Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

show that $$ \cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)\\ \text{ if } AB=BA $$

the question gives a hint to use: $$ \sin(A)=\frac{1}{2i}(e^{iA}-e^{-iA})\\ \cos(A)=\frac{1}{2}(e^{iA}+e^{-iA}) $$

share|improve this question
1  
Have you tried multiplying them and using the fact that if $A,B$ commute, then $e^{t (A+B)} = e^{tA} e^{tB} = e^{tB} e^{tA}$ ($t$ is $\pm i$ above)? –  copper.hat Oct 1 '12 at 21:20

2 Answers 2

Using the hint, and we'd like to apply the identity $e^{A+B}=e^Ae^B$ to get $$\cos(A+B)=1/2(e^{i(A+B)}+e^{-i(A+B)})=1/2(e^{iA}e^{iB}+e^{-iA}e^{-iB})=$$ $$1/4[(e^{iA}e^{iB}+e^{-iA}e^{-iB}+e^{iA}e^{-iB}+e^{-iA}e^{iB})+(e^{iA}e^{iB}+e^{-iA}e^{-iB}-e^{iA}e^{-iB}-e^{-iA}e^{iB})]=$$ $$=\cos(A)\cos(B)-\sin(A)\sin(B)$$ So all we've got to prove is that identity when $AB=BA$. This is done as follows:

Observe that in the power series expansion $$I+A+B+1/2!(A+B)^2+1/3!(A+B)^3+...$$ of $e^{A+B}$, if we expand out the powers of $A+B$ there are only a finite number of terms of any fixed degree $n$, where $n$ is the sum of the power of $A$ and the power of $B$. Similarly, in the product of power series $$e^Ae^B=(I+A+(1/2!)A^2+...)(I+B+(1/2!)B^2+...)$$ we can't get any more terms of degree $n$ once we're multiplying terms of degree $n+1$ or higher from either multiplicand.

So verifying $e^{A+B}=e^Ae^B$ reduces to finite verifications that the part of each power series of degree $n$ equals that part of the other. For instance, the degree 0 parts are both $I$, since in the product the only two matrices I can multiply to get $I$ are $I$ from the left and $I$ from the right. Let's look at a higher degree, say, 3. We have $$(1/3!)(A+B)^3=(1/3!)(A^3+AB^2+BAB+B^2A+A^2B+ABA+BA^2+B^3)$$ while the degree-3 part of $e^Ae^B$ is $$(1/3!)IB^3+(1/2!)AB^2+(1/2!)A^2B+(1/3!)A^3I$$ Now we see why the hypothesis $AB=BA$ matters: in general these two expressions aren't equal at all! But using commutation we can collect, for instance, $AB^2,BAB,$ and $B^2A$ all together, and we see that in fact the two power series do agree on their degree-3 parts.

To carry out the proof for general $n$ is essentially just to point out that since $A$ and $B$ commute, the standard binomial theorem applies for $(A+B)^n$, and that in the expansion of $e^Ae^B$ we get the binomial coefficients.

share|improve this answer

$$ \cos(A+B)=\frac{1}{2}(e^{i(A+B)}+e^{-i(A+B)})\\ $$ and $$ \cos(A)\cos(B)=\frac{1}{4}(e^{i(A+B)}+e^{i(A-B)}+e^{i(B-A)}+e^{-i(A+B)})\\ $$ and $$ \sin(A)\sin(B)=\frac{-1}{4}(e^{i(A+B)}-e^{i(A-B)}-e^{i(B-A)}+e^{-i(A+B)})\\ $$ so $$ \therefore \cos(A)\cos(B)-\sin(A)\sin(B)=\frac{1}{2}e^{i(A+B)}+\frac{1}{2}e^{-i(A+B)} $$

share|improve this answer
    
You seem to be assuming $\mathrm e^{A+B}=\mathrm e^A\mathrm e^B$. That's true, but I think part of the idea of this exercise is to show why. –  joriki Oct 2 '12 at 12:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.