How to Use the First Identity to Prove the Last Step?

[Manasan3010]

Homework Statement

Prove, $tan(\frac{\pi}{12})=2-\sqrt{3}$

Relevant Equations

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I am stuck at the last step.How to prove the last? Any Help?

 

[BvU]

How to prove the last?

No idea.

Any Help?

Notice that $2{\pi\over 12 } = {\pi\over6}$ and for that one you know the value of the tangent

Spoiler

$1\over\sqrt 3$

 

[Manasan3010]

No idea.
Notice that 2π12=π62π12=π6 and for that one you know the value of the tangent

Spoiler

1/2

Where are you getting 2*pi/12 from? in which line are you putting it in?

Will my answer be wrong If I simply substitute and simplify the values in the first step itself(Using trigonometry Table). Will my warks get deduct?

 

[BvU]

? You have nothing to substitute

 

[Manasan3010]

? You have nothing to substitute

Where are you getting 2*pi/12 from? in which line are you putting it in?

 

[BvU]

I'm not 'getting' it from anywhere. The exercise asks for a tangent of an angle that I don't know the value of the tangent for. But I realize I do know the tangent of twice that angle and I try to make a link between the two ...

Looking at the original post: what is the grey stuff ? Some template you are forced to use ?

Note: I had to edit the spoiler in post #2 ( )

 

[Manasan3010]

I try to make a link between the two ...

How/Where are you making the link?
BTW the template is Wolfram Mathematica Notebook Template

 

[BvU]

If you don't know how to make the link between the tangent of an angle and the tangent of the half angle, you cannot reasonably be presented with this exercise, so I suppose you do know...

 

[HallsofIvy]

You don't just start writing an equation arbitrarily. What reason did you have to immediately write "$tan(\theta)= \frac{sin(\theta)}{cos(\theta)}$"? It's true, of course, but so are many other equations. Writing $tan(\pi/12)= \frac{sin(\pi/12)}{cos(\pi/12)}$ doesn't help because we don't, immediately, know $sin(\pi/12)$ or $cos(\pi/12)$!

Instead, you should notice that $\frac{\pi}{12}= \frac{1}{2}\frac{\pi}{ 6}$. We know that $sin(\pi/6)= \frac{1}{2}$ and $cos(\pi/6)= \frac{\sqrt{3}}{2}$ and we have half-angle formulas.

Think about those things before you start writing equations!

 

[Manasan3010]

Thank I've finally found the identity and solved. They didn't teach me the second or third one. They only taught me the first one. Can i derive the answer using the first Identity?

 

[Mark44]

Thank I've finally found the identity and solved. They didn't teach me the second or third one. They only taught me the first one. Can i derive the answer using the first Identity?
View attachment 247659

Yes, you can use the first identity. This is probably what @BvU was thinking of when he suggested working with $\frac \pi 6$.

Since both $\frac \pi 6$ and $\frac \pi {12}$ are in the first quadrant, their tangents are positive, so you should use the pos. square root in the formula.