A geometry problem by Jus Jaisinghani

Geometry Level 4

4 cos ⁡ 2 0 ∘ − 3 cot ⁡ 2 0 ∘ = ? \large 4\cos20^\circ-\sqrt{3}\cot20^\circ= \, ?

Calculate the above without using a calculator.


The answer is -1.

Jus Jaisinghani by Jus Jaisinghani , 21, India

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1 solution

Jus Jaisinghani
Jul 7, 2016
  • 4 cos ⁡ 2 0 ∘ − 3 cot ⁡ 2 0 ∘ 4\cos20^\circ-\sqrt{3}\cot20^\circ = 4 cos ⁡ 2 0 ∘ 4\cos20^\circ - 3 cos ⁡ 2 0 ∘ s i n 2 0 ∘ \frac {\sqrt{3}\cos20^\circ}{sin20^\circ}
  • = 4 sin ⁡ 2 0 ∘ cos ⁡ 2 0 ∘ − 3 cos ⁡ 2 0 ∘ sin ⁡ 2 0 ∘ \frac{4\sin20^\circ \cos20^\circ-\sqrt{3}\cos20^\circ}{\sin20^\circ} = 2 sin ⁡ 4 0 ∘ − 3 cos ⁡ 2 0 ∘ sin ⁡ 2 0 ∘ \frac{2\sin40^\circ-\sqrt{3}\cos20^\circ}{\sin20^\circ}
  • Now ,
  • sin ⁡ 2 0 ∘ − 3 cos ⁡ 2 0 ∘ = 2 ⋅ ( cos ⁡ 6 0 ∘ sin ⁡ 2 0 ∘ − sin ⁡ 6 0 ∘ cos ⁡ 2 0 ∘ ) = − 2 sin ⁡ 4 0 ∘ \sin20^\circ-\sqrt{3}\cos20^\circ=2\cdot(\cos60^\circ\sin20^\circ-\sin60^\circ\cos20^\circ)=-2\sin40^\circ
  • thus, sin ⁡ 2 0 ∘ − 3 cos ⁡ 2 0 ∘ = − 2 sin ⁡ 4 0 ∘ \sin20^\circ-\sqrt{3}\cos20^\circ=-2\sin40^\circ

  • 3 cos ⁡ 2 0 ∘ = 2 sin ⁡ 4 0 ∘ + s i n 2 0 ∘ \sqrt{3}\cos20^\circ=2\sin40^\circ+sin20^\circ

  • Substituting in above equation we get,

  • 4 cos ⁡ 2 0 ∘ − 3 cot ⁡ 2 0 ∘ 4\cos20^\circ-\sqrt{3}\cot20^\circ = 2 sin ⁡ 4 0 ∘ − ( 2 sin ⁡ 4 0 ∘ + s i n 2 0 ∘ ) sin ⁡ 2 0 ∘ \frac{2\sin40^\circ-(2\sin40^\circ+sin20^\circ)}{\sin20^\circ}
  • = − sin ⁡ 2 0 ∘ sin ⁡ 2 0 ∘ = − 1 =\frac{-\sin20^\circ}{\sin20^\circ}=\boxed{-1}
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