Expressing $12\sin( \omega t - 10)$ in cosine form

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

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.

• 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. Aug 26, 2014 at 15:14
• 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). Aug 26, 2014 at 15:14

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 $$\sin x=\cos(90°-x)=\cos(x-90°).$$
• 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? Aug 26, 2014 at 16:40
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).
In your case, let $x=\omega t -10°$, then $\sin (\omega t -10°) = \cos(\omega t -10°-90°)$.
• Which is $\sin\left(x\right)=\cos\left(x-90\right)$ here Aug 26, 2014 at 15:15