Write $\cos(2t) + \sqrt3 \sin(2t)$ in the form $C\cos(w(t-t_0))$

Question

Problem

Write $\cos(2t) + \sqrt3 \sin(2t)$ in the form $C\cos(w(t-t_0))$

Attempts

I'm woefully under-practiced in the art of rewriting trigonometric expressions. I've tried looking through the Wikipedia article listing trigonometric identities, but I'm coming up short. Any help and/or solutions appreciated!

Answer 1

Hint use angle sum and difference identities for trignonometric functions: $$C\cos(w(t-t_0))=C(\cos(wt)\cos(wt_0)+\sin(wt)\sin(wt_0))$$ $$=C\cos(wt_0)\cos(wt)+C\sin(wt_0)sin(wt)$$

Compare this expression (for $w=2$) with $$\cos(2t)+\sqrt{3}\sin(2t)$$

After comparing both representations, we can conclude

$$1=C\cos(2t_0)$$ $$\sqrt{3}=C\sin(2t_0)$$

Square both equations and add them to get $$1+3=C^2(\cos^2(2t)+\sin^2(2t))=C^2$$

In the last step we used $\cos^2(2t)+\sin^2(2t)=1$. So $C=\pm2$. Now plug this result into $1=C\cos(2t_0)$ and solve for $t_0$.

Answer 2

Hint: $$\cos(2t)+\sqrt{3}\sin(2t)=2\left(\frac{1}{2}\cos(2t)+\frac{\sqrt{3}}{2}\sin(2t)\right).$$ Can you take it from here? Note that there is an angle $\theta$ so that $\cos\theta=1/2$ and $\sin\theta=\frac{\sqrt{3}}{2}$ simultaneously.

Answer 3

HINT: use that $$a\cos(x)+b\sin(x)=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos(x)+\frac{b}{\sqrt{a^2+b^2}}\sin(x)\right)$$

Answer 4

If coefficients of sin, cos are a, b we divide by $ \sqrt{a^2+b^2}$ to merge them.

$$\cos(2t)+\sqrt{3}\sin(2t)=2*[\frac{1}{2}\cos(2t)+\frac{\sqrt{3}}{2}\sin(2t)].$$ It is in the form $2 [ \cos \pi/3 \cos 2t + ....] $ which can be handled further as cos of a combined angle.

Answer 5

We would like to rewrite $\cos 2t +\sqrt{3}\sin 2t$ in this form $$\cos\alpha\cos 2t-\sin\alpha\sin2t$$ that way, we could simplify the expression to $\cos(\alpha + 2t)$.

Following this model, we would have $\cos\alpha=1$ and $\sin\alpha=-\sqrt{3}$. At this point, we find a contradiction in our argument because $$1 = \cos^2\alpha + \sin^2\alpha = 1^2+(-\sqrt{3})^2=4$$

To fix the problem, we need to scale by a factor of $\sqrt{4}$, so lets define $\alpha$ as $\cos\alpha = \frac{1}{2}$ and $\sin\alpha=-\frac{\sqrt{3}}{2}$. Now we have $$\cos^2\alpha + \sin^2\alpha = (\frac{1}{2})^2+(-\frac{\sqrt{3}}{2})^2 = \frac{1+3}{4}=1$$ so we have that $$\cos\alpha\cos 2t-\sin\alpha\sin2t = \frac{1}{2}\cos 2t +\frac{\sqrt{3}}{2}\sin 2t$$ scaling this expression by a factor of $2$, or $\sqrt{1^2+\sqrt{3}^2}$, we have our original expression back $$2(\frac{1}{2}\cos 2t +\frac{\sqrt{3}}{2}\sin 2t)=\cos 2t +\sqrt{3}\sin{2t}$$ Now we find alpha, and plug it into our expression (notice the acrobatics with the signs) $$\begin{array}{lll} 2(\frac{1}{2}\cos 2t +\frac{\sqrt{3}}{2}\sin 2t)&=&2(\cos(-\frac{\pi}{3})\cos 2t - \sin(-\frac{\pi}{3})\sin 2t)\\ &=& 2\cos(2t -\frac{\pi}{3})\\ &=&2\cos(2(t - \frac{\pi}{6})) \end{array}$$