SAT1000 - P275

Algebra Level 3

Without using calculator , Find cos ⁡ 10 ° tan ⁡ 20 ° + 3 sin ⁡ 10 ° tan ⁡ 70 ° − 2 cos ⁡ 40 ° \dfrac{\cos 10 \degree}{\tan 20 \degree}+\sqrt{3} \sin 10 \degree \tan 70 \degree - 2 \cos 40 \degree .

Let A A denote the answer. Submit ⌊ 1000 A ⌋ \lfloor 1000A \rfloor .

Have a look at my problem set: SAT 1000 problems


The answer is 2000.

Alice Smith by Alice Smith , 20, China

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1 solution

Chew-Seong Cheong
Jun 20, 2020

X = cos ⁡ 1 0 ∘ tan ⁡ 2 0 ∘ + 3 sin ⁡ 1 0 ∘ tan ⁡ 7 0 ∘ − 2 cos ⁡ 4 0 ∘ = cos ⁡ 1 0 ∘ + 3 sin ⁡ 1 0 ∘ tan ⁡ 2 0 ∘ − 2 cos ⁡ 4 0 ∘ = 2 sin ⁡ 4 0 ∘ tan ⁡ 2 0 ∘ − 2 cos ⁡ 4 0 ∘ = 4 sin ⁡ 2 0 ∘ cos ⁡ 2 0 ∘ sin ⁡ 2 0 ∘ cos ⁡ 2 0 ∘ − 2 cos ⁡ 4 0 ∘ = 4 cos ⁡ 2 2 0 ∘ − 2 cos ⁡ 4 0 ∘ = 2 cos ⁡ 4 0 ∘ + 2 − 2 cos ⁡ 4 0 ∘ = 2 \begin{aligned} X & = \frac {\cos 10^\circ}{\tan 20^\circ} + \sqrt 3 \sin 10^\circ \tan 70^\circ - 2 \cos 40^\circ \\ & = \frac {\cos 10^\circ + \sqrt 3 \sin 10^\circ}{\tan 20^\circ} - 2 \cos 40^\circ \\ & = \frac {2 \sin 40^\circ}{\tan 20^\circ} - 2 \cos 40^\circ \\ & = \frac {4 \sin 20^\circ \cos 20^\circ}{\frac {\sin 20^\circ}{\cos 20^\circ}} - 2 \cos 40^\circ \\ & = 4 \cos^2 20^\circ - 2 \cos 40^\circ \\ & = 2 \cos 40^\circ + 2 - 2 \cos 40^\circ \\ & = \boxed 2 \end{aligned}

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