This is the data set in question, it is a set for average month temperatures of air in a Croatian town,the town name is Osijek: \begin{bmatrix} (1)Jan &(2)Feb &(3)Mar &(4)Apr &(5)May &(6)Jun \\ -0.9^{\circ}C&1.1^{\circ}C &6.1^\circ C &11.4^{\circ}C &16.4^{\circ}C &19.7 ^{\circ}C \\ (7)Jul&(8)Aug &(9)Sep &(10)Oct &(11)Nov &(12)Dec \\ 21.5^{\circ}C&20.7 ^{\circ}C &16.7 ^{\circ}C &11.2 ^{\circ}C &5.6^{\circ}C &1.3^{\circ}C \end{bmatrix}
We are given the function which we need to solve using this data set, $$f(x)=A\sin(\omega x+\varphi)+D$$ (I am not quite sure if the D at the end of equation should be there) We are also given that P, which is the period of this function is $P=12$. We are to find all the unknown variables, $A, \omega,\varphi,D$. Also we are to draw this function. I've been trying to solve this system but to no avail. If anyone can help, or suggest a course of action, it would be much appreciated. Thank You All
UPDATE: I've managed to solve this by finding some connections between the data in the table and the formula. I got this A-is the amplitude, that is the maximum height of the wave, the way to calculate this is $A=\frac{(max \ value\ of \ data)-(min\ data\ value)}{2}=\frac{21.5-(-0.9)}{2}=11.2$, the period formula is $P=\frac{2\pi}{\omega}$, which gives, since $P=12$, that $\omega=\frac{\pi}{6}$. Also for $D$ the formula goes, $D=\frac{max\ value+ min\ value}{2}=10.3$ . Now, we almost have everything. I just inserted in the function formula at the beginning all these numbers then calculated $\varphi$ from that $$f(1)=-0.9=11.2*sin(\frac{\pi}{6}*1+\varphi)+10.3\Rightarrow$$ $$-1=sin(\frac{\pi}{6}+\varphi) / sin^{-1}$$ $$sin^{-1}(-1)=\frac{\pi}{6}+\varphi\Rightarrow$$ From this we get that $\varphi=\frac{4\pi}{3}$. Now, the function is as follows $\Rightarrow$ $$f(x)=11.2*sin(\frac{\pi}{6}*x+\frac{4\pi}{3})+ 10.3$$
It turned out in my notebook, the one I drew, pretty much the same as in the plotting program.
First fill in this table (with sines and cosines in radians, or else use degrees but put $180$ instead of $\pi$ where you see $\pi$ below): $$ \begin{array}{cccc} \text{month} & \text{temperature} & \cos(\pi\times\text{month}/6) & \sin(\pi\times\text{month}/6) \\ \hline 1 & -0.9 & 0.866 & 0.500 \\ 2 & \phantom{+}1.1 & 0.500 & 0.866 \\ 3 & \phantom{+}6.1 & 0.000 & 1.000 \\ 4 & \phantom{+}11.4 & \cdots & \cdots \\ 5 & \cdots & \cdots & \cdots \\ \vdots \end{array} $$ Then use linear regression software to fit the model $$ \text{temperature} = \text{constant} + E\cos(\cdots) + F\sin(\cdots) $$ with the data from the table above.
The use a trigonometric identity: $$ E\cos(\omega x) + F\sin(\omega x) = \sqrt{E^2 + F^2} \, \sin(\omega x + \varphi) $$ with $\sin\varphi= E/\sqrt{E^2+F^2}$ and $\cos\varphi = F/\sqrt{E^2+F^2}.$
Note that $\omega x$ must go from $0$ to $2\pi\text{ radians}$ or to $180^\circ$ as $x$ goes from $0$ to $12,$ so $\omega = \dfrac\pi6 \text{ radians}$ or $\omega = \dfrac{180^\circ} 6.$