# Taking derivatives of Trig Functions

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!

• Tip: many standard functions have their own commands. So type \sin in stead of sin. – gebruiker May 24 '18 at 8:56
• Done, thanks! What's the difference between \sin and sin though? – user472288 May 24 '18 at 8:57
• @user472288 It's formulated as a mathematical function showing up as $\sin$ instead of $sin$. – Rebellos May 24 '18 at 8:59
• sin is interpreted as three seperate letters: s i n. \sin is interpreted as the sine fuction. There are typsetting differences as you can see. – gebruiker May 24 '18 at 9:02
• Just on a point of terminology (as it's a bug bear for me :-) ) you don't have an equation, what you have is an expression which requires evaluation. – Kevin May 24 '18 at 9:10

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.