In agree with you that most of the time, there isn't a step-by-step approach you can take to a problem—you have to be creative. However, there are a few things to note:
- The problems given in school often have 'nice' solutions that can be found by using the techniques and tricks taught in class. This will motivate students, who know that they are able to get to the answer if they persevere
- Being able to spot things is rather satisfying
For example, here was a problem that I was struggling with today:
Assuming only that $\sin^2\theta+\cos^2\theta=1$, show that $\sin\theta\cos\theta\leq\frac{1}{2},$
There are a number of things that make this problem interesting to me:
- I know that there is a true result which can be found. Moreover, I know that it can be found using what I've learnt so far
- If I do connect the dots, then it is extremely satisfying, as it shows the link between different things I have learnt
Here, the key to finding the solution was realising that
\begin{align}
(\sin\theta+\cos\theta)^2&=\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta \\
&=1+2\sin\theta\cos\theta
\end{align}
This links the concept of adding trigonometric functions with multiplying them—a key ingredient in finding the solution! When I saw that this was the key, it did feel very satisfying. At first, it seemed like the identity $\sin^2\theta+\cos^2\theta=1$ was unrelated to the product of $\sin\theta$ and $\cos\theta$, but it suddenly had become obvious what that relationship was. Rearranging the equation, we have
$$
\sin\theta\cos\theta=\frac{(\sin\theta+\cos\theta)^2-1}{2}
$$
And, I got stuck again. I say this because most of the solutions given in textbooks make the process of finding the solution seem like a simple, mechanical process. Well it is if you already have the answer in front of you! For everyone else, though, you just have to keep on trying. There was actually an elegant solution that didn't require calculus, but here is what I did:
The maximum of $(\sin\theta+\cos\theta)^2$ is when the derivative of $\sin\theta+\cos\theta$ equals $0$:
$$
\cos\theta-\sin\theta=0
\implies \cos\theta=\sin\theta
$$
This occurs when $\theta=\frac{\pi}{4}$. Hence, the maximum of $\sin\theta+\cos\theta=\sqrt{2}$, and substituting this back in gives the desired result. Using calculus, I got to the same answer that a clever rearrangement did. The fact that there were two equally valid approaches to this problem is also very reassuring. It shows that as long as we are doing good maths, and looking out for ways to simplify problems, then it is definitely possible to get to the answer, even if our solutions aren't mechanical. It's OK to get stuck, it's OK for your first attempt to be messy, and it's OK to not spot what you can do to simplify a problem. If you keep trying, the payoff is immense.
Best Answer
Any time you have a difference of squares, $$a^2 - b^2$$ it can be factored as $$a^2-b^2 = (a+b)(a-b).$$
Here you have a difference of squares, with $a=\sin x+1$ and $b=\sin x -1$, so you can factor it this way, and then simplify to see what you get. Note that this factorization is completely independent of the fact that you are dealing with trigonometric expressions; it's just simple algebra.