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# C3 Trig Question

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1. 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.
2. (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.
3. (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?
4. (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

Alternatively, you should be able to recognize the periodicity of tan x from its graph.
5. (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.
6. (Original post by Zacken)
....
oh right so for sin and cos it would be 2pi and -2pi etc?
7. (Original post by SirRaza97)
oh right so for sin and cos it would be 2pi and -2pi etc?
Bang on.
8. (Original post by Zacken)
Bang on.

Edit:
E.g.

sin(pi/3) = sin(2pi/3) so, pi/3 = 2pi/3 + 2npi

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