# Simple factorisation

1. Nov 15, 2008

### kasse

I want $$\frac{dI}{d \theta} = 0$$ when $$I = 0.8 \cdot 10^{-5}[\frac{273}{16} + \frac{17}{2}cos(kd sin \theta) + 2cos(2kd sin \theta)]$$

If I've calculated correctly, this means that $$-8.5sin(8.1sin \theta)8.1cos \theta - 32.4sin(16.2sin \theta)cos \theta = 0$$. Can somebody help me factorise this?

2. Nov 15, 2008

### joeyar

If only the stuff in brackets is the argument of the first sine of each term, you could get away with dividing through by $$cos \theta$$ to leave you with

$$-8.5sin(8.1sin \theta)8.1 - 32.4sin(16.2sin \theta) = 0$$

which is equivalent to
$$-68.85sin(8.1sin \theta) - 32.4sin(16.2sin \theta) = 0$$

3. Nov 15, 2008

### joeyar

Then recognising 16.2sinθ = 2(8.1sinθ) we can apply the sin double angle formula (sin2α = 2cosαsinα) on the second term