Need help solving this problem of finding an expression for $y=f(x)+g(x)$ if
$$g(x) = 4 \sin(3x)$$ $$f(x) = 2 \cos(3x)$$
Also determine the minimum value $y=f(x)+g(x)$ can have.
I have re-written $2\cos(3x)$ to $2\sin((\pi/2)-3x)$ but I don't know if that is what I should do.
On
Compare $$ R\sin(A+3x)=R(\sin(A)\cos(3x)+\cos(A)\sin(3x)) $$ with your $$y = 2\cos(3x)+4\sin(3x)$$
From this $$ R\sin(A)=2 \\ R\cos(A)=4 $$
This gives you an alternative expression for $y$.
The $\sin$ function varies between $-1$ and $+1$ so your new expression varies between $-R$ and $+R$.
On
If you're looking for the minimum that's pretty easy since you're dealing with trig functions which produce simple derivatives...
Check out the graph on https://www.desmos.com/calculator
This sounds like a question for something for school, so they probably taught you a method for finding the minimum. Perhaps you're supposed to find the derivative and solve for ${dy\over dx}=0$?
In any case, you'll probably be expected to use whatever method they taught you.
On
**Hint*:
The standard way consists in noting that if in the expression $a\sin 3x+b\cos3 x$, $\;;a^2+b^2=1$, we can set $\, a=\cos\varphi,\;b=\sin\varphi$ for some $\varphi$ and thus use the addition formula to rewrite the expression a $\sin(3x+\varphi)$.
Based on this observation, in the general case, we rewrite \begin{align} a\sin 3x+b\cos3 x&=\sqrt{a^2+b^2}\biggl(\underbrace{\frac a{\sqrt{a^2+b^2}}}_{=\,a'}\sin 3x+\underbrace{\frac b{\sqrt{a^2+b^2}}}_{=\,b'}\cos 3x\biggr)\\[1ex] &=\sqrt{a^2+b^2}\sin(3x+\varphi),\quad\text{where}\enspace\begin{cases}\cos\varphi=a',\\\sin\varphi=b'. \end{cases} \end{align} This system of trigonometric equations *implies, but is not equivalent to $ \tan\varphi=\dfrac{b'}{a'}=\dfrac ba $. So all we know is that $$\varphi\equiv\arctan\frac ba\mod \pi,$$ the exact value of $\varphi \mod 2\pi$ depending on the signs o $a$ and $b$.
On
You can work this out without expanding the triple angles.
Let $t(x)=4\sin3x+2\cos3x$. Now as you know, the minimum of function $t$ is $t'(x)=0$, thus: $$\begin{align} t(x)=4\sin3x+2\cos3x&\\ t'(x)=12\cos3x-6\sin3x&=0\\ 12\cos3x&=6\sin3x\\ 2\cos3x&=\sin3x\\\tag{1} \bigl(2\cos3x&=\sin3x\bigr)^2\\ 4\cos^23x&=\sin^23x\\ 4\cos^23x&=1-\cos^23x\\ 4\cos^23x+\cos^23x&=1\\ 5\cos^23x&=1\\ \cos^23x&=\frac15\\ \end{align}$$ $$\bbox{x\Rightarrow\pm\frac13\cos^{-1}\pm\sqrt{\frac15}+2\pi K \ \mathbb{for} \ K\in\mathbb{Z}}$$ Another way would be to note that $(1)$ can be written as: $$\begin{align} \frac{\sin3x}{2\cos3x}&=1\\ \frac12\tan3x&=1\\ \tan3x&=2\\ \end{align}$$
$$\bbox{x\Rightarrow\frac13\tan^{-1}2+2\pi K\ \mathbb{for} \ K\in\mathbb{Z}}$$ $$x\Rightarrow \frac13\biggl(-\pi+\tan^{-1}2\biggr)+2\pi K \ \mathbb{for} \ K\in\mathbb{Z}$$
Either way, you'll get the same result.
Needless to say, the maximum and minimum values of $t(x)$ are defined by: $$t_{max}=t\Biggl(\frac13\cos^{-1}+\sqrt{\frac15}\Biggr)=4\sin\biggl(\cos^{-1}+\sqrt{\frac15}\biggr)+2\cos\biggl(\cos^{-1}+\sqrt{\frac15}\biggr)=4.47$$ $$t_{min}=t\Biggl(-\frac13\cos^{-1}-\sqrt{-\frac15}\Biggr)=4\sin\biggl(-\cos^{-1}-\sqrt{\frac15}\biggr)+2\cos\biggl(-\cos^{-1}-\sqrt{\frac15}\biggr)=-4.47$$
Hint With $\alpha=3x$ The range of expression $$a\sin\alpha+b\cos\alpha$$ Is $$[-\sqrt {a^2+b^2},\sqrt {a^2+b^2}]$$
And for solving $$\cos\alpha=2\sin\alpha$$ Try squaring both sides, use $\sin^2\alpha=1-\cos^2\alpha$ And then reject the unwanted solutions obtained after solving the above obtained equation