how do you verify the equation, $$(\sin 3\theta \div \sin \theta)-(\cos3\theta \div \cos \theta)=2$$, is an idenity?

Question

anonymous · Answer

$$\sin 3\theta = 3\sin \theta - 4 \sin ^{3}\theta $$
$$\cos 3\theta = 4 \cos ^{3}\theta - 3 \cos \theta$$

L.H.S: - 4times$$\cos ^{2}\theta - \sin ^{2}\theta = -4$$  and +6 = 2

anonymous · Answer

wow! i have no idea where you got any of that.

anonymous · Answer

all we have learned thus far is double and half angles could u explain to me in those terms?

anonymous · Answer

Try to read the Trigonometry identities its simple don't worry,All you have to do is learn some of the simple identities like sin2(theta) cos2(theta) sin3(theta) cos3(theta). look at this list: http://en.wikipedia.org/wiki/List_of_trigonometric_identities

anonymous · Answer

ok, we are suppose to figure it in terms of double angles could u explain in terms of double angles?

anonymous · Answer

You don't really have to worry about 'you should only do in double angles' coz it doesn't make a sense converting the sin3(theta) into sin2(theta) all you have do bother about is  under which form you gona derive the sin3(theta) because you eventually have to divide the numerator with the denominator , which is a sin(theta) here.  So convert the numerator term so that you can able to go away with denominator part in the equation and finally come up with a constant to show the equality true.