So I have the equation
$\frac{d}{dx}[\cos(2x)-3\sin(x)-1]$
What I did was simplify $\cos(2x)$ to $1-2\sin^2(x)$ by applying the double angle formula.
Here are my steps:
$\frac{d}{dx}[1-2\sin^2(x)-3\sin(x)-1]$
Simplified =$\frac{d}{dx}[-2\sin^2(x)-3\sin(x)]$
Apply chain rule for $-2\sin^2(x)$ = $-4\sin(x)\cos(x)$
Derivative of $-3\sin(x)$ =$-3\cos(x)$
Final result = $-4\sin(x)\cos(x)-3\cos(x)$
The correct answer is $-2\sin(2x)-3\cos(x)$.
I have a feeling I shouldn't be using the double angle formula but I'm not sure why? Simplifying using algebra is allowed before differentiating/integrating as far as I know.
Any help is appreciated!
Why use a double angle formula when it complicates your differentiation by having a square and a product ?
Simply enough :
$$[\cos(2x)-3\sin(x)-1]' = (\cos(2x))' - (3\sin(x))'-(1)' $$
$$=$$
$$-2\sin(2x) - 3\cos(x)$$
Regarding your computation, it is correct though and by the double angle formula substitution that you used, the first term of your expression, using the trigonometric identity $\sin(2x) = 2\sin(x)\cos(x)$ , is :
$$-4\sin(x)\cos(x) = -2(2\sin(x)\cos(x)) = -2\sin(2x)$$
Thus, this means that you have computed the correct result, but it depends on knowing the trigonometric identity to simplify its form.