Question

Prove: cos3x/cosx=1-4sin^2x

Answer 1

I got up to cos3xsecx=1-4sin^2x

Answer 2

Mmm no we don't want that secant ^^
We want to apply the Sum Angle Formula for Cosine to the top,
And then the Double Angle Formula for Cosine after that.

\[\cos3x=\cos(2x+x)\]
Recall our Sum Angle Formula for Cosine:\[\large \cos(\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta\]

Answer 3

Okay, thanks.

Answer 4

What would I do after that @zepdrix ?

Answer 5

\[\cos2x \cos x-\sin2xsinx=cosx-4\sin^2xcosx\]

Answer 6

Ok looks good so far!

From here, let's factor a cosx out of each term.

Answer 7

How would I factor a cosx out of a sinx?

Answer 8

Or did you just mean the cos?

Answer 9

The cosine is not `inside` of the sine function, it's being multiplied by it.

Let \(\large a=\cos x\)
Then we have \(\large a-4a \sin^2x\)
Factor an a of out of each term :)

Answer 10

cos3x = cos(x) cos (2x) - sin(x) sin (2x) = cos(x) (cos² (x) - sin²(x)) - sin(x) (sinx cosx +cos x sin x) = cos(x) (1 - sin² (x) - sin²(x)) - 2 sin² x cos x = cos x (1 - 4 sin² x)

Dividing by cos x gives:

cos 3x / cos x = 1 - 4 sin² x

Formulae we used in the calculation:

cos (a+b) = cos a cos b - sin a sin b
sin (a+b) = sin a cos b + cos a sin b
cos² a = 1 - sin² a

Answer 11

I mean out of the left side. The right side I understand.

Answer 12

Yes I meant the left side also :o

Answer 13

Where is the sin being multiplied by cos on the left side?

Answer 14

\[\large \cos x-4\sin^2x(\cos x) \qquad = \qquad \cos x-4(\cos x)\sin^2x\]

Multiplication is commutative, so you can do it in any order.
So I just moved the cosine over, so it's a little easier to read.

Answer 15

I'm talking about the other side of the equation (cos2xcosx-sin2xsinx)

Answer 16

Oohhhh I see what happened, you multiplied the cosine over to the right?
Ok ok ok I misunderstood, I thought you worked out the double angle formulas already.

If you're trying to show an identity is true, do not do anything to the other side, we want to keep that cosx in the denominator.

Ok ok, let's see.

Answer 17

We want to apply these two formulas:\[\large \cos2x=1-2\sin^2x\]
\[\large \sin2x=2\sin x \cos x\]

Answer 18

Wait, so we don't do anything to the cos3x/cosx?

Answer 19

We can't move any of it to the right side.

Answer 20

I thought we could multiply by cosx on both sides.

Answer 21

No :o

Answer 22

So we have cos3x=cosx-4sinx^2cosx

Answer 23

If you're "proving" something, it means you start with one side, and end with the other side.

You don't manipulate both sides and meet them half way.

Answer 24

Wouldn't the results be the same either way?

Answer 25

Tehcnically speaking.

Answer 26

It would, yes :) But the correct process of working through proofs is kind of important.

Answer 27

Oh, then everything I know is a lie. :P

Answer 28

XD

Answer 29

Okay then, so we leave it at cos3x/cosx?

Answer 30

Yes :) Don't move anything to the right.
We have to use trig identities to `change it`.
So here's what we've done so far.
\[\large \frac{\cos3x}{\cos x}=1-4\sin^2x\]

\[\large \frac{\cos2x \cos x-\sin2x \sin x}{\cos x}=1-4\sin^2x\]

Answer 31

\[\large \color{royalblue}{\cos2x=1-2\sin^2x}\]\[\large \color{purple}{\sin2x=2\sin x \cos x}\]We're going to use these two identities to change things around.\[\large \frac{\color{royalblue}{\cos2x} \cos x-\color{purple}{\sin2x} \sin x}{\cos x}=1-4\sin^2x\]

Understand what to do next? :o

Answer 32

OH!

Answer 33

Okay, so it becomes 1-2sin^2xcosx-2sinxcosxsinx=1-4in^2x which then becomes 1-2sin^2xcosx-3sinxcosx=1-4sin^2x

Answer 34

Right?

Answer 35

Mmm your math looks like a little funky.
Should use brackets when we apply this first identity,\[\large \frac{(\color{royalblue}{1-2\sin^2x}) \cos x-\color{purple}{2\sin x \cos x} \sin x}{\cos x}=1-4\sin^2x\]

Answer 36

Okay, it looks better on paper, :P

Answer 37

So we are left with 1-2sin^2x-2sin^2x?

Answer 38

Oh you divided out the cosine from the denominator?
Yah sounds good!

Answer 39

Okay, just making sure, and is my end correct?

Answer 40

Or would it be 3sin instead of 2sin^2x?

Answer 41

The second one, not the frist 2sin^2x.

Answer 42

So it looks like you got it to this point:
\[\large 1-2\sin^2x-2\sin^2x \qquad=\qquad 1-4\sin^2x\]

Which is correct.
Understand what to do next? :)

Answer 43

\[\large 2\sin x \cos x \sin x \qquad=\qquad 2\sin^2x \cos x\]Just in case there was any confusion on that part.

Answer 44

Wait, I don't get it.

Answer 45

Don't get what? That last multiplication step?

Answer 46

What did you multiply? Sorry, I had to do the dishes.