Simplify in sum and difference equation

[aisha]

I was never taught how to do this I was just given problems with no solutions so I don't know what to do can someone please 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})

Help please quick!

 

[xanthym]

It says simplify the following
\cos(\pi+x) + \cos (\pi-x)

and

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

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


~~

 

[aisha]

Holy moly that looks really confusing I don't understand what is going on and that doesn't look very familiar from the unit ...

 

[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...

 

[Curious3141]

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

That, BTW, should be :

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

 

[dextercioby]

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

Daniel.

 

[Curious3141]

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

Daniel.

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

 

[dextercioby]

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

daniel.

 

[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?

 

[dextercioby]

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

Daniel.

 

[aisha]

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

Daniel.

See I don't 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 don't know what to do with them

 

[dextercioby]

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

Daniel.

 

[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 don't 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 please help me understand how to do these problems I don't have a clue I know how to evaluate sin(75 degrees) but that's about it because it uses special triangles but these problems are totally confusing for me .

 

[dextercioby]

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

Daniel.

 

[aisha]

ahhhhhhhh meany ur a smarty pant that's y u know how to do this if ud explain maybe i'd get it too

 

[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.

 

[aisha]

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.

addition and subtraction formulas for cos and sin?

 

[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.