A2 Level Trigonometry

I'm stuck on 21 & 22, you have to prove it using compound and double angle formulae and trig identities and magic. If someone knows how to do it I would be eternally grateful. Thanks for your help.

(edited 9 years ago)

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Reply 1

Old_Simon

Ok for #21.

Express the whole thing in terms of cos theta.

Reply 2

1374786

Original post by Old_Simon

Ok for #21.

Express the whole thing in terms of cos theta.

How do you do that?

Reply 3

Old_Simon

Original post by kcorbins

How do you do that?

Well sin^4 theta = sin^2 theta times sin^2 theta. So what is that in terms of cos ?

(this is the very first trig identity you are meant to know).

Reply 4

1374786

Original post by Old_Simon

Well sin^4 theta = sin^2 theta times sin^2 theta. So what is that in terms of cos ?

(this is the very first trig identity you are meant to know).

Oh 1-cos^2, then what do you do?

Reply 5

Old_Simon

Original post by kcorbins

Oh 1-cos^2, then what do you do?

Well rewrite it now using that identity and have another think about the next step.

Reply 6

1374786

Original post by Old_Simon

Well rewrite it now using that identity and have another think about the next step.

Multiply it out?

Reply 7

Old_Simon

Original post by kcorbins

Multiply it out?

yup

now think about the rhs too.

Reply 8

atsruser

I think for 21 the most direct approach is to note that:

a^2+b^2 = (a+b)^2-2ab

and use a suitable choice of

a,b

to form the LHS of the identity. The RHS then simplifies to something that can be worked into the desired form via the double angle formulae.

Reply 9

1374786

Original post by Old_Simon

yup

now think about the rhs too.

So you get 1 - 2cos^2(-) + 2cos^4(-), I don't know what to think about next.

Reply 10

Old_Simon

Original post by atsruser

I think for 21 the most direct approach is to note that:

a^2+b^2 = (a+b)^2-2ab

and use a suitable choice of

a,b

to form the LHS of the identity. The RHS then simplifies to something that can be worked into the desired form via the double angle formulae.

That is really neat thank you.

Reply 11

1374786

Original post by Old_Simon

That is really neat thank you.

I haven't tried that technique, but it's a proof question so you need to go from one equation to the next using proofs, how does this do that?

Reply 12

BobLoblawLawBlog

Original post by kcorbins

I haven't tried that technique, but it's a proof question so you need to go from one equation to the next using proofs, how does this do that?

I definitely agree with atsruser with regards to 21. Although you do have to be good at recognising when you can use the identities.

Spoiler

For 22, have you come across de Moivres theorem? If not, then you will have to do it the long way. So use the relevant identities on

\frac{sin(3\theta)}{cos(3\theta)}

. Once you have this, you need to divide both the top and bottom by something to get it in the required form.

(edited 9 years ago)

Reply 13

Old_Simon

Original post by kcorbins

I haven't tried that technique, but it's a proof question so you need to go from one equation to the next using proofs, how does this do that?

Yes but going from one to the other is the proof. That is what you need to do.
The guy gave you a fantastic direct route.

Reply 14

1374786

, ok I understand this, what next? turn the first bit into 1?

Reply 15

BobLoblawLawBlog

Original post by kcorbins

, ok I understand this, what next? turn the first bit into 1?

Yeah, so the first thing that stands out is that the first bit is 1. Then, can you notice any similarities between the second bit and the expression for