trig help for vector calc class

**lauren2988** · October 29th 2008, 03:03 PM

A’= A(cosx)^2 + 2Bcosxsinx + C(sinx)^2
C’= A(sinx)^2 – 2Bcosxsinx + C(cosx)^2

I need to prove that (A’)(C’) is equal to AC – B^2

If tried multiplying both together but I cant seem to find the right trig identities or how to simplify it.
any steps or hints would be Great

**Opalg** · October 30th 2008, 02:31 AM

Originally Posted by lauren2988

A’= A(cosx)^2 + 2Bcosxsinx + C(sinx)^2
C’= A(sinx)^2 – 2Bcosxsinx + C(cosx)^2

I need to prove that (A’)(C’) is equal to AC – B^2

If tried multiplying both together but I cant seem to find the right trig identities or how to simplify it.
any steps or hints would be Great

I believe that there is something missing here, and that the correct identity should be $A'C'-B'^2 = AC-B^2$ , where $B' = -A\sin x\cos x + B(\cos^2x-\sin^2x) + C\sin x\cos x$ .

To prove it, check that $\begin{bmatrix}A'&B'\\B'&C'\end{bmatrix} = \begin{bmatrix}\phantom{-}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix} \begin{bmatrix}A&B\\B&C\end{bmatrix} \begin{bmatrix}\cos x&-\sin x\\\sin x&\phantom{-}\cos x\end{bmatrix}$ , and take the determinant of both sides.

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