SAT1000 - P277

Geometry Level pending

Without using calculator , find 4 cos ⁡ 50 ° − tan ⁡ 40 ° 4 \cos 50 \degree - \tan 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 1732.

Alice Smith by Alice Smith , 20, China

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2 solutions

David Vreken
Jun 25, 2020

4 cos ⁡ 50 ° − tan ⁡ 40 ° 4 \cos 50° - \tan 40°

= 4 sin ⁡ 40 ° − tan ⁡ 40 ° = 4 \sin 40° - \tan 40°

= 4 sin ⁡ 40 ° − sin ⁡ 40 ° cos ⁡ 40 ° = 4\sin 40° - \frac{\sin 40°}{\cos 40°}

= 2 cos ⁡ 40 ° ( 2 sin ⁡ 40 ° cos ⁡ 40 ° − 1 2 sin ⁡ 40 ° ) = \frac{2}{\cos 40°}(2 \sin 40° \cos 40° - \frac{1}{2} \sin 40°)

= 2 cos ⁡ 40 ° ( sin ⁡ 80 ° − 1 2 sin ⁡ 40 ° ) = \frac{2}{\cos 40°}(\sin 80° - \frac{1}{2} \sin 40°)

= 2 cos ⁡ 40 ° ( sin ⁡ 100 ° − 1 2 sin ⁡ 40 ° ) = \frac{2}{\cos 40°}(\sin 100° - \frac{1}{2} \sin 40°)

= 2 cos ⁡ 40 ° ( sin ⁡ ( 60 ° + 40 ° ) − 1 2 sin ⁡ 40 ° ) = \frac{2}{\cos 40°}(\sin (60° + 40°) - \frac{1}{2} \sin 40°)

= 2 cos ⁡ 40 ° ( sin ⁡ 60 ° cos ⁡ 40 ° + cos ⁡ 60 ° sin ⁡ 40 ° − 1 2 sin ⁡ 40 ° ) = \frac{2}{\cos 40°}(\sin 60° \cos 40° + \cos 60° \sin 40° - \frac{1}{2} \sin 40°)

= 2 cos ⁡ 40 ° ( 3 2 cos ⁡ 40 ° + 1 2 sin ⁡ 40 ° − 1 2 sin ⁡ 40 ° ) = \frac{2}{\cos 40°}(\frac{\sqrt{3}}{2} \cos 40° + \frac{1}{2} \sin 40° - \frac{1}{2} \sin 40°)

= 2 cos ⁡ 40 ° ( 3 2 cos ⁡ 40 ° ) = \frac{2}{\cos 40°}(\frac{\sqrt{3}}{2} \cos 40°)

= 3 = \sqrt{3}

Therefore, A = 3 A = \sqrt{3} and ⌊ 1000 A ⌋ = 1732 \lfloor 1000A \rfloor = \boxed{1732} .

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The value of the given expression after simplification is tan ⁡ 60 ° = 3 = 1.73205080... \tan 60\degree =\sqrt 3=1.73205080... .

So, the answer is 1732 \boxed {1732} .

0 Helpful 0 Interesting 0 Brilliant 2 Confused

What simplication is required?

Pi Han Goh - 11 months, 3 weeks ago

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