# Homework Help: Deriving double-angle formula

1. Oct 14, 2009

### Jules18

1. The problem statement, all variables and given/known data

2. Relevant equations

My notes say:

Cos(x+a) = CosxCosa - SinxSina

3. The attempt at a solution

When I differentiated each term (using the product rule twice on the right side) I got:

(d/dx) { Sin(x+a) = SinxCosa + CosxSina }

. Cos(x+a) = -SinxSina + CosaCosx + CosxCosa - SinaSinx

. Cos(x+a) = 2CosxCosa - 2SinxSina

Which is not right ... I'm pretty sure I'm just making a stupid mistake with my product rule, but I've tried it a couple times and I keep doing the same thing.

Does anyone see why I'm getting 2's in front of everything when I shouldn't be?

2. Oct 14, 2009

### Bohrok

You are differentiating with respect to x; that means that cos(a) and sin(a) are just constants. Differentiate again, treat them as constants, and you'll get the correct sum formula for cos(x + a). Double-angle formula would be for cos2x.

3. Oct 14, 2009

### Jules18

hmm ... I don't know if you're right.
I've seen the double angle formula with two separate variables.
Plus it shouldn't matter whether it's a constant or a variable, the formula should still work out if the end result is supposed to be in terms of both x and a.

4. Oct 15, 2009

### Office_Shredder

Staff Emeritus
Re-word your question. Prove that for any constant a,

cos(x+a)=cos(x)cos(a)-sin(x)sin(a)

Do you see how this is equivalent?

Basically, the error that you made is that even if a is a variable, not a constant, it's independent of x as an input. So when you differentiate with respect to x, you treat the a as constant. This is something covered in Calculus III classes usually (multivariable calc)

5. Oct 15, 2009

### Bohrok

Yes, you do use two different values for the sum formulas. However, you need to differentiate to get the sum formula for cosine, and you are using d/dx which is differentiating with respect to x. Did you notice that when you differentiated the left side, d/dx sin(x + 1), you treated a as a constant? Although a is a constant, you can let it be any value when using the formula, just like when you choose any value for x.

6. Oct 15, 2009

### blake knight

My friends, first of all "x" & "a" can be both variable or can be a variable "x" and a constant "a" in the expression: Sin(x+a) = SinxCosa + CosxSina. To prove, try solving:
Sin(30+5) = Sin30Cos5 + Cos30Sin5 and youll find out the formula works. Now, starting from the fact that they can both variable or a constant and a variable, we could think of two possible solutions to this problem which of course will lead to the same result: Cos(x+a) = CosxCosa - SinxSina.

7. Oct 15, 2009

### blake knight

Lets try the first assumption, i.e., x is a variable and a as a constant:

Sin(x+a) = SinxCosa + CosxSina--->differentiate both sides with respect to x.
you'll get: cos(x+a) = [sinx*(0) + cosa*cosx] + [cosx*(0) + sina*(-sinx)]-->notice the derivative of cosa = 0 because the derivative of a constant is 0, likewise the derivative of sina.

Simplifying, we get: Cos(x+a) = CosxCosa - SinxSina.--->which solves the problem.

8. Oct 15, 2009

### blake knight

Now assuming that both "x" and "a" are variables, can you try solving the problem? I have the solution already, I just wanna see how you do it.

9. Oct 18, 2009

### Jules18

Okay, I see what you guys mean.
... I tried it again and it worked out fine: I got

Cos(x+a) = CosaCosx - SinaSinx

^_^
whew