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What are the different ways one can predict the sign of Trigonometric functions. The function may contain combination of different functions.

let us suppose I have a function f(x)= sin(2x) - cos(3x)

Now I want to predict the sign of this function in the interval [0,2π].

Is there any mechanical(doing some math calculations) approach or it is a matter of fact that you have to remember the graphs of both the functions and have a good idea of what will be the result (without actually calculating any thing).

However I can predict the nature of the function f(x) using this graph [pink-->Resultant || green-->cos(3x) || violet-->sin(2x)] enter image description here]1

please admit if you have some shortcut or any trick to quickly predict the sign. sorry for my bad English :(

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You could use the product-sum identities.

In particular, let's try this identity: $$ \cos(a) - \cos(b) = -2 \sin\left(\frac{a+b}{2}\right) \sin\left(\frac{a-b}{2}\right). $$ You have sine and cosine rather than two sines, but we can fix that using the fact that $\sin(\theta) = \cos\left(\frac12\pi - \theta\right).$ Write \begin{align} \sin(2x) - \cos(3x) &= \cos\left(\tfrac12\pi - 2x\right) - \cos(3x) \\ &= -2 \sin\left(\frac{\frac12\pi - 2x + 3x}{2}\right) \sin\left(\frac{\frac12\pi - 2x - 3x}{2}\right) \\ &= 2 \sin\left(\tfrac12 x + \tfrac14\pi\right) \sin\left(\tfrac52 x - \tfrac14\pi\right). \end{align}

So $\sin(2x) - \cos(3x)$ will change sign at each value of $x$ such that either $\sin\left(\tfrac12 x + \tfrac14\pi\right) = 0$ or $\sin\left(\tfrac52 x - \tfrac14\pi\right) = 0.$ Those equations should be relatively easy to solve. Then evaluate $\sin(2x) - \cos(3x)$ at any other point ($x=0$ is an easy one in this case) to find the signs in each interval.

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The solution that follows is the only way I can think of that solves your problem. It really doesn't look to be too useful though.

If $f(x)$ is continuous on $\mathbb R$, then a sign change will only occur about the points, $\xi$, where $f(\xi) = 0$.

If $f(x) = \sin 2x - \cos 3x$, then you need to solve

\begin{align} f(x) &= 0\\ \sin(2x) &= \cos(3x) \\ 2 \sin(x) \cos(x) &= \cos^3(x) - 3 \sin^2(x) \cos(x) \\ 2 \sin(x) &= \cos^2(x) - 3 \sin^2(x) \\ 2 \sin(x) &= 1 - 4 \sin^2(x) \\ \sin x &\in \left\{ -\dfrac 14 - \dfrac{\sqrt 5}{4}, -\dfrac 14 + \dfrac{\sqrt 5}{4} \right\} \\ \sin x &\in \{ \sin(-54^\circ),\; \sin(18^\circ) \}\\ x &\in 360^{\circ}n +\{18^\circ, 162^\circ, 234^\circ, 306^\circ\} \end{align}

Since the period of $f(x)$ is $360^\circ$, then

f(x) is negative if $x \pmod{360^\circ} \in (0^\circ, 18^\circ) \cup (162^\circ, 234^\circ) \cup (306^\circ, 360^\circ)$

f(x) is positive if $x \pmod{360^\circ} \in (18^\circ, 162^\circ) \cup (234^\circ, 306^\circ)$

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  • $\begingroup$ ah.. there is no easy way! $\endgroup$ – Pushkar Soni Dec 29 '16 at 16:51

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