One of the exam papers I had asked me to sketch $\cos{\theta}=\frac{1}{\sqrt2}\sin\alpha+\frac{1}{2\sqrt2}\cos\alpha$ with $\theta$ and $\alpha$ as the axises and for $0<\alpha<2\pi$.
I've tried converting $\frac{1}{\sqrt2}\sin\alpha+\frac{1}{2\sqrt2}\cos\alpha$ into a single trig term which was $\sqrt{\frac{5}{8}}\cos(\alpha-\arctan(2))$, using this I've found the maxima to be at $\alpha=\arctan(2)$ and $\alpha=\arctan(2)+\pi$ for the domain but that's as far as I can go. What else do I need to find to sketch this graph?
All you need to do is recognise that $\sin x+\cos x$ is just another sinusoidal wave - like wave superposition from physics. So maybe plot key points, like the y-intercept and the turning points, but recognise that it is just another wavy thing so interpolating between key points shouldn't be so hard (in a sketch of course).
Turning points of a sinusoidal occur when the derivative is zero, so the turning points will solve the equation:
$$0=\frac{1}{\sqrt{2}}\cos(\alpha)-\frac{1}{2\sqrt{2}}\sin(\alpha)\\\sin^2(\alpha)=4\cos^2(\alpha)\\3\cos^2(\alpha)+(\cos^2(\alpha)-\sin^2(\alpha))=0\\3\cos^2(\alpha)+(2\cos^2(\alpha)-1)=0\dots$$You can solve the rest from here and find turning points. Note: you must be careful with squaring/square rooting! The $\alpha$ must satisfy both $\cos(\alpha)=\pm\frac{1}{\sqrt{5}}$ and $\sin(\alpha)=\pm\frac{2}{\sqrt{5}}$, which can be seen by substituting them into the original derivative equation (as they must sum to zero), so what you must check after finding some candidate $\alpha$ is whether $\sin(\alpha)=2\cos(\alpha)$.
To find the y-intercept, just set $\alpha=0$ and you get $\cos(\theta)=\frac{1}{2\sqrt{2}}$ and while there are infinitely many solutions to this, just pick the principal branch of $\arccos$ and you'll be fine sketching this. For reference, here is the curve in Desmos:
The blue lines are the turning points, (I asked Desmos to solve $\sin(x)=2\cos(x)$, but Desmos has the advantage of being able to numerically solve equations, which the sketcher can't, so you first have to find solutions to the above equations and then check for $\sin=2\cos$) and the green lines are the solutions to $\cos(\theta)=\frac{1}{2\sqrt{2}}$. The purple is one of the original curves.
The second picture shows what I mean about infinitely many solutions - lots of green and purple!
And lastly I'll show what I meant about being careful with solutions and squares/square rooting... the red and blue lines are the plus/minus solutions to $\sin(x)=\pm\frac{2}{\sqrt{5}}$ - notice how some of the plus, some of the minus are turning points and some are not. This is why the secondary condition $\sin(x)=2\cos(x)$ is necessary to ensure they coincide and actually find the turning points!
