# Trigonometry

### Trigonometry

Can you solve this limit please?
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### Re: Trigonometry

Ouch! That's not easy!

The first thing I would try is just setting x= 0 in the given function. Unfortunately that makes both numerator and denominator 0 so the fraction is "undetermined" and we need to use another method to find the limit. The next thing I would try is to "rationalize the denominator". The denominator is $$\sqrt{1+ sin^2(2x)}- \sqrt{1+ sin^2(x)}$$ so multiply both numerator and denominator by $$\sqrt{1+ sin^2(2x)}+ \sqrt{1+ sin^2(x)}$$.

$$\frac{x tan(3x)\left(\sqrt{1+ sin^2(2x)}+ \sqrt{1+ sin^2(x)}\right)}{sin^2(2x)- sin^2(x)}$$.

Now use the trig identities $$sin(2x)= 2sin(x)cos(x)$$ so that $$sin^2(2x)- sin^2(x)= 4sin^2(x)cos^2(x)- sin^2(x)= sin^2(x)(4cos^2(x)- 1)$$ and $$sin^3(x)= \frac{3}{4}sin(x)- \frac{1}{4}sin(3x)$$ (I got that by looking up an identity for sin(3x))

HallsofIvy

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