Half-angle Values | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Use the double-angle identity in reverse

The same formula works with $A=x/2$ . Rearranging the cosine forms gives the squares of the half-angle values.

\begin{gathered}\cos^2\frac x2=\frac{1+\cos x}{2}\\ \sin^2\frac x2=\frac{1-\cos x}{2}\end{gathered}

A square does not determine a sign. Halve the given interval for $x$ first, then choose the positive or negative square root from the quadrant of $x/2$ .

Worked example

Given cos x = −7/25 and 180° < x < 270°, find sin(x/2) and cos(x/2).

The half angle lies between $90^\circ$ and $135^\circ$ . Sine is positive and cosine is negative.

\begin{gathered}\sin^2\frac x2=\frac{1+7/25}{2}\\=\frac{16}{25}\\\cos^2\frac x2=\frac{1-7/25}{2}\\=\frac9{25}\end{gathered}

Answer

\begin{gathered}\sin\frac x2=\frac45\\\cos\frac x2=-\frac35\end{gathered}

Check $2\cos^2(x/2)-1=18/25-1=-7/25$ .

Your turnPractice

Given $\cos x=7/25$ and $3\pi/2<x<2\pi$ , find $\tan(x/2)$ exactly.

Show worked answer

Now $3\pi/4<x/2<\pi$ , still quadrant II. Sine is $3/5$ and cosine is $-4/5$ .

Answer

-\frac34

A tangent half angle can give two algebraic candidates

If only tangent is known, put $t=\tan(x/2)$ into $\tan x=2t/(1-t^2)$ . Solve the quadratic, then apply the half-angle quadrant. Do not assume the half angle has the original angle's sign pattern.

Worked example

Given tan x = 3/4 and 180° < x < 270°, find tan(x/2).

\begin{gathered}\frac34=\frac{2t}{1-t^2}\\ 3t^2+8t-3=0\\(3t-1)(t+3)=0\\t=\frac13\quad\text{or}\quad t=-3\end{gathered}

The half angle is in quadrant II, so tangent is negative. Neither candidate makes $1-t^2$ zero, but only the negative one satisfies the interval.

Answer

\tan(x/2)=-3

Your turnPractice

If the same $\tan x=3/4$ is given with $0^\circ<x<90^\circ$ , which candidate is valid? Explain.

Show worked answer

$t=1/3$ , because $0^\circ<x/2<45^\circ$ , where tangent is positive. The interval changes the answer even though $\tan x$ is unchanged.

Use the double-angle identity in reverse

The same formula works with $A=x/2$ . Rearranging the cosine forms gives the squares of the half-angle values.

\begin{gathered}\cos^2\frac x2=\frac{1+\cos x}{2}\\ \sin^2\frac x2=\frac{1-\cos x}{2}\end{gathered}

A square does not determine a sign. Halve the given interval for $x$ first, then choose the positive or negative square root from the quadrant of $x/2$ .

Worked example

Given cos x = −7/25 and 180° < x < 270°, find sin(x/2) and cos(x/2).

The half angle lies between $90^\circ$ and $135^\circ$ . Sine is positive and cosine is negative.

\begin{gathered}\sin^2\frac x2=\frac{1+7/25}{2}\\=\frac{16}{25}\\\cos^2\frac x2=\frac{1-7/25}{2}\\=\frac9{25}\end{gathered}

Answer

\begin{gathered}\sin\frac x2=\frac45\\\cos\frac x2=-\frac35\end{gathered}

Check $2\cos^2(x/2)-1=18/25-1=-7/25$ .

Your turnPractice

Given $\cos x=7/25$ and $3\pi/2<x<2\pi$ , find $\tan(x/2)$ exactly.

Show worked answer

Now $3\pi/4<x/2<\pi$ , still quadrant II. Sine is $3/5$ and cosine is $-4/5$ .

Answer

-\frac34

A tangent half angle can give two algebraic candidates

If only tangent is known, put $t=\tan(x/2)$ into $\tan x=2t/(1-t^2)$ . Solve the quadratic, then apply the half-angle quadrant. Do not assume the half angle has the original angle's sign pattern.

Worked example

Given tan x = 3/4 and 180° < x < 270°, find tan(x/2).

\begin{gathered}\frac34=\frac{2t}{1-t^2}\\ 3t^2+8t-3=0\\(3t-1)(t+3)=0\\t=\frac13\quad\text{or}\quad t=-3\end{gathered}

The half angle is in quadrant II, so tangent is negative. Neither candidate makes $1-t^2$ zero, but only the negative one satisfies the interval.

Answer

\tan(x/2)=-3

Your turnPractice

If the same $\tan x=3/4$ is given with $0^\circ<x<90^\circ$ , which candidate is valid? Explain.

Show worked answer

$t=1/3$ , because $0^\circ<x/2<45^\circ$ , where tangent is positive. The interval changes the answer even though $\tan x$ is unchanged.