Range of trigonometric functions I would like to know if there is a simple approach to find the range of functions in the form:
$$\sin x\sin2x$$ 
$$\cos x\cos3x$$
$$\sin 2x\cos 4x$$
For example, finding the range of a function in the form:
$$a\cos\theta + b\sin\theta$$ is simple (the minimum value is $-\sqrt{a^2 + b^2}$ while the maximum value is $\sqrt{a^2 + b^2}$.

We have always $ a^2\leq a^2+b^2\  \Rightarrow\  |a|\leq\sqrt{a^2+b^2} \Rightarrow \frac{|a|}{\sqrt{a^2+b^2}}\leq 1\Rightarrow-1\leq\frac{a}{\sqrt{a^2+b^2}}\leq1$ so there exists a $\alpha $ that  $ \cos\alpha=\frac{a}{\sqrt{a^2+b^2}}$. Beside  We know that $\sin\alpha=\pm\sqrt{1-\cos^2\alpha}=\pm\sqrt{1-(\frac{a}{\sqrt{a^2+b^2}})^2}=\pm\frac{b}{\sqrt{a^2+b^2}}$. Now let be  $y=a cos \theta+b\sin \theta $ then $\frac{y}{\sqrt{a^2+b^2}}=\frac{a}{\sqrt{a^2+b^2}}\sin \theta+\frac{b}{\sqrt{a^2+b^2}}\cos \theta=\cos\alpha\sin \theta\pm\sin\alpha\cos \theta=\sin (\theta\pm\alpha)$. Because of $-1\leq\sin (\theta\pm\alpha)\leq 1$ then $-1\leq\frac{y}{\sqrt{a^2+b^2}}\leq 1$. Hence $-\sqrt{a^2+b^2}\leq a\cos \theta+b\sin \theta\leq \sqrt{a^2+b^2}$. 
