How Do You Convert a Trigonometric Expression into a Sine Equation Form?

[karush]

$$-2 \sqrt{3}\cos\left({\theta}\right)+6\sin\left({\theta}\right) $$

Convert to

$$A\sin\left({B\left[\theta-C\right]}\right)+D$$

I couldn't find an example how to do this coversion

 

[Ackbach]

$$-2 \sqrt{3}\cos\left({\theta}\right)+6\sin\left({\theta}\right) $$

Convert to

$$A\sin\left({B\left[\theta-C\right]}\right)+D$$

I couldn't find an example how to do this coversion

I would recommend switching to a single cosine first, then use an angular shift to get the sine. Does that help?

 

[Deveno]

Do you know that cosine and sine are $\dfrac{\pi}{2}$ (or 90 degrees) out-of-phase?

 

[karush]

$$a\cos\left({\theta}\right)+b\sin\left({\theta}\right)=R\cos\left({\theta-\alpha}\right)$$

 

[Ackbach]

$$a\cos\left({\theta}\right)+b\sin\left({\theta}\right)=R\cos\left({\theta-\alpha}\right)$$

Right. So you can find $R$ and $\alpha$ from the usual rectangular-to-polar equations. Then, if you recall that $\sin(\theta+\pi/2)=\cos(\theta)$, as Deveno mentioned earlier, you can finish.

 

[karush]

$$R=4\sqrt{3}$$
$$ \alpha=\frac{2\pi}{3}$$
$$4\sqrt{3}\sin\left({\theta}-\frac{\pi}{6}\right)$$

 

[Greg]

Alternatively,

\(\displaystyle 6\sin x-2\sqrt3\cos x=c\sin(x+d)\)

\(\displaystyle c\sin\left(\frac{\pi}{2}+d\right)=6\)

\(\displaystyle c\sin(\pi+d)=2\sqrt3\)

\(\displaystyle c\cos d=6\)

\(\displaystyle c\sin d=-2\sqrt3\)

\(\displaystyle \tan d=-\frac{1}{\sqrt3}\Rightarrow d=-\frac{\pi}{6}\)

\(\displaystyle c\sin\left(\frac{\pi}{2}-\frac{\pi}{6}\right)=6\Rightarrow c=4\sqrt3\)

\(\displaystyle 6\sin x-2\sqrt3\cos x=4\sqrt3\sin\left(x-\frac{\pi}{6}\right)\)

 

[karush]

Well that save $\cos\left({\theta}\right)$ to
$\sin\left({\theta}\right)$ conversion
How do know to use + or -.

 

[Greg]

Can you be more specific?