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Volgvan
4 years ago
13

Tan 3A in terms of tan​

Mathematics
1 answer:
Zanzabum4 years ago
8 0

Here's the sum rule for the tangent function:

\tan(a+b)=\dfrac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}

In the special case a=b, this becomes the double angle formula:

\tan(a+a)=\tan(2a)=\dfrac{\tan(a)+\tan(a)}{1-\tan(a)\tan(a)}=\dfrac{2\tan(a)}{1-\tan^2(a)}

In your case, you case use the sum rule once:

\tan(3a)=\tan(2a+a)=\dfrac{\tan(2a)+\tan(a)}{1-\tan(2a)\tan(a)}

And use it again, in the special case of the double angle:

\dfrac{\dfrac{2\tan(a)}{1-\tan^2(a)}+\tan(a)}{1-\dfrac{2\tan(a)}{1-\tan^2(a)}\tan(a)}

We can obvisouly simplify this expression a lot: let's deal with the numerator and denominator separately: the numerator is

\dfrac{2\tan(a)}{1-\tan^2(a)}+\tan(a) = \dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-\tan^2(a)}

and the denominator is

1-\dfrac{2\tan(a)}{1-\tan^2(a)}\tan(a) = \dfrac{1-\tan^2(a)-2\tan^2(a)}{1-\tan^2(a)} = \dfrac{1-3\tan^2(a)}{1-\tan^2(a)}

So, the fraction is

\dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-\tan^2(a)}\cdot \dfrac{1-\tan^2(a)}{1-3\tan^2(a)} = \dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-3\tan^2(a)}

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