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C3 Trig Question

SirRaza97

Hello could I have some help on trying to find the second possible solution for this question part c.

This is what I have done:

I thought of using cast but wasn't sure how to go about it.

Reply 1

ghostwalker

Original post by SirRaza97

I thought of using cast but wasn't sure how to go about it.

If tan a = tan b, then a = b +n pi, where n is any integer.

Use that and rearrrange to get theta = ..., then see what values you can get.

Reply 2

SirRaza97

Original post by ghostwalker

If tan a = tan b, then a = b +n pi, where n is any integer.

Use that and rearrrange to get theta = ..., then see what values you can get.

Woah where did this spooky rule come from?

Reply 3

davros

Original post by SirRaza97

Woah where did this spooky rule come from?

Er, you should have a section in your book that deals with the general solutions of things like sin A = sin B, cos A = cos B and tan A = tan B :smile:

Alternatively, you should be able to recognize the periodicity of tan x from its graph.

Reply 4

Zacken

Original post by SirRaza97

...

You seem to be under the impression that if you have tan a = tan b, then it must be only the case that a = b.

Whilst it is true that a = b does make the equation true, you should realise that tangent is a pi-periodic function, so if I wanted to, I could say that tan a = tan (b + pi). Since tan(b+pi) = tan(b).

Otherwise, I would say that tan 1 = tan (1 + pi) and you'd sit there and tell me: "Hey! That means 1 = 1 + pi, i.e: 0 = pi. (illimunati)". You can clearly see that that is nonsense.

So, if you have tan a = tan b, then you could have a = b, you could have a = b + pi, you could have a = b - pi, you could have a = b + pi + pi, you could have a = b - pi - pi, etc... onto infinite amounts of adding or subtracting pi's.