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This is my attempt: $$ \begin{align} & \phantom{={}}\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B) \\[8pt] & = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B) - \cos(A)\sin(B)) \\[8pt] & = (\sin(A)+\sin(B))(\cos(B)+\cos(A))(\sin(A)-\sin(B))(\cos(B)-\cos(A)) \\[8pt] & = (\sin(A)+\sin(B))^2(\cos(B)-\cos(A))^2 \end{align} $$

But now I can't get rid of the cosines. How can I get rid of them?

Michael Hardy
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Kat
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  • your first equality is wrong. – mathemagician Dec 10 '13 at 00:41
  • @mathemagician: isn't it the addition identities? – Kat Dec 10 '13 at 00:42
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    @Kat : You should put $=$ between things already known to be equal, not between things you're trying to prove are equal, except when you write things like "We need to prove that...." etc. – Michael Hardy Dec 10 '13 at 00:49
  • Start with your second displayed line. Multiply out. The cross terms will cancel, and you should get $\sin^2 A\cos^2B-\cos^2 A\sin^2B$. Replace the $\cos^2 x$ by $1-\sin^2 x$. – André Nicolas Dec 10 '13 at 00:51
  • @MichaelHardy: What do you mean? Like I should start with sin(A + B)sin(A -B) = ... ? – Kat Dec 10 '13 at 00:55
  • You could start either there or with $\sin^2 A-\sin^2 B=\cdots$, but "$=$" should appear between things already known to be equal. – Michael Hardy Dec 10 '13 at 00:59
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    It might sound pedantic, but really, you absolutely must start with things you know to be true! Going from something true to what you want is a proof. Going from what you want to something true is easy: $1 + 1 = 3 \implies 0 (1 + 1) = 0(3) \implies 0 = 0$. One would hope this doesn't imply that $1 + 1 = 3$... – Henry Swanson Dec 10 '13 at 01:03

5 Answers5

3

You got off to a good start: $$ \sin(A+B)\sin(A-B) = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B)-\cos(A)\sin(B)) $$ This is of the form $(x+y)(x-y)$ so $$ \sin(A+B)\sin(A-B) = \sin^2(A)\cos^2(B)-\cos^2(A)\sin^2(B) $$ Eliminate the cosines (since $\sin^2(x)+\cos^2(x)=1$, so $\cos^2(x)=1-\sin^2(x)$) and expand: $$ \begin{align} \sin(A+B)\sin(A-B) &= \sin^2(A)(1-\sin^2(B))-(1-\sin^2(A))\sin^2(B)\\ &= \sin^2(A)-\sin^2(A)\sin^2(B)-\sin^2(B)+\sin^2(A)\sin^2(B)\\ &=\sin^2(A)-\sin^2(B) \end{align} $$ as required.

Rick Decker
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$x^2-y^2=(x+y)(x-y)$, we have $$\sin^2A-\sin^2B\\ =(\sin A+\sin B)(\sin A-\sin B)\\ =2\sin\frac{A+B}2\cos\frac{A-B}2\cdot2\cos\frac{A+B}2\sin\frac{A-B}2\\ =2\sin\frac{A+B}2\cos\frac{A+B}2\cdot2\sin\frac{A-B}2\cos\frac{A-B}2\\ =\sin(A+B)\sin(A-B)$$

ziang chen
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You wrote: $$ \begin{align} & \phantom{={}}\sin^2(A) - \sin^2(B) \overset{(1)}{=} \sin(A + B)\sin(A -B) \\[8pt] & = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B) - \cos(A)\sin(B)) \\[8pt] & \overset{(2)}{=} (\sin(A)+\sin(B))(\cos(B)+\cos(A))(\sin(A)-\sin(B))(\cos(B)-\cos(A)) \\[8pt] & \overset{(3)}{=} (\sin(A)+\sin(B))^2(\cos(B)-\cos(A))^2 \end{align} $$ I've put numbers above things that are dubious or wrong.

The problem with the $(1)$ is that that is something you want to prove, not something you already know.

The problem with $(2)$ is that if you expand that product of four sums, you get $16$ terms, and if you expand the thing before it, you get a sum of only $8$ terms. You haven't shown that they're equal.

The problem with $(3)$ is that if you multiply $\cos B+\cos A$ by $\cos B-\cos A$, you get $\cos^2 B-\cos^2 A$, not $(\cos B-\cos A)^2$, and a similar thing applies to the sines.

I think some others have posted some things you can try instead.

Michael Hardy
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HINT:

Expand the right-hand side using the addition identities. You'll end up with:

$$\sin^2A\cos^2B - \sin^2B\cos^2A$$

Then use the basic trigonometric identity.

Stefan4024
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$2\sin(x)\sin(y) = \cos(x-y)-\cos(x+y)$

Let $ x = A-B , y = A+B$

Therfore you have , RHS

$$0.5(\cos(2B)-\cos(2A))$$

Now $\cos(2X) = 1-2\sin^2(X)$

Use this identity to put in the previously written question . You will get it equal to LHS

abkds
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