So, last minute my teacher posted something saying to study double-angle formulas for our derivative test tomorrow. So in the back of the book it shows three things for $\cos x$
- $2 \cos^2 x$
- $1-2\sin^2 x$
- $\cos^2 x - \sin^2 x$
So I am not sure which one of these to use, and do I just find the derivative of them afterwards?
$\endgroup$2 Answers
$\begingroup$Use whichever you want as
$$\frac{d(1-2\sin^2x)}{dx}=\frac{d(1)}{dx}-2\frac{d(\sin^2x)}{d(\sin x)}\cdot\frac{d(\sin x)}{dx}=0-2(2\sin x)\cos x=-2\sin2x$$
Similarly, $$\frac{d(2\cos^2x-1)}{dx}=\cdots=2(2\cos x)(-\sin x)$$
Can you try $$\frac{d(\cos^2x-\sin^2x)}{dx}?$$
$\endgroup$ $\begingroup$Note that
$cos(2x) = cos(x+x) = cos(x)cos(x) - sin(x)sin(x) = cos^2(x) - sin^2(x)$.
The derivative is ...
${d \over dx}(cos^2(x)-sin^2(x))$
$={d \over dx}cos^2(x) - {d \over dx}sin^2(x)$
$=-2cos(x)sin(x) - 2sin(x)cos(x)$
$= -4sin(x)cos(x)$
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