A geometry problem by Mayank Srivastava

Geometry Level 3

What is the value of

( cot 7 0 + 4 cos 7 0 ) 2 ? ( \cot 70 ^ \circ + 4 \cos 70 ^ \circ ) ^ 2 ?


The answer is 3.00.

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2 solutions

Chew-Seong Cheong
Feb 28, 2015

cot 7 0 + 4 cos 7 0 = k cos 7 0 sin 7 0 + 4 cos 7 0 = k cos 7 0 + 4 sin 7 0 cos 7 0 = k sin 7 0 cos 7 0 + 2 sin 14 0 = k sin 7 0 cos 7 0 k sin 7 0 = 2 sin 14 0 cos 7 0 k sin 7 0 = 2 cos 13 0 cos 7 0 k sin 7 0 = 2 ( cos 6 0 cos 7 0 sin 6 0 sin 7 0 ) cos 7 0 k sin 7 0 = 2 ( 1 2 cos 7 0 3 2 sin 7 0 ) cos 7 0 k sin 7 0 = cos 7 0 3 sin 7 0 k = 3 \begin{aligned} \cot {70^\circ} + 4\cos {70^\circ} & = \sqrt{k} \\ \dfrac {\cos {70^\circ}}{\sin {70^\circ}} + 4\cos {70^\circ} & = \sqrt{k} \\ \cos {70^\circ} + 4\sin {70^\circ} \cos {70^\circ} & = \sqrt{k} \sin {70^\circ} \\ \cos {70^\circ} + 2\sin {140^\circ} & = \sqrt{k} \sin {70^\circ} \\ \cos {70^\circ} - \sqrt{k} \sin {70^\circ} & = - 2\sin {140^\circ} \\ \cos {70^\circ} - \sqrt{k} \sin {70^\circ} & = 2 \cos {130^\circ} \\ \cos {70^\circ} - \sqrt{k} \sin {70^\circ} & = 2 ( \cos {60^\circ} \cos {70^\circ} - \sin {60^\circ} \sin {70^\circ} ) \\ \cos {70^\circ} - \sqrt{k} \sin {70^\circ} & = 2 \left( \dfrac{1}{2} \cos {70^\circ} - \dfrac{\sqrt{3}}{2} \sin {70^\circ} \right) \\ \cos {70^\circ} - \sqrt{k} \sin {70^\circ} & = \cos {70^\circ} - \sqrt{3} \sin {70^\circ} \\ \implies k & = \boxed{3} \end{aligned}

Absolutely Correct and really Brilliant Solution Sir

Mayank Srivastava - 6 years, 3 months ago

How id you change -2Sin140 to 2Cos130?

Hafizh Ahsan Permana - 6 years, 3 months ago

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-sin140=-sin40=-cos 50=cos130

Chew-Seong Cheong - 6 years, 3 months ago

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Can u please explain again? I didnt understand it. Thanks

Yousuf Khan - 6 years, 3 months ago

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@Yousuf Khan Okay.

First, we know that: sin 14 0 = sin ( 18 0 14 0 ) = sin 4 0 \quad \sin{140^\circ} = \sin{(180^\circ -140^\circ)} = \sin{40^\circ} .

Then: sin 4 0 = cos ( 9 0 4 0 ) = cos 5 0 \quad \sin{40^\circ} = \cos{(90^\circ -40^\circ)} = \cos{50^\circ}

Finally: cos 5 0 = cos ( 18 0 5 0 ) = cos 13 0 \quad - \cos{50^\circ} = \cos{(180^\circ -50^\circ)} = \boxed { \cos{130^\circ}}

You can use a calculator to confirm that they are equal. I have done that with a spreadsheet.

Chew-Seong Cheong - 6 years, 3 months ago

Sin(270-130)= -Cos130

Venkat Praveen - 6 years, 3 months ago

That's call brilliant..

Yogesh Kurmi - 6 years, 3 months ago

That's absolutely Brilliant!

Anandhu Raj - 5 years, 9 months ago

that is mind blowing

A Former Brilliant Member - 4 years, 8 months ago

please show me which is formula you have used?

Varsha Anare - 6 years, 3 months ago

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Two formulas: Double Angle Formula 1. sin2θ = 2sinθcosθ Angle Addition Formula 2. cos(θ±φ)=cosθcosφ∓sinθsinφ

Wookho Chiang - 3 years, 2 months ago
Louis W
Mar 11, 2015

Prepare for a trig identity overload. Most of this will deal with what is inside the parentheses, and will be squared at the very end. All of this is performed in degrees.

Trig Identity 1: cot θ = cos θ sin θ \cot\theta = \frac{\cos\theta}{\sin\theta}

cot 70 + 4 cos 70 = cos 70 sin 70 + 4 cos 70 = cos 70 + 4 sin 70 cos 70 sin 70 \cot70 + 4\cos70 = \frac{\cos70}{\sin70} + 4\cos70 = \frac{\cos70 + 4\sin70 \cos70}{\sin70}

Trig Identity 2: sin 2 θ = 2 sin θ cos θ \sin2\theta = 2\sin\theta\cos\theta

Trig Identity 3: cos ( 90 θ ) = sin θ \cos(90-\theta) = \sin\theta

= cos ( 90 20 ) + 2 ( 2 sin 70 cos 70 ) sin 70 = sin 20 + 2 sin 140 sin 70 =\frac{\cos(90-20) + 2(2\sin70\cos70)}{\sin70} = \frac{\sin20 + 2\sin140}{\sin70} = sin 20 + sin 140 + sin 140 sin 70 = \frac{\sin20 + \sin140 + \sin140}{\sin70}

Trig Identity 4: sin θ + sin ϕ = 2 sin ( θ + ϕ 2 ) cos ( θ ϕ 2 ) \sin\theta+\sin\phi=2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2})

Unit Circle: cos 60 = 1 2 \cos60=\frac{1}{2}

Unit Circle: cos 30 = 3 2 \cos30=\frac{\sqrt{3}}{2}

= 2 sin 80 cos 60 + sin 140 sin 70 = sin 80 + sin 140 sin 70 =\frac{2\sin80\cos60+\sin140}{\sin70}=\frac{\sin80+\sin140}{\sin70}

= 2 sin 110 cos 30 sin 70 = 3 sin 110 sin 70 =\frac{2\sin110\cos30}{\sin70}=\frac{\sqrt{3}\sin110}{\sin70}

Trig Identity 5: sin ( 180 θ ) = sin θ \sin(180-\theta)=\sin\theta

= 3 sin ( 180 70 ) sin 70 = 3 sin 70 sin 70 = 3 =\frac{\sqrt{3}\sin(180-70)}{\sin70}=\frac{\sqrt{3}\sin70}{\sin70}=\sqrt{3}

This was all squared, therefore ( 3 ) 2 = 3 (\sqrt{3})^{2}=3\space\space\space\Box

Brilliant One Sir

Mayank Srivastava - 6 years, 2 months ago

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