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I was asked to find if the function f(t) = sin ɷt - cos ɷt is periodic.

I solved it the following way :-

it's of the form sin x - sin y = 2 cos (x +y/2) sin (x-y/2)

sinɷt - sin (Π/2-ɷt) = 2 cos (Π/4) sin (2ɷt -Π/2)

= 2 cos (Π/4) 1/2 sin(ɷt -Π/4)

=cos(Π/4) sin (ɷt -Π/4)

= root 2 sin (ɷt -Π/4)

I have the feeling that I've done something wrong in the last few steps. And is it possible to superimpose 2 functions (sin and cos here) with different values of angles for eg here cos is pi/4 and sin (ɷt -Π/4)

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    $\begingroup$ Use $\sin A-\cos A=\sqrt2\sin\left(A-\dfrac\pi4\right)$ $\endgroup$ Commented Dec 22, 2017 at 7:02
  • $\begingroup$ I was following the steps mentioned in my textbook, thanks I could really use this formula but I wanted to know if the steps I followed were correct :) $\endgroup$
    – Jacob P.J
    Commented Dec 22, 2017 at 7:03
  • $\begingroup$ We have reached at the same expression $\endgroup$ Commented Dec 22, 2017 at 7:04
  • $\begingroup$ Is that to say I could superimpose two trig functions with different angles? $\endgroup$
    – Jacob P.J
    Commented Dec 22, 2017 at 7:05
  • $\begingroup$ See en.wikibooks.org/wiki/Trigonometry/… $\endgroup$ Commented Dec 22, 2017 at 7:06

1 Answer 1

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There are quite a number of mistakes in your derivation. I really have no idea how you actually ended up with a correct final answer.

We start with \begin{align} f(t) &= \sin \omega t - \cos \omega t\\ &=1\cdot\sin \omega t +(-1)\cdot \cos \omega t \end{align}

Suppose we can write $f(t)$ in the form $A\sin(\omega t+\phi)$.

We have \begin{align} f(t) &=A\sin(\omega t+\phi)\\ &=A\cos(\phi)\sin(\omega t)+A\sin(\phi)\cos(\omega t) \end{align}

Since $\sin$ and $\cos$ are orthogonal (don't bother about this term if you don't get it), we can compare coefficients.

So $$A\cos(\phi)=1 \tag{1}$$ and $$A\sin(\phi)=-1 \tag{2}$$

Using $(1)^2+(2)^2$, we have \begin{align} A^2(\cos^2(\phi)+\sin^2(\phi))&=1^2+(-1)^2\\ A^2=2\\ A=\sqrt2 \end{align} for positive $A$.

Then we know that So $$\cos(\phi)=\frac{1}{\sqrt2}$$ and $$\sin(\phi)=-\frac{1}{\sqrt2}$$

Hence, with $\frac{(1)}{(2)}$, $$\tan\phi=-1$$ so $$\phi=-\frac{\pi}{4}$$

Therefore we have $$f(t) =\sqrt2\sin\left(\omega t-\frac{\pi}{4}\right)$$ which is indeed periodic.


Your understanding of 'superposition' is already correct. At every $x$ value, you add the $y$ value of each individual curve together to get another curve.

enter image description here

So for this diagram, the yellow curve, and the red curve add up to form the blue curve.


Let me label your original work this way for easy reference:

\begin{align} \sin\omega t - \sin \left(\frac{\pi}{2}-\omega t\right) &= 2 \cos \left(\frac{\pi}{4}\right) \sin \left(2\omega t -\frac{\pi}{2}\right)\tag{1}\\ &= 2 \cos \left(\frac{\pi}{4}\right) \cdot \frac12 \sin\left(\omega t -\frac{\pi}{4}\right)\tag{2}\\ &=\cos\left(\frac{\pi}{4}\right) \sin \left(\omega t -\frac{\pi}{4}\right)\tag{3}\\ &= \sqrt2 \sin \left(\omega t -\frac{\pi}{4}\right)\tag{4}\\ \end{align}

$(1)$ is correct. There's no problem with that.

$(2)$ is wrong. Note that $$\sin \left(2\omega t -\frac{\pi}{2}\right)\neq\frac12 \sin\left(\omega t -\frac{\pi}{4}\right)$$ In general, $$\sin 2x =2\sin x \cos x\neq 2\sin x \neq \frac12 \sin x$$

$(3)$ and $(4)$ on their own are correct but a major mistake is already made at $(2)$.

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  • $\begingroup$ Thank you so much sir, but I still have this one doubt what does it mean to superpose? Is it adding two trig functions for the same angle ? $\endgroup$
    – Jacob P.J
    Commented Dec 22, 2017 at 10:45
  • $\begingroup$ Could you tell me where I went wrong in my derivation? I liked your approach but it'd be really helpful to know what went wrong :) $\endgroup$
    – Jacob P.J
    Commented Dec 22, 2017 at 10:58
  • $\begingroup$ @JacobP.J Hi, sorry I just came back from something. Is it okay if I do this tomorrow? It's kinda of late in my time zone right now. $\endgroup$ Commented Dec 22, 2017 at 17:11
  • $\begingroup$ its k :) thank u $\endgroup$
    – Jacob P.J
    Commented Dec 23, 2017 at 16:02

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