What happened to 1 5 ∘ 15^{\circ} ?

Geometry Level 4

The expression ( 1 tan ⁡ 5 + cot ⁡ 5 ) ( 1 tan ⁡ 25 + cot ⁡ 25 ) ( 1 tan ⁡ 35 + cot ⁡ 35 ) ( 1 tan ⁡ 45 + cot ⁡ 45 ) \left(\dfrac{1}{\tan 5 + \cot 5}\right)\left(\dfrac{1}{\tan 25 + \cot 25}\right)\left(\dfrac{1}{\tan 35 + \cot 35}\right)\left(\dfrac{1}{\tan 45 + \cot 45}\right) can be expressed as p q \dfrac{p}{q} for positive coprime integers p , q p,q . Find p + q p+q .

Details and Assumptions \text{Details and Assumptions}

All angles are in degrees.


The answer is 129.

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Daniel Liu by Daniel Liu , 21, USA

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1 solution

Daniel Liu
Apr 21, 2014

Since we are dealing with 1 tan ⁡ θ + cot ⁡ θ \dfrac{1}{\tan\theta+\cot\theta} for some θ \theta , let's try to simplify that.

First let's focus on the denominator, tan ⁡ θ + cot ⁡ θ \tan\theta+\cot\theta first. We can set tan ⁡ θ = sin ⁡ θ cos ⁡ θ \tan\theta=\dfrac{\sin\theta}{\cos\theta} and cot ⁡ θ = cos ⁡ θ sin ⁡ θ \cot\theta=\dfrac{\cos\theta}{\sin\theta} . We are left with sin ⁡ θ cos ⁡ θ + cos ⁡ θ sin ⁡ θ = sin ⁡ 2 θ + cos ⁡ 2 θ sin ⁡ θ cos ⁡ θ = 1 sin ⁡ θ cos ⁡ θ \begin{aligned}\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}&=\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\\ &= \dfrac{1}{\sin\theta\cos\theta}\end{aligned}

Thus, we see that tan ⁡ θ + cot ⁡ θ = 1 sin ⁡ θ cos ⁡ θ \tan\theta+\cot\theta=\dfrac{1}{\sin\theta\cos\theta} (given that θ ≠ 90 n \theta \ne 90 n for any integer n n ).

Therefore 1 tan ⁡ θ + cot ⁡ θ = sin ⁡ θ cos ⁡ θ \dfrac{1}{\tan\theta+\cot\theta}=\sin\theta\cos\theta . We can further simplify sin ⁡ θ cos ⁡ θ \sin\theta\cos\theta into 1 2 sin ⁡ 2 θ \dfrac{1}{2}\sin 2\theta . Thus, we arrive at the conclusion that 1 tan ⁡ θ + cot ⁡ θ = 1 2 sin ⁡ 2 θ \boxed{\dfrac{1}{\tan\theta+\cot\theta}=\dfrac{1}{2}\sin 2\theta}

We need to find the value of ( 1 2 sin ⁡ 10 ) ( 1 2 sin ⁡ 50 ) ( 1 2 sin ⁡ 70 ) ( 1 2 sin ⁡ 90 ) \left(\dfrac{1}{2}\sin 10\right)\left(\dfrac{1}{2}\sin 50\right)\left(\dfrac{1}{2}\sin 70\right)\left(\dfrac{1}{2}\sin 90\right) . First, simplify this to 1 16 sin ⁡ 10 sin ⁡ 50 sin ⁡ 70 \dfrac{1}{16}\sin10\sin50\sin70 . There dos not seem to be an easy relation between 10 , 50 , 70 10,50,70 so check the cosine versions to see if there is a relation. Indeed, we can see a relation: sin ⁡ 10 sin ⁡ 50 sin ⁡ 70 = cos ⁡ 20 cos ⁡ 40 cos ⁡ 80 \sin10\sin50\sin70=\cos20\cos40\cos80 . Many of you should recognize this curious identity as equal to 1 8 \dfrac{1}{8} (proof will be in a comment on this solution). Thus, our original expression is equal to 1 16 ⋅ 1 8 = 1 128 \dfrac{1}{16}\cdot \dfrac{1}{8}=\dfrac{1}{128} , and our answer is 1 + 128 = 129 1+128=\boxed{129} .

Problem writer's note: Even with a 1 tan ⁡ 15 + cot ⁡ 15 \dfrac{1}{\tan15+\cot15} term, the problem still is easy to solve. I excluded it to add some asymmetry, which seems to intimidate people more. ⌣ ¨ \ddot\smile

25 Helpful 0 Interesting 0 Brilliant 0 Confused

Proof of cos ⁡ 20 cos ⁡ 40 cos ⁡ 80 \cos20\cos40\cos80 Identity:

We think of double angle formulas when we see the doubled angles. However, using cos ⁡ 2 θ = 2 cos ⁡ 2 θ − 1 \cos 2\theta = 2\cos^2\theta -1 certainly will not work, as it creates a huge mess. What else can we do?

Well, there still is one other double angle identity: sin ⁡ 2 θ = 2 sin ⁡ θ cos ⁡ θ \sin 2\theta = 2\sin\theta\cos\theta . We can see that cos ⁡ θ = sin ⁡ 2 θ 2 sin ⁡ θ \cos\theta=\dfrac{\sin 2\theta}{2\sin\theta} . Thus, cos ⁡ 20 cos ⁡ 40 cos ⁡ 80 = ( sin ⁡ 160 2 sin ⁡ 80 ) ( sin ⁡ 80 2 sin ⁡ 40 ) ( sin ⁡ 40 2 sin ⁡ 20 ) = 1 8 ( sin ⁡ 160 sin ⁡ 20 ) \cos20\cos40\cos80 = \left(\dfrac{\sin160}{2\sin80}\right)\left(\dfrac{\sin80}{2\sin40}\right)\left(\dfrac{\sin40}{2\sin20}\right)=\dfrac{1}{8}\left(\dfrac{\sin160}{\sin20}\right) . Note that sin ⁡ 160 = sin ⁡ 20 \sin160 = \sin20 , so we are left with 1 8 \boxed{\dfrac{1}{8}}

Daniel Liu - 7 years, 1 month ago

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You could have instead used the well known identity:

sin ⁡ θ sin ⁡ ( π 3 − θ ) sin ⁡ ( π 3 + θ ) = 1 4 sin ⁡ ( 3 θ ) \displaystyle \sin\theta \sin\left(\frac{\pi}{3}-\theta\right)\sin\left(\frac{\pi}{3}+\theta\right)=\frac{1}{4}\sin(3\theta)

Pranav Arora - 7 years, 1 month ago

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I didn't know that identity! Thanks for sharing :)

Daniel Liu - 7 years, 1 month ago

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@Daniel Liu – Your welcome and thanks for the problems you share on Brilliant. :)

Here are two analogous versions for cos and tan, you may find these useful sometime:

cos ⁡ θ cos ⁡ ( π 3 − θ ) cos ⁡ ( π 3 + θ ) = 1 4 cos ⁡ ( 3 θ ) \displaystyle \cos\theta \cos\left(\frac{\pi}{3}-\theta\right)\cos \left(\frac{\pi}{3}+\theta\right)=\frac{1}{4}\cos(3\theta)

tan ⁡ θ tan ⁡ ( π 3 − θ ) tan ⁡ ( π 3 + θ ) = tan ⁡ ( 3 θ ) \displaystyle \tan\theta \tan\left(\frac{\pi}{3}-\theta\right) \tan\left(\frac{\pi}{3}+\theta\right)=\tan(3\theta)

Pranav Arora - 7 years, 1 month ago

i can't latex (hi daniel this is mathtastic :D) and I basically did the same thing as you but I introduced symmetry into it so I said let x be what we want then 8x=sin10sin30sin50sin70=sin10cos20cos40*1/2 then multiply both sides by 4cos10 to get 32xcos10=2sin10cos10cos20cos40 and using double angle we have 32xcos10=sin20cos20cos40 and multiply by 2 and apply double angle 64xcos10=sin40cos40 multiply by 2 and apply double angle 128cos10x=sin80 divide by cos10 and use sin(x)=cos(90-x) so 128x=1 and x=1/128

Nathan Ramesh - 7 years, 1 month ago

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hey you can use daum equation editor

Rishabh Jain - 7 years ago

thnk for solution..

Max B - 7 years, 1 month ago

i did the same way vohoo and nice problem

will jain - 6 years, 7 months ago

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