Question

a.cos@ + b.sin@ = c then (a.sin@ + b.cos@)^2=

Answer 1

\[\begin{align}(a\sin\theta+b\cos\theta)^2&=a^2\sin^2\theta+2ab\sin\theta\cos\theta+b^2\cos^2
\theta\\
&=a^2(1-\cos^2\theta)+2ab\sin\theta\cos\theta +b^2(1-\sin^2\theta)\\
&=-a^2\cos^2\theta+2ab\sin\theta\cos\theta-b^2\sin^2\theta+a^2+b^2\\
&=-(a\cos\theta-b\sin\theta)^2+a^2+b^2
\end{align}\]
If that's really \(a\cos\theta+b\sin\theta=c\), then I don't know what tot do next.

Answer 2

The answer given is \[a ^{^{2}} + b ^{2}- c ^{2}\]

Answer 3

Then what is given should be \(a\cos\theta-b\sin\theta=c\), not +.

Answer 4

Yes I think you are right.