# Simplify in sum and difference equation

1. Feb 25, 2005

### aisha

I was never taught how to do this I was just given problems with no solutions so I dont know what to do can someone plz help me I have two questions which are similar maybe if I get help with one I can solve the other one on my own.

It says simplify the following
$$\cos(\pi+x) + \cos (\pi-x)$$
for this one i expanded cos into the brackets and simplified and got 2cospi is this correct?

and

$$\cos (\frac {7\pi} {10}) \cos (\frac {\pi} {5}) +\sin (\frac {7\pi} {10}) \sin (\frac {\pi} {5})$$

Last edited: Feb 25, 2005
2. Feb 25, 2005

### xanthym

From standard trig identities:
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
cos(-e) = cos(e)
sin(-g) = -sin(g)

cos(Pi + x) + cos(Pi - x) = cos(Pi)cos(x) - sin(Pi)sin(x) + cos(Pi)cos(-x) - sin(Pi)sin(-x) =
= (-1)cos(x) - (0)sin(x) + (-1)cos(x) + (0)sin(x) =
= (-2)cos(x)

cos(7*Pi/10)cos(Pi/5) + sin(7*Pi/10)sin(Pi/5) = cos{(7*Pi/10) - (Pi/5)} =
= cos{Pi/2} =
= 0

~~

3. Feb 25, 2005

### aisha

Holy moly that looks really confusing I dont understand what is going on and that doesnt look very familiar from the unit .....

4. Feb 26, 2005

### dextercioby

$$\cos(\pi+x)=\cos\pi\cos x-\sin\pi\sin x=-\cos x$$

$$\cos (\pi-x)=\cos\pi\cos x+\sin\pi\sin x=-\cos x$$

Add them and see what u get.

As for the last,u have to use the identity:

$$\cos a\cos b+\sin a\sin b=\cos(a-b)$$

See who's "a" & who's "b"...

Daniel.

EDIT:The typo was because of a missing space between \cos and b...

Last edited: Feb 26, 2005
5. Feb 26, 2005

### Curious3141

That, BTW, should be :

$$\cos a\cos b+\sin a\sin b=\cos(a-b)$$

6. Feb 26, 2005

### dextercioby

Check the code by clicking on the formula,it (the "cos b") was there...

Daniel.

7. Feb 26, 2005

### Curious3141

Yes, I know it was, but you forgot the space.

8. Feb 26, 2005

### dextercioby

You got me...Yes,as the Americans say :"S*** happens".Today i've had a lousy day...:yuck:

daniel.

9. Feb 26, 2005

### aisha

$$\cos(\pi+x)=\cos\pi\cos x-\sin\pi\sin x=-\cos x$$

$$\cos (\pi-x)=\cos\pi\cos x+\sin\pi\sin x=-\cos x$$

What are u doing to these pies and xs to get -cosx? How do u get that?

10. Feb 26, 2005

### dextercioby

There's no "pie" there...:tongue2: I used the fact that
$$\cos \pi =-1$$
$$\sin \pi =0$$

Daniel.

11. Feb 26, 2005

### aisha

See I dont know that wish I was as smart as you, ahhhhh so what do I do for the other question?? This is so confusing, I hate these sum and difference identities I dont know what to do with them

12. Feb 26, 2005

### dextercioby

What about the other one...?In post #2 it was solved...

Daniel.

13. Feb 26, 2005

### aisha

$$\cos (\frac {7\pi} {10}) \cos (\frac {\pi} {5}) +\sin (\frac {7\pi} {10}) \sin (\frac {\pi} {5})$$

here is the question in post 2 the person wrote $$cos [(\frac {7\pi}{10}) -(\frac {\pi} {5})] = \cos (\frac {\pi} {2}) = 0$$

How come I dont know what is going on what is happening where are these numbers comming from how do I do this on my own?

So the final simplified answer for this problem is 0? Can someone plz help me understand how to do these problems I dont have a clue I know how to evaluate sin(75 degrees) but thats about it because it uses special triangles but these problems are totally confusing for me . :yuck:

Last edited: Feb 26, 2005
14. Feb 26, 2005

### dextercioby

So what...?That's perfectly correct...

Daniel.

15. Feb 26, 2005

### aisha

:tongue2: ahhhhhhhh meany ur a smarty pant thats y u know how to do this if ud explain maybe i'd get it too :grumpy:

16. Feb 26, 2005

### dextercioby

Label
$$\frac{7\pi}{10}=a$$
and
$$\frac{\pi}{5}=b$$
And then compute
$$\cos a\cos b+\sin a\sin b$$

using addition and subtraction formulas for cosine and sine.

Daniel.

17. Feb 26, 2005

### aisha

addition and subtraction formulas for cos and sin?

18. Feb 26, 2005

### dextercioby

My dear god,the formulas:
$$\cos (a-b),\cos(a+b),\sin(a-b),\sin(a+b)$$

Daniel.

P.S.I'm going to bed now,so be a nice girl and don't make anything bad till i return.