1
$\begingroup$

$$ 12\sin( \omega t - 10) $$

I understand how it's solved when using the graphical method, however I'm having trouble understanding something about the trigonometric identities method.

The solution in the text book goes like this (It wants positive amplitudes) : (All angles are in degrees)

$$ 12\cos( \omega t - 10 - 90) $$

$$ 12\cos( \omega t - 100) $$

I know that in order to convert from sine to cosine angle you either add or subtract $90$ degrees. What I don't understand is whether I should add or subtract to get the equivalent with positive amplitude. The way I approach this is that I imagine the graph where $+\cos \omega t$ is the positive $x$-axis, $-\cos \omega t$ is the negative $x$-axis, $+\sin \omega t$ is the negative $y$-axis and $-\sin \omega t$ is the positive $y$-axis.

Since I want to change from positive amplitude sine to positive amplitude cosine I add $90$ degrees. But apparently that is incorrect. Please explain this to me.

$\endgroup$
  • $\begingroup$ Maybe this Wikipedia article might help you to understand the sin and cosine phases en.wikipedia.org/wiki/Cosine#Unit-circle_definitions There are nice graphs on the right. $\endgroup$ – Matthias Aug 26 '14 at 15:14
  • $\begingroup$ You're applying the transformation wrong. The typical relation is $\sin(\alpha)=\cos(90°-\alpha)$, so you have to substract the original argument to 90 (and you're doing the opposite). $\endgroup$ – cjferes Aug 26 '14 at 15:14
1
$\begingroup$

The identities you can use are: \begin{align} \sin x&=\cos(90°-x)\\ \cos x&=\cos(-x) \end{align} Therefore $$ \sin(\omega t-10°)=\cos(90°-(\omega t-10°))= \cos(100°-\omega t)=\cos(\omega t-100°). $$ Of course, you could also directly use $$ \sin x=\cos(90°-x)=\cos(x-90°). $$

$\endgroup$
  • $\begingroup$ What about $-10\cos(\omega t+50)$ to sine? $-10\sin(90-(\omega t+50)$ becomes $-10\sin(40-\omega t)$ becomes $10\sin(\omega t -40)$ correct? $\endgroup$ – Osama Qarem Aug 26 '14 at 16:40
  • $\begingroup$ @OsamaKerem Yes, it's correct. $\endgroup$ – egreg Aug 26 '14 at 16:52
1
$\begingroup$

Look at the graphs of $\cos,\sin$, you can see that $\cos x = \sin (x+ {\pi \over 2})$ or the equivalent $\sin x = \cos (x- {\pi \over 2})$ (use the appropriate change in degrees if you prefer).

enter image description here

In your case, let $x=\omega t -10°$, then $\sin (\omega t -10°) = \cos(\omega t -10°-90°)$.

$\endgroup$
  • $\begingroup$ Which is $\sin\left(x\right)=\cos\left(x-90\right)$ here $\endgroup$ – Matthias Aug 26 '14 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.