ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Stick on October 21, 2012, 12:01:10 pm

Title: Proving a trigonometric identity
Post by: Stick on October 21, 2012, 12:01:10 pm: So I have a test tomorrow on advanced circular functions and I'm stumped by the following proving problem in the chapter review:

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

I could see DOPS and an addition formula in the LHS so here is where I went:

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

$=(\sin A\cos B)^2-(\cos A\sin B)^2$

$=(\sin A\cos B-\cos A\sin B)(\sin A\cos B+\cos A\sin B)$

$=\sin(A-B)\sin(A+B)$

I didn't know where to go from here so I then started from the RHS using DOPS:

$RHS: \sin^2 A-\sin^2 B$

$=(\sin A-\sin B)(\sin A+\sin B)$

Again, I wasn't sure where to go from here, but my method is suggesting that $\sin(A-B)\sin(A+B)=(\sin A-\sin B)(\sin A+\sin B)$ which I don't think is true. What am I doing wrong? :S
Title: Re: Proving a trigonometric identity
Post by: b^3 on October 21, 2012, 12:10:00 pm: You're overcomplicating it Stick, its just a simple use of $\sin^{2}(x)+\cos^{2}(x)=1$ :P

$\begin{alignedat}{1}\mathrm{LHS} & =\sin^{2}A\cos^{2}B-\cos^{2}A\sin^{2}B \\ & =\sin^{2}A\left(1-\sin^{2}B\right)-\left(1-\sin^{2}A\right)\sin^{2}B \\ & =\sin^{2}A-\sin^{2}A\sin^{2}B-\sin^{2}B+\sin^{2}A\sin^{2}B \\ & =\sin^{2}A-\sin^{2}B \\ & =\mathrm{RHS} \\ \therefore\sin^{2}A\cos^{2}B-\cos^{2}A\sin^{2}B & =\sin^{2}A-\sin^{2}B \end{alignedat}$
Title: Re: Proving a trigonometric identity
Post by: Stick on October 21, 2012, 12:46:47 pm: FML. I knew it. Thanks, b^3. :) I suck at proving. :P
Title: Re: Proving a trigonometric identity
Post by: paulsterio on October 21, 2012, 08:57:58 pm: Quote from: Stick on October 21, 2012, 12:01:10 pm
$=(\sin A\cos B-\cos A\sin B)(\sin A\cos B+\cos A\sin B)$

$=\sin(A-B)\sin(A+B)$

You'll get better with experience, but from here, I can see that it's not going right. Why? Because the RHS has $sin(A)$ and $sin(B)$ , it doesn't have any $A+B$ or $A-B$

So only use double/sum/difference angle formulae when they help you get to what you want (i.e. the RHS).