cos 20) / (cos 40 cos 80) - (1/cos 20) ..

(cos 20) / (cos 40 cos 80) - (1/cos 20) ...

Take LCM and add both ...

(cos²20 - cos 40 cos 80) / (cos 20 cos 40 cos 80) ...

I am gonna solve the denominator seperately, ...

cos 20 cos 40 cos 80 ..

Multiply and divide by 2 sin 20 ..

(2 sin 20 cos 20 cos 40 cos 80) / (2 sin 20) ..

2 sin A cos A = sin 2A ..

(sin 40 cos 40 cos 80) / (2 sin 20) .. . Multiply and divide by 2 ...

(2 sin 40 cos 40 cos 80) / (4 sin 20) ...

(sin 80 cos 80) / (4 sin 20) ....

Again 2, ....

sin 160 / (8 sin 20) ....

sin (180 - 20) = sin 20 ....

sin 160 = sin 20 .....

= 1/8 ...

So, denominator is equal to 1/8 ...

Our expression becomes....,

8 (cos²20 - cos 40 cos 80) ....

Take one 2 inside and use 2 cos A cos B = cos ((A + B)/2) + cos ((A - B)/2) .....

4 (2cos²20 - (cos 120 + cos 40)) ....

2cos²A = 1 + cos 2A ....

4 (1 + cos 40 - cos 120 - cos 40) .....

4 (1 - cos 120) ....

4 (1 - (-1/2)) ....

4 (3/2) = 6....

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