Can you please explain me how it generate the general forms for the trigonometric values..?

i:e for " tan(θ)=tan(α), the general solution is : θ=nπ+α,n∈Z. "

( I'm asking about the how they derivative this general equation ": θ=nπ+α,n∈Z " for tan(θ) s )

Thank u. Urgent help is needed

Question

Can you please explain me how it generate the general forms for the trigonometric values..?

i:e for    " tan(θ)=tan(α), the general solution is : θ=nπ+α,n∈Z. "

( I'm asking about the how they derivative  this general equation ": θ=nπ+α,n∈Z " for tan(θ) s )

Thank u. Urgent help is needed

e.mccormick · Answer

In this case, you are describing any coterminal angle and any angle 180 degrees off of the coterminal angle.  Bacically, any angle that would have the same result as a trig ratio.

e.mccormick · Answer

In the first and thir quadrants, the tangent is +.  In the 2nd and 4th it is negative:

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e.mccormick · Answer

So there is always going to be some angle in the quadrants with the same sign that ends up as being the same ratio just by having the same numbers but with sign changes that cancel out.

For example:
$\dfrac{1}{2}=\dfrac{-1}{-2}$ and $\dfrac{-1}{2}=\dfrac{1}{-2}$

jamierox4ev3r · Answer

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e.mccormick · Answer

So when you say this:
$ \tan(\theta )=\tan(\alpha )$, the general solution is : $\theta =n\pi + \alpha ,n\in Z$ the math part on the right is read as, theta equals enn times pi plus alpha for all values of enn that are an integer.

anonymous · Answer

@e.mccormic And @jamierox4ev3r thank you for your explanations.. can you further explain me how they got this general solution  ": θ=nπ+α,n∈Z "

e.mccormick · Answer

Well, do you understand what I am saying in the text version of that?

anonymous · Answer

@e.mccormick  yes really  .. I feel easy now  :)

e.mccormick · Answer

Take the unit circle.  Let me look at the tangent of $\dfrac{\pi}{3}$.  This is $\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}$.  That simplifies to $\sqrt{3}$.

Well, if I add 180 degrees or $\pi$ radians I land at $\dfrac{4\pi}{3}$.  That has a tangent of $\dfrac{-\dfrac{\sqrt{3}}{2}}{-\dfrac{1}{2}}$ which again simplifies to $\sqrt{3}$.

If I keep adding or subtracting amounts in increments of $\pi$ radians I will always land on one of those two angles on a coterminal angle.  Every coterminal angle has the same trig ratios.

That is what thus is all about:  An equation that describes all coterminal angles that have the same trig ratios.

e.mccormick · Answer

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