Trigonometry

Part (i).

The key idea is that the period of a sum of two trig functions is the lowest common multiple of their individual periods.

The function $\cos (2x+\frac{\pi }{3})$ has period $\frac{2\pi }{2}=\pi$. The function $\sin \left(\frac{3x}{2}-\frac{\pi }{4}\right)$ has period $\frac{2\pi }{3/2}=\frac{4\pi }{3}$.

So the period of $\mathrm{f}(x)$ is $\mathrm{lcm}\left(\pi ,\frac{4\pi }{3}\right)=4\pi$.

Part (ii).

The goal is to write $\mathrm{f}(x)=0$ as an equation of the form $\cos (A)=\cos (B)$, which is then easy to solve systematically.

Notice that $\cos \left(\frac{\pi }{2}+\theta \right)=-\sin \theta$, so $\sin \left(\frac{3x}{2}-\frac{\pi }{4}\right)=-\cos \left(\frac{\pi }{2}+\frac{3x}{2}-\frac{\pi }{4}\right)=-\cos \left(\frac{3x}{2}+\frac{\pi }{4}\right)$.

Thus $\mathrm{f}(x)=0$ becomes:$$\cos \left(2x+\frac{\pi }{3}\right)=\cos \left(\frac{3x}{2}+\frac{\pi }{4}\right).$$Using $\cos (A)=\cos (B)⟺A=2n\pi \pm B$, we get two families:

Case 1: $2x+\frac{\pi }{3}=2n\pi +\frac{3x}{2}+\frac{\pi }{4}$, which gives $-\frac{x}{2}=2n\pi -\frac{\pi }{12}$, so $x=-4n\pi +\frac{\pi }{6}$.

In $[-\pi ,\pi ]$: take $n=0$ to get $x=\frac{\pi }{6}$, and $n=1$ to get $x=-\frac{23\pi }{6}$ (out of range). So just $x=\frac{\pi }{6}$... wait — re-examining with the correct sign: $x=\frac{\pi }{6}-4n\pi$. For $n=0$: $x=\frac{\pi }{6}$.

Case 2: $2x+\frac{\pi }{3}=2n\pi -\frac{3x}{2}-\frac{\pi }{4}$, which gives $\frac{7x}{2}=2n\pi -\frac{7\pi }{12}$, so $x=\frac{4n\pi }{7}-\frac{\pi }{6}$.

In $[-\pi ,\pi ]$: $n=0$ gives $x=-\frac{\pi }{6}$; $n=1$ gives $x=\frac{4\pi }{7}-\frac{\pi }{6}=\frac{17\pi }{42}$; $n=-1$ gives $x=-\frac{4\pi }{7}-\frac{\pi }{6}=-\frac{31\pi }{42}$; $n=2$ gives $x=\frac{8\pi }{7}-\frac{\pi }{6}=\frac{41\pi }{42}$; $n=-2$ gives $x=-\frac{8\pi }{7}-\frac{\pi }{6}$ which is less than $-\pi$.

The four solutions in $[-\pi ,\pi ]$ are:$$x=-\frac{31\pi }{42},x=-\frac{\pi }{6},x=\frac{\pi }{6},x=\frac{17\pi }{42},x=\frac{41\pi }{42}.$$Wait — from Case 1 with $n=0$ we get $x=\frac{\pi }{6}$, and from Case 2 with $n=0$ we also get $x=-\frac{\pi }{6}$. The value $x=-\frac{\pi }{6}$ actually appears in both Case 1 (with $n=0$ giving $x=\frac{\pi }{6}$) and Case 2. In fact, checking directly: when $x=-\frac{\pi }{6}$, both $\cos (2x+\frac{\pi }{3})=\cos (0)=1$ and $\sin (\frac{3x}{2}-\frac{\pi }{4})=\sin (-\frac{\pi }{2})=-1$, giving $\mathrm{f}(-\frac{\pi }{6})=0$. Since $x=-\frac{\pi }{6}$ is a root appearing in two separate families, it is a repeated root — the curve touches the $x$-axis at $x=-\frac{\pi }{6}$ without crossing it.

Part (iii).

Since $-1\le \cos (\cdot )\le 1$ and $-1\le \sin (\cdot )\le 1$, the maximum value of $\mathrm{f}(x)$ is $1+1=2$. So $\mathrm{f}(x)=2$ forces both terms to simultaneously reach their maximum:$$\cos \left(2x+\frac{\pi }{3}\right)=1\text{and}\sin \left(\frac{3x}{2}-\frac{\pi }{4}\right)=1.$$From the first: $2x+\frac{\pi }{3}=2k\pi$, so $x=k\pi -\frac{\pi }{6}$. In $[0,2\pi ]$: $k=1$ gives $x=\frac{5\pi }{6}$, $k=2$ gives $x=\frac{11\pi }{6}$.

From the second: $\frac{3x}{2}-\frac{\pi }{4}=\frac{\pi }{2}+2m\pi$, so $x=\frac{\pi }{2}+\frac{4m\pi }{3}$. In $[0,2\pi ]$: $m=0$ gives $x=\frac{\pi }{2}$, $m=1$ gives $x=\frac{11\pi }{6}$.

The only value in $[0,2\pi ]$ satisfying both simultaneously is $x=\frac{11\pi }{6}$.

Part (i).

Rewrite the equation using $\sin \varphi =\cos (90^{\circ }-\varphi )$, so $\cos \theta =\sin 4\theta$ becomes $\cos \theta =\cos (90^{\circ }-4\theta )$. Two angles share the same cosine when they differ by a full revolution or are negatives of each other, giving two families:$$5\theta =360n^{\circ }+90^{\circ }⟹\theta =18^{\circ },90^{\circ },162^{\circ },$$$$3\theta =360n^{\circ }-90^{\circ }⟹\theta =30^{\circ },150^{\circ }.$$The complete solution set in $(0^{\circ },180^{\circ })$ is $\{18^{\circ },30^{\circ },90^{\circ },150^{\circ },162^{\circ }\}$.

To pin down $\sin 18^{\circ }$, let $s=\sin 18^{\circ }$ and use the double-angle formula twice to expand $\sin 4\theta$:$$\sin 4\theta =2\sin 2\theta \cos 2\theta =2(2s\cos \theta )(1-2s^2)=4s\cos \theta (1-2s^2).$$Since $18^{\circ }\ne 90^{\circ }$, we have $\cos 18^{\circ }\ne 0$, so dividing both sides of $\cos \theta =4s\cos \theta (1-2s^2)$ by $\cos \theta$ gives the cubic $8s^3-4s+1=0$. This factors as $(2s-1)(4s^2+2s-1)=0$. The root $s=\frac{1}{2}$ corresponds to $\theta =30^{\circ }$, not $18^{\circ }$, so the relevant equation is $4s^2+2s-1=0$, giving $s=\frac{-1\pm \sqrt{5}}{4}$. Since $18^{\circ }$ is acute, $s>0$, and therefore:$$\sin 18^{\circ }=\frac{\sqrt{5}-1}{4}.$$Part (ii).

Substitute $\sin 2x=2\sin x\cos x$ to write ${\sin }^{2}2x=4{\sin }^{2}x(1-{\sin }^{2}x)$. Setting $s=\sin x$, the equation $4s^2+1=4{\sin }^{2}2x$ becomes:$$16s^4-12s^2+1=0.$$This is a quadratic in $s^2$: $s^2=\frac{12\pm \sqrt{144-64}}{32}=\frac{3\pm \sqrt{5}}{8}$. The crucial observation is that $\frac{3\pm \sqrt{5}}{8}={\left(\frac{\sqrt{5}\pm 1}{4}\right)}^2$, so extracting the square root yields:$$\sin x=\pm \frac{\sqrt{5}+1}{4}\text{or}\sin x=\pm \frac{\sqrt{5}-1}{4}.$$In the required form $p+q\sqrt{5}$ these are $\frac{1}{4}\pm \frac{1}{4}\sqrt{5}$ and $-\frac{1}{4}\pm \frac{1}{4}\sqrt{5}$.

Part (iii).

The equation ${\sin }^{2}3\alpha +{\sin }^{2}5\alpha ={\sin }^{2}6\alpha$ can be linked to part (ii) by writing $\sin 6\alpha =2\sin 3\alpha \cos 3\alpha$, so ${\sin }^{2}6\alpha =4{\sin }^{2}3\alpha (1-{\sin }^{2}3\alpha )$. Rearranging:$${\sin }^{2}5\alpha =4{\sin }^{2}3\alpha -4{\sin }^{2}3\alpha \cdot {\sin }^{2}3\alpha -{\sin }^{2}3\alpha ={\sin }^{2}3\alpha (4-4{\sin }^{2}3\alpha -1).$$A cleaner approach: suppose $\sin 5\alpha =2\sin 3\alpha \cos 3\alpha =\sin 6\alpha$... that forces specific angle relations. Instead, notice that if $3\alpha$ and $5\alpha$ take values from the solution list in part (i), the equation is satisfied. Testing $\alpha =6^{\circ }$: $3\alpha =18^{\circ }$ and $5\alpha =30^{\circ }$, so $6\alpha =36^{\circ }$, giving ${\sin }^{2}18^{\circ }+{\sin }^{2}30^{\circ }={\sin }^{2}36^{\circ }$. From part (i), $\sin 18^{\circ }=\frac{\sqrt{5}-1}{4}$ and setting $x=18^{\circ }$ in part (ii)'s equation (which is satisfied by $\sin 18^{\circ }$ and $\sin 30^{\circ }=\frac{1}{2}$) confirms this identity holds. So $\alpha =6^{\circ }$ is one solution.

For the second, test $\alpha =66^{\circ }$: $3\alpha =198^{\circ }$ and $5\alpha =330^{\circ }$, so ${\sin }^{2}198^{\circ }={\sin }^{2}18^{\circ }$ and ${\sin }^{2}330^{\circ }={\sin }^{2}30^{\circ }=\frac{1}{4}$; also $6\alpha =396^{\circ }=36^{\circ }$, so ${\sin }^{2}6\alpha ={\sin }^{2}36^{\circ }$. The equation holds by the same identity. Therefore the two values are $\boxed{\alpha =6^{\circ }}$ and $\boxed{\alpha =66^{\circ }}$.

Part (i). The key insight is that $\sin y=\sin x$ does not force $y=x$. The general solution of $\sin {\theta }_1=\sin {\theta }_2$ splits into two families: ${\theta }_1={\theta }_2+2k\pi$ (same angle) or ${\theta }_1=\pi -{\theta }_2+2k\pi$ (supplementary angles). Restricting to $-\pi ⩽x,y⩽\pi$, the relevant lines are:$$y=x,y=\pi -x,y=-(\pi +x).$$Each is a straight-line segment; the sketch shows three diagonal lines crossing the square $[-\pi ,\pi {]}^2$.

Part (ii). Starting from $\sin y=\frac{1}{2}\sin x$, differentiate implicitly with respect to $x$:$$\cos y\cdot y^{\prime }=\frac{1}{2}\cos x⟹y^{\prime }=\frac{\cos x}{2\cos y}.$$In the first quadrant, $\cos y=\sqrt{1-{\sin }^{2}y}=\sqrt{1-\frac{1}{4}{\sin }^{2}x}$, so:$$y^{\prime }=\frac{\cos x}{2\sqrt{1-\frac{1}{4}{\sin }^{2}x}}=\frac{\cos x}{\sqrt{4-{\sin }^{2}x}}.$$Differentiate again. Write $f=\cos x$ and $g=(4-{\sin }^{2}x{)}^{1/2}$; by the quotient rule:$$y^{\prime \prime }=\frac{-\sin x\cdot g-\cos x\cdot g^{\prime }}{g^2}.$$Now $g^{\prime }=\frac{1}{2}(4-{\sin }^{2}x{)}^{-1/2}\cdot (-2\sin x\cos x)=\frac{-\sin x\cos x}{g}$, so:$$y^{\prime \prime }=\frac{-\sin x\cdot g-\cos x\cdot \left(\frac{-\sin x\cos x}{g}\right)}{g^2}=\frac{-\sin x}{g^3}(g^2-{\cos }^{2}x)\cdot \frac{1}{g^0}.$$Expanding: $g^2-{\cos }^{2}x=(4-{\sin }^{2}x)-{\cos }^{2}x=4-{\sin }^{2}x-1+{\sin }^{2}x=3$. Therefore:$$y^{\prime \prime }=\frac{-3\sin x}{(4-{\sin }^{2}x{)}^{3/2}}.$$For the sketch on $[0,\frac{\pi }{2}]\times [0,\frac{\pi }{2}]$: at $(0,0)$, $y^{\prime }=\frac{1}{2}$ (slope less than $1$, so the curve lies below $y=x$). Since $y^{\prime \prime }<0$ throughout, the curve is concave. It ends at $(\frac{\pi }{2},\frac{\pi }{6})$ (since $\sin y=\frac{1}{2}$ there).

For the full square $[-\pi ,\pi {]}^2$, symmetry extends the picture. The curve is symmetric under $(x,y)\mapsto (-x,-y)$ (rotational symmetry about $O$) and under reflection in $x=\frac{\pi }{2}$ (giving the piece from $\frac{\pi }{2}$ to $\pi$, which bends back down). The result is a closed loop of four arcs, one in each quadrant pair.

Part (iii). Use the identity $\cos y=\sin (\frac{\pi }{2}-y)$. The equation $\cos y=\frac{1}{2}\sin x$ becomes $\sin (\frac{\pi }{2}-y)=\frac{1}{2}\sin x$, which is the same equation as in (ii) but with $y$ replaced by $\frac{\pi }{2}-y$. The graph is therefore the graph from (ii) translated vertically by $+\frac{\pi }{2}$ (i.e. shifted upwards by $\frac{\pi }{2}$), then reflected: equivalently, it is obtained by applying the transformation $y\mapsto \frac{\pi }{2}-y$ to the (ii) sketch, which reflects it in the horizontal line $y=\frac{\pi }{4}$.