Multiple-Angle Trig Equations | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Give the whole inside angle a name

$\sin(2x)$ means take sine of the angle $2x$ , not twice $\sin x$ . Set $u=2x$ and solve for $u$ first.

Solve $\sin(2x)=1/2$ for $0^\circ\le x\le360^\circ$ .

Multiply both ends of the interval by 2 as well:

0^\circ\le u\le720^\circ

The line is sin u = 1/2. There are four intersections across two repeats.

There are two complete sine cycles, not one. The first gives 30° and 150°; add 360° for the next:

u=30^\circ,\ 150^\circ,\ 390^\circ,\ 510^\circ

Now divide every angle by 2.

Answer

x=15^\circ,\ 75^\circ,\ 195^\circ,\ 255^\circ

Changing the interval first prevents missed solutions. If the inside angle is $x/2$ , halve the interval first, then multiply each final inside-angle solution by 2.

Include any shift in the interval

For $\sin(2x+\pi/6)=1/2$ on $0\le x\le\pi$ , set $u=2x+\pi/6$ . Apply both operations to the endpoints:

\frac{\pi}{6}\le u\le\frac{13\pi}{6}

In that interval, the sine solutions are:

u=\frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{13\pi}{6}

Undo the addition before undoing the multiplication:

x=\frac{u-\pi/6}{2} \quad\Rightarrow\quad x=0,\ \frac{\pi}{3},\ \pi

Both endpoints are included. If the multiplier is negative, evaluate the inside angle at both endpoints and put the smaller value first. For example, $u=\pi-2x$ on $0\le x\le\pi$ runs from $\pi$ down to $-\pi$ , so solve over $-\pi\le u\le\pi$ .

Your turn

Your turnPractice

Find all exact solutions of $\cos(2x-\pi/3)=-1/2$ for $0\le x<\pi$ .

Show worked answer

Set $u=2x-\pi/3$ . Then $-\pi/3\le u<5\pi/3$ . Cosine is −1/2 at $u=2\pi/3$ and $u=4\pi/3$ in this interval.

x=\frac{u+\pi/3}{2}

Answer

x=\frac{\pi}{2},\ \frac{5\pi}{6}

Both are in the original interval. Substitution returns inside angles $2\pi/3$ and $4\pi/3$ , whose cosine is −1/2.

Give the whole inside angle a name

$\sin(2x)$ means take sine of the angle $2x$ , not twice $\sin x$ . Set $u=2x$ and solve for $u$ first.

Solve $\sin(2x)=1/2$ for $0^\circ\le x\le360^\circ$ .

Multiply both ends of the interval by 2 as well:

0^\circ\le u\le720^\circ

The line is sin u = 1/2. There are four intersections across two repeats.

There are two complete sine cycles, not one. The first gives 30° and 150°; add 360° for the next:

u=30^\circ,\ 150^\circ,\ 390^\circ,\ 510^\circ

Now divide every angle by 2.

Answer

x=15^\circ,\ 75^\circ,\ 195^\circ,\ 255^\circ

Changing the interval first prevents missed solutions. If the inside angle is $x/2$ , halve the interval first, then multiply each final inside-angle solution by 2.

Include any shift in the interval

For $\sin(2x+\pi/6)=1/2$ on $0\le x\le\pi$ , set $u=2x+\pi/6$ . Apply both operations to the endpoints:

\frac{\pi}{6}\le u\le\frac{13\pi}{6}

In that interval, the sine solutions are:

u=\frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{13\pi}{6}

Undo the addition before undoing the multiplication:

x=\frac{u-\pi/6}{2} \quad\Rightarrow\quad x=0,\ \frac{\pi}{3},\ \pi

Your turn

Your turnPractice

Find all exact solutions of $\cos(2x-\pi/3)=-1/2$ for $0\le x<\pi$ .

Show worked answer

Set $u=2x-\pi/3$ . Then $-\pi/3\le u<5\pi/3$ . Cosine is −1/2 at $u=2\pi/3$ and $u=4\pi/3$ in this interval.

x=\frac{u+\pi/3}{2}

Answer

x=\frac{\pi}{2},\ \frac{5\pi}{6}

Both are in the original interval. Substitution returns inside angles $2\pi/3$ and $4\pi/3$ , whose cosine is −1/2.