#Trigonometry-3

Geometry Level 3

The value of cot ⁡ θ cot ⁡ 2 θ − cot ⁡ 2 θ cot ⁡ 3 θ − cot ⁡ 3 θ cot ⁡ θ \cot \theta \cot 2\theta -\cot 2\theta \cot 3\theta -\cot 3\theta \cot \theta is:

0 s i n θ sin \theta c o s θ cos \theta 1

Vilakshan Gupta by Vilakshan Gupta , 19, India

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2 solutions

Maximos Stratis
Jun 5, 2017

The value of the expression for θ = π 6 \theta=\frac{π}{6} is 1.
But sin ⁡ π 6 = 1 2 \sin{\frac{π}{6}}=\frac{1}{2} and cos ⁡ π 6 = 3 2 \cos{\frac{π}{6}}=\frac{\sqrt{3}}{2}
Therefor the only option left is:
cot ⁡ θ cot ⁡ 2 θ − cot ⁡ 2 θ cot ⁡ 3 θ − cot ⁡ 3 θ cot ⁡ θ = 1 \boxed{\cot{\theta}\cot{2\theta}-\cot{2\theta}\cot{3\theta}-\cot{3\theta}\cot{\theta}=1}

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This is how to multiple guess the answer from given options. It does not solve the problem.

Marta Reece - 4 years ago

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You are right Sir. But still i thought i could post the way i used to find the result.

maximos stratis - 4 years ago

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The limitations of the software are partly to blame for the fact that we can't always make the problems in such a fashion that shortcuts and guesses won't get people to the right answers. The number of "solvers" reported is usually much higher than the number of actual solvers. On the other hand, it gets people to where they can see the real solutions, and without a penalty. So maybe it serves a purpose, too.

Marta Reece - 4 years ago
Marta Reece
Jun 5, 2017

Using expressions cot ⁡ 2 θ = cot ⁡ 2 θ − 1 2 cot ⁡ θ \cot2\theta=\dfrac{\cot^2\theta-1}{2\cot\theta} and cot ⁡ 3 θ = cot ⁡ 2 θ cot ⁡ θ − 1 cot ⁡ 2 θ + cot ⁡ θ \cot3\theta=\dfrac{\cot2\theta \cot\theta-1}{\cot2\theta+\cot\theta} .

These can be easily derived from sin ⁡ ( α + β ) = sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta and cos ⁡ ( α + β ) = cos ⁡ α cos ⁡ β − sin ⁡ α sin ⁡ β \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta

The rest is basic algebra.

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