trigonometry - Express $-sqrt{3} sin x - cos x$ in the form of $R cos(x+theta)$

• How to express $-\sqrt{3} \sin x - \cos x$ in the form of $R \cos(x+\theta)$, where $R>0$ and $\theta \in [0^{\circ}, 360^{\circ} ]$ ?




• Hence, how to find the maximum value of $y=1+2\sqrt{3} \sin x + 2 \cos x$ and the corresponding value of $x$ for $x \in [0^{\circ}, 360^{\circ} ]$ ?

Any hints?

Answer

Your goal in such problems is to express the linear combination $$a\cos\phi +b\sin\phi,$$ with $a,b,\phi\in \mathrm R$ in the form $$R(\sin\psi\cos\phi+\cos\psi\sin\phi),$$ which can be written as $$R\sin(\psi+\phi).$$ You may also use the form of a cosine, whichever is more convenient. If you want a cosine, just note that $\sin (\psi+\phi)=\cos(π/2-\psi-\phi).$

In any case, let's continue. Can we always do this? Yes. First, because we know that $|\sin\psi|,\,|\cos\psi|\le 1,$ we can always achieve this by factoring out $|(a,b)|=\sqrt{a^2+b^2},$ since for all real $a,b$ we have that $a,\,b\le \sqrt{a^2+b^2}.$ If we write $R=\sqrt{a^2+b^2},$ then our linear combination becomes $$R\left(\frac aR\cos\phi+\frac bR\sin\phi\right),$$ so that we may take $\sin\psi =\frac aR$ and $\cos\psi=\frac bR.$ Now we may write $$a\cos\phi +b\sin\phi=R\sin(\psi+\phi),$$ where we know how to calculate $R$ and $\psi.$

To apply this to your specific problem, note that $a=-1$ and $b=-\sqrt 3.$ You should be able to continue to find $R$ and $\psi$ now. Then you can write your expression as $$R\sin(\psi+\phi)=R\cos(π/2-\psi-\phi).$$

For the second part, note that $|a+b+c|\le |a|+|b|+|c|,$ and combine this with the facts that $|\sin y|,\,|\cos y|\le 1$ to bound the second function above. Then determine whether it achieves that bound in the interval $[0,2π].$