Hi im trying to rewrite an equation but i can't see how the answer given is obtained which annoys me. The answer states that $\sin(\alpha_1 \cdot \cos(\omega \cdot t)=\alpha_1 \cdot \cos(\omega \cdot t)$ for $\alpha_1\gg1$?
How is this possible when $|\sin(x)|=1$?
you are correct about it not being possible for $a_1 >> 1$. But from what I'm seeing, the given solution assumes $a_1 << 1$, which makes more sense.
Here we're using small angle approximation, which takes advantage of the fact that $\sin(x)$ can be written in the following way (Taylor Expansion of the sine function): $$\sin(x) = x - {x^3 \over 3!} + {x^5 \over 5!} - \dots$$
Now we know that for large exponents, the power of a number that's smaller than $1$ gets smaller and smaller, as opposed to numbers greater than $1$. We use this to our advantage by saying that for a very very small number $x$, any power of $x$ higher or equal to $3$ does not affect the result in any meaningful way. Hence we get the famous small angle approximation:
$$\sin(x) = x$$
Now coming back to your question, we can then see that for a very small $a_1$ we get the following result: $$\sin(a_1\cos(\omega t)) = a_1\cos(\omega t)$$ Hope that helps :)