Trig Problem

[matthew042]

I have been given the following problem:
(tan(?/12))(tan(2?/12))(tan(3?/12))(tan(4?/12))(tan(5?/12))

I have been able to simplify it down to:
(tan(?/12))(tan(5?/12))
from the unit circle.

I know the answer is one but I am stuck on getting there. I cannot use the half angle formula. Can only use complementary angle identities, supplementary angle identites, opposite angle identities, and phase shift identities.

sin(90°-?) = cos?
cos(90°-?) = sin?
tan(90°-?) = cot?

sin(90°+?) = cos?
cos(90°+?) = -sin?
tan(90°+?) = -cot?

sin(180°-?) = sin?
cos(180°-?) = -cos?
tan(180°-?) = -tan?

sin(180°+?) = -sin?
cos(180°+?) = -cos?
tan(180°+?) = tan?

Any help will be appreciated. Thank you!

 

[galactus]

You've got it down to

\(\displaystyle tan(\pi/12)tan(5\pi/12)\)

Rewrite as:

\(\displaystyle \frac{sin(\frac{\pi}{12})}{cos(\frac{\pi}{12})}\cdot\frac{sin(\frac{5\pi}{12})}{cos(\frac{5\pi}{12})}\)

Now, use the identity \(\displaystyle cos\theta=sin(\frac{\pi}{2}-\theta)\)

Then, \(\displaystyle cos(\frac{5\pi}{12})=sin(\frac{\pi}{12})\)

Continue with the other and you will see it. They will cancel and all you're left with is 1.

You have the worst already done.

Good job.

 

[Deleted member 4993]

Another way - sort of:

\(\displaystyle tan(\theta) \ = \ cot(\frac{\pi}{2}-\theta)\)

\(\displaystyle tan(\theta) \ * \ tan(\frac{\pi}{2}-\theta) \ = 1\)

\(\displaystyle tan(\frac{\pi}{12}) \ * \ tan(\frac{\pi}{2}-\frac{\pi}{12}) \ = 1\)

 

[soroban]

Hello, matthew042!

\(\displaystyle \text{Simplify: }\:\tan(\tfrac{\pi}{12})\tan(\tfrac{2\pi}{12})\tan(\tfrac{3\pi}{12})\tan(\tfrac{4\pi}{12}\tan(\tfrac{5\pi}{12})\)

\(\displaystyle \text{I have been able to simplify it down to: }\:\tan(\tfrac{\pi}{12})\tan(\tfrac{5\pi}{12})\) from the unit circle.

\(\displaystyle \text{I know the answer is }1\text{, but I am stuck on getting there.}\)

\(\displaystyle \text{You have: }\:\tan(\tfrac{\pi}{12})\tan(\tfrac{5\pi}{12})\)

\(\displaystyle \text{You missed one identity: }\:\tan\theta \:=\:\cot(\tfrac{\pi}{2}-\theta)\)

. . \(\displaystyle \text{Hence: }\:\tan(\tfrac{5\pi}{12}) \:=\:\cot(\tfrac{\pi}{2} - \tfrac{5\pi}{12}) \;=\;\cot(\tfrac{\pi}{12})\)

\(\displaystyle \text{Therefore: }\:\tan(\tfrac{\pi}{12})\tan(\tfrac{5\pi}{12}) \;=\;\tan(\tfrac{\pi}{12})\cot(\tfrac{\pi}{12}) \:=\:1\)