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 $\sqrt{a^2 + b^2}$.
 A: It's more interesting when it is not the obvious upper bound $1$. I will take the example $\sin 3x \cos 5 x$. It does not reach the value $1$, so we have some work to do. Let's find an implicit equation for the curve $C \colon \{(\sin(3t), \cos (5t))\ \mid \ t\in [0, 2 \pi]\}$  ( http://en.wikipedia.org/wiki/Lissajous_curve). Skipping some details, it is the curve with equation 
$$-1 + 25 x^2 - 200 x^4 + 560 x^6 - 640 x^8 + 256 x^{10} + 9 y^2 - 
 24 y^4 + 16 y^6=0$$
It has to do with the Chebyshev polynomials. In fact, a point $(x,y)$ in $\mathbb{R}^2$ is of the form $(\sin (3t), \cos(5t))$ if and only if $1 = P(x) + Q(y)$ where $P(\sin( \alpha))= \sin^2 (5 \alpha)$ and $Q(\cos(\beta)) = \cos^2(3 \beta)$. (the only if is clear, since $\sin^2(15 t) + \cos^2(15 t) = 1$). So it is not that hard to get the implicit form for the curve $C$. 
So now we need to find 
$$\max x y \ \text{ where }\  -1 + 25 x^2 - 200 x^4 + 560 x^6 - 640 x^8 + 256 x^{10} + 9 y^2 - 
 24 y^4 + 16 y^6=0$$
We omit the calculations using Lagrange multipliers. It turns out that the maximum $M$ is the largest root of the equation 
$$1073741824\, t^8-1644167168\, t^6+656998400\, t^4-52537500 \,t^2+84375=0$$ $M= 0.96410...$
Well, at least setting the Lagrange multiplier problem in general is not that hard. Solving it is a different thing.

It is not clear to me whether a general easy formula for this maximum exists for general $m$, $n$ for $\max \sin(m t) \cos (n t )$. Maybe a general method, not a general formula that is easy to apply. 
Alternatively one writes 
$$\sin 3t \cos 5 t = \frac{1}{2}( \sin 8t - \sin 2t)$$
Reduce to an equivalent problem: maximize  $ \sin 4u - \sin u$. Even this one is not straightforward. Certainly the derivative is easy to calculate but the maximal value is again the solution of an equation of degree $8$. Perhaps the advantage is that one can find the solution from the graphs of $\sin u$, $\sin 4u$. 

$\bf{Added:}$. If one looks for the maximum value of $\sin mt \cos nt$, it's enough to consider the case $m,n$ relatively prime. Then, even it the maximal value is not $1$, it will get closer to one with the increase of $\max (m,n)$. It is intuitive since the Lissajous curves tend to fill up the square. It would be interesting to investigate how close to $1$ one gets as $\max(m,n) \to \infty$. It appears that number theory, more precisely - rational approximation, appears. 
A: i don't know if this what you are looking for. 
take for example the fitst function $y = f(x) = \sin x \sin 2x.$ you have product of two periodic functions of periods $\pi$ and $2\pi.$ so the product is $\pi$-periodic. you can think of 
$y = \sin x \sin 2x = A\sin 2x $ where the amplitude $A = \sin x$ not constant but slowly varying compared to the faster varying $\sin 2x$ in fact the graph of $f(x)$ is between the envelopes $y = \pm \sin x$ you will see the local maximum of $f$ a little to the right of $\pi/4,$ local max of $\sin 2x$ and the local min a little to the left of $3\pi/4,$ local min of $\sin 2x.$
all this will be much clearer if you can sketch the graphs of $y = f(x), y = \pm \sin x.$ 
A: 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 $-\sqrt{a^2+b^2}\leq a\cos \theta+b\sin \theta\leq \sqrt{a^2+b^2}$. 
