how do I calculate the limit of this trigonometric function?

Question

$$ \lim_{t\to 0} \frac{-\sin(-4t)+\sin(-4t) \cos(-6t)}{5t^2}$$

any clues to solve it?

Answer 1

Hint:

First simplify the expression to $$\frac{\sin(4t)-\sin(4t) \cos(6t)}{5t^2}=\frac{\sin(4t)(1-\cos(6t)}{5t^2}$$ Next, you have the standard limit $$\frac{1-\cos x}{x^2}\xrightarrow[x\to 0]{}\frac12.$$

Answer 2

Write your term in the form $$\frac{\sin(4t)(1-\cos^2(6t))}{1+\cos(6t)}$$ and this $$\frac{\sin(4t)\sin^2(6t)}{1+\cos(6t)}$$