# Why does 2(sin(a)*cos(a))= sin(2a) ?

1. Dec 3, 2008

### Georgepowell

I think it does.... have I got that right?

If I have then why does it? is there a simple explanation?

2. Dec 3, 2008

### mgb_phys

Yes it does.
It's easy to prove from the sum formula, sin (a + b)=sin a cos b + cos a sin b
Of course now you have to prove that!

3. Dec 4, 2008

### Georgepowell

Go on...

How would I do that? I don't need a solid proof, just a simple explanation would do...

Thanks

4. Dec 4, 2008

### D H

Staff Emeritus
Set b=a.

5. Dec 4, 2008

### Georgepowell

Yeh I got that part, What I wanted was for someone to tell me why

sin(a + b) = sin(a)cos(b) + sin(b)cos(a) ?

6. Dec 4, 2008

### D H

Staff Emeritus
One way to derive the sine and cosine of the sum (or difference) of pair of angles is via Euler's identity:

$$\exp(i\theta) = \cos\theta + i\sin\theta$$

Setting $\theta=a+b$,

\aligned \exp(i(a+b)) &= \exp(ia)\exp(ib) \\ & = (\cos a + i\sin a)(\cos b + i\sin b) \\ &= (\cos a \cos b - \sin a \sin b) + i(\sin a \cos b + \cos a \sin b)) \\ &= \cos(a+b) + i\sin(a+b) \endaligned

Equating real and imaginary parts,

\aligned \cos(a+b) &= \cos a \cos b - \sin a \sin b \\ \sin(a+b) &= \sin a \cos b + \cos a \sin b \endaligned

7. Dec 4, 2008

### Georgepowell

Can you go over this step please? Thanks

8. Dec 4, 2008

### nicksauce

By using $$e^{i\theta}=\cos{\theta} + i\sin{\theta}$$
the LHS becomes
$$\cos{a+b} + i\sin{a+b}$$.

9. Dec 4, 2008

### Georgepowell

haha, ok... I think this is getting well out of my depth. Unless there is a more simple explanation, I'm outahere.

Anyway - Thanks for all your help everyone.

10. Dec 4, 2008

### HallsofIvy

Assuming that you are using the "circle" definition of sine and cosine: that the coordinates of a point a distance t along the circumference of the unit circle from (1, 0) are (cos(t), sin(t)), then you can use argue that the straight line distance from (1, 0) to (cos(t), sin(t)) and from (cos(s), sin(s)) to (cos(s+t), sin(s+t)) are the same because because they both cut an arc of length t.
$$\sqrt{(cos(t)- 1)^2+ sin^2(t)}= \sqrt{(cos(s)-cos(s+t))^2+ (sin(s)- sin(s+t))^2}$$
Squaring both sides and multiplying everything out,
$$cos^2(t)- 2cos(t)+ 1+ sin^2(t)= cos^2(t)- 2cos(s)cos(s+t)+ cos^2(s+t)+ sin^2(s)- 2sin(s)sin(s+t)+ sin^2(s+ t)$$
The $cos^2(t)$ and $sin^2(t)$ on each side cancel and $cos^2(s+t)+ sin^2(s+ t)= 1$ which cancels the "1" on the left. That leaves
-2cos(t)= -2 cos(s)cos(s+t)- 2sin(s)sin(s+t) so cos(t)= cos(s)cos(s+t)+ sin(s)sin(s+t).

Let a= s, b= s+t. Then t= b- a and we have cos(b-a)= cos(a)cos(b)+ sin(a)sin(b).

Replacing a by -a and using the fact that cos(-a)= cos(a) while sin(-a)= -sin(a), we have cos(a+ b)= cos(a)cos(b)- sin(a)sin(b).

Use the fact that $sin(x)= cos(\pi/2- x)$ to change to a formula for sin(a+b).

That's the proof that is used in most pre-calculus classes.

11. Dec 4, 2008

### Georgepowell

Thanks, that's what I wanted. I understand it now.

I don't like memorising formulas without understanding them, and my teachers teach the formulas and not the explanations. Which I don't like.

I suppose I'm not even meant to know this formula yet, but I'm interested.

12. Dec 4, 2008

### tiny-tim

Hi Georgepowell!

Draw a right-angled triangles PQR and PQS with R = S = 90º and QPR = A + B and QPS = A (so SPR = SQR = B).
Draw the perpendiculars from S to PR and QR.

Then, if PQ = 1, QR = sin(A+B) …

and you can work out the rest!

13. Dec 4, 2008

### Georgepowell

Is this what you meant?

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14. Dec 4, 2008

### HallsofIvy

But how do you apply that, say, to A= 50 degrees, B= 60 degrees?

15. Dec 5, 2008

### tiny-tim

Yes … and now draw the perpendiculars from S to PR and QR

(and note that the first one will be parallel to QR).

(and the circle isn't needed)

16. Dec 5, 2008

### maze

A braindead way to prove any trig identity is to write the identity in terms of complex exponentials and simplify as much as possible until 0=0 is reached. Then the reverse of your steps is the proof of the identity.

For example,
$$2 sin(a) cos(a) - sin(2a) = 0$$

$$2\left(\frac{e^{i a}-e^{-i a}}{2 i}\right)\left(\frac{e^{i a}+e^{-i a}}{2}\right) - \frac{e^{i 2a}-e^{-i 2a}}{2 i} = 0$$

$$\frac{1}{2 i}\left(e^{i a}e^{i a} + e^{i a}e^{-i a} - e^{-i a}e^{i a} - e^{-i a}e^{-i a}\right) - \frac{1}{2 i}\left(e^{i 2a}-e^{-i 2a}\right) = 0$$

$$\frac{1}{2 i}\left(e^{i 2a} - e^{-i 2a}\right) - \frac{1}{2 i}\left(e^{i 2a}-e^{-i 2a}\right) = 0$$

$$0 = 0$$

Now a proper proof would run these steps in reverse.

Last edited: Dec 5, 2008
17. Dec 5, 2008

### HallsofIvy

I believe that was what mbg_phys did in the first several responses but GeorgePowell was not familiar with the exponential form of sine and cosine and could not follow.

18. Dec 5, 2008

### tiny-tim

keep it real

Yeah keep it real!

Support CAMREG …

the CAMpaign for REal Geometry!

19. Dec 5, 2008

### Georgepowell

That's a good simple proof that I understand, but doesn't it only prove it for:

2A + B = 90 ?

**note: In the picture, I changed the letters, but it is the same concept. And the circle was just to help me get the right-angles right**

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20. Dec 5, 2008

### tiny-tim

Hi Georgepowell!

(can't see your new diagram yet)

No, I think that it works for any A + B < 90º …

and even for A + B ≥ 90º provided both A and B < 90º if you put R on the other side of the circle.
Yeah, the letters don't matter …

it's always best to use whatever letters you find easiest.

hmm … now you mention it, my own diagram (which you can't see would have been a lot better if i'd started off with a circle )​