# Addition and subtraction formula, proving identities

## Main Question or Discussion Point

The question states, "Prove the identity."
(1) cos(x + y) cos(x-y) = cos(^2)x – sin(^2)y.
Should i start off using the addition and subtraction formulas for the LHS, and breaking down the perfect square for the RHS? If not or if so, how would I go about solving this problem? Step by step.

(2) sin(x + y + z) = sinx cosy cosz + cosx siny cosz + cosx cosy sinz - sinx siny sinz.
I started this off using the addition formula for the LHS, but I ended up splitting the LHS into two separate identities: sin(x + y) sin(y + z). How do I approach this problem?

tiny-tim
Homework Helper
(1) cos(x + y) cos(x-y) = cos(^2)x – sin(^2)y.
Should i start off using the addition and subtraction formulas for the LHS, and breaking down the perfect square for the RHS?
Hi cheab14!

No … don't start on both sides and try and make them meet in the middle!

You're not building a tunnel!

Just stick to one side, and try to make it equal the other.

In this case, do the LHS, exactly the way you suggested.
(2) sin(x + y + z) = sinx cosy cosz + cosx siny cosz + cosx cosy sinz - sinx siny sinz.
I started this off using the addition formula for the LHS, but I ended up splitting the LHS into two separate identities: sin(x + y) sin(y + z).
Show us how you did that.

hello tiny-tim!!

for (1) i worked out the LHS like so:
(cosxcosy - sinxsiny)(cosxcosy + sinxsiny)
which ended up being: cos(^2)x cos(^2)y - sin(^2)x sin(^2)y. And then I'm stuck.

for (2) I'm just lost
-I have: sinxcosy + cosxsiny + sinycosz + cosysinz for the LHS which makes no sense to me

tiny-tim
Homework Helper
for (1) i worked out the LHS like so:
(cosxcosy - sinxsiny)(cosxcosy + sinxsiny)
which ended up being: cos(^2)x cos(^2)y - sin(^2)x sin(^2)y.
Hi cheab14!

Well, you know the answer has only cos(^2)x and sin(^2)y.

So just rearrange the sin(^2)x and cos(^2)y in terms of cos(^2)x and sin(^2)y !

(Alternatively, you could have used the rule cosAcosB = (cos(A+B) + cos(A-B))/2.)
I have: sinxcosy + cosxsiny + sinycosz + cosysinz for the LHS which makes no sense to me
How did you get that?

Just use sin(A+B) = … , with A = x+y and B = z.

ok thank u much!!

symbolipoint
Homework Helper
Gold Member
A specific graph can be created from which the addition formula can be very understandably derived. Try an internet search since I cannot remember what books, or site showed it. You might also find this in an old Calculus book (but not using any calculus) written by Anton.

tiny-tim
Homework Helper
… all together now …

You can also sing it to the tune of The Battle Hymn of the Republic:

Sin A plus B equals sin A cos B plus cos A sin B

Cos A plus B equals cos A cos B minus sin A sin B

Sin A minus B equals sin A cos B minus cos A sin B

CosAminusB equals cos A cos B plus sin A sin B!

It's your patriotic duty to learn it!

lol ok...i needed a way to remember those formulas...but i have two other questions: one involving proving identities...n the other involving expressions:
(1)prove the identity:
tan(x-y) + tan(y-z) + tan(z-x) = tan(x-y) tan(y-z)tan(z-x)
for this i started using the subraction formula on the LHS, but then that didn't seem to get me any closer. Then I decided to use tan in terms of sin and cos, now I don't what to do.

(2)write the expression in terms of sine only:
(the square root of 3)(sinx) + cosx.
At first I said that the square root of 3 was tan(pi/3) and then broke tan down into sin and cos components still multiplying that function by sinx; then i used the double angle formula for cosx which gave me the square root of of:(1-sin(^2)x) for cosx. Then I didn't know where to go from there.

tiny-tim
Homework Helper
Hi cheab14!

(1) To save on typing, I'll put tanx = X, tany = Y, tanz = Z.

Then LHS x (1+XY)(1+YZ)(1+ZX)
= (X-Y)(1+YZ)(1+ZX)
+ (Y-Z)(1+XY)(1+ZX)
+ (Z-X)(1+XY)(1+YZ);

you need to show that the only items which don't cancel are those with three letters;
well, the 1s obviously cancel;
and then … ?

(2) When it says " in terms of sine only", that means sine and numbers! - so you don't have to convert the √3.