Exact Values with Compound Angles | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Split into known angles, or recognise the expansion

For an exact value, choose standard angles that add or subtract to the required input. Keep fractions and surds; a rounded calculator value is only a check.

Worked example

Find sin 15° exactly.

\begin{gathered}\sin15^\circ=\sin(45^\circ-30^\circ)\\ =\frac{\sqrt2}{2}\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\frac12\\ =\frac{\sqrt6-\sqrt2}{4}\end{gathered}

Use the same identities backwards when products already match a formula. The two individual angles need not be standard.

\begin{gathered}\cos52^\circ\cos22^\circ\\{}+\sin52^\circ\sin22^\circ\\=\cos(52^\circ-22^\circ)\\=\frac{\sqrt3}{2}\end{gathered}

Your turnPractice

Find $\tan75^\circ$ and $\sin18^\circ\cos42^\circ+\cos18^\circ\sin42^\circ$ exactly.

Show worked answer

\begin{gathered}\tan75^\circ=\frac{1+1/\sqrt3}{1-1/\sqrt3}\\ =\frac{\sqrt3+1}{\sqrt3-1}=2+\sqrt3\end{gathered}

Use $75^\circ=45^\circ+30^\circ$ ; multiplying numerator and denominator by $\sqrt3+1$ removes the surd denominator. The second expression is $\sin60^\circ=\sqrt3/2$ .

Find missing ratios without finding the angles

If a question gives sine or cosine as a fraction, use $\sin^2A+\cos^2A=1$ to find the other magnitude. Use the quadrant to choose its sign before substitution.

Worked example

Given sin A = 5/13 with A obtuse, and cos B = 4/5 with B acute, find sin(A + B).

Obtuse means $90^\circ<A<180^\circ$ , so cosine is negative. Acute means between $0^\circ$ and $90^\circ$ , so both ratios are positive.

\begin{gathered}\cos A\\=-\sqrt{1-(5/13)^2}\\=-\frac{12}{13}\\\sin B=\sqrt{1-(4/5)^2}\\=\frac35\\\sin(A+B)\\=\frac5{13}\frac45-\frac{12}{13}\frac35\end{gathered}

Answer

-\frac{16}{65}

No approximate angle was needed. The answer lies between −1 and 1, as a sine value must.

Your turnPractice

Using the same $A,B$ , find $\cos(A-B)$ exactly.

Show worked answer

\begin{gathered}\cos(A-B)\\=-\frac{12}{13}\frac45+\frac5{13}\frac35\end{gathered}

Answer

-\frac{33}{65}

Split into known angles, or recognise the expansion

For an exact value, choose standard angles that add or subtract to the required input. Keep fractions and surds; a rounded calculator value is only a check.

Worked example

Find sin 15° exactly.

\begin{gathered}\sin15^\circ=\sin(45^\circ-30^\circ)\\ =\frac{\sqrt2}{2}\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\frac12\\ =\frac{\sqrt6-\sqrt2}{4}\end{gathered}

Use the same identities backwards when products already match a formula. The two individual angles need not be standard.

\begin{gathered}\cos52^\circ\cos22^\circ\\{}+\sin52^\circ\sin22^\circ\\=\cos(52^\circ-22^\circ)\\=\frac{\sqrt3}{2}\end{gathered}

Your turnPractice

Find $\tan75^\circ$ and $\sin18^\circ\cos42^\circ+\cos18^\circ\sin42^\circ$ exactly.

Show worked answer

\begin{gathered}\tan75^\circ=\frac{1+1/\sqrt3}{1-1/\sqrt3}\\ =\frac{\sqrt3+1}{\sqrt3-1}=2+\sqrt3\end{gathered}

Use $75^\circ=45^\circ+30^\circ$ ; multiplying numerator and denominator by $\sqrt3+1$ removes the surd denominator. The second expression is $\sin60^\circ=\sqrt3/2$ .

Find missing ratios without finding the angles

If a question gives sine or cosine as a fraction, use $\sin^2A+\cos^2A=1$ to find the other magnitude. Use the quadrant to choose its sign before substitution.

Worked example

Given sin A = 5/13 with A obtuse, and cos B = 4/5 with B acute, find sin(A + B).

Obtuse means $90^\circ<A<180^\circ$ , so cosine is negative. Acute means between $0^\circ$ and $90^\circ$ , so both ratios are positive.

\begin{gathered}\cos A\\=-\sqrt{1-(5/13)^2}\\=-\frac{12}{13}\\\sin B=\sqrt{1-(4/5)^2}\\=\frac35\\\sin(A+B)\\=\frac5{13}\frac45-\frac{12}{13}\frac35\end{gathered}

Answer

-\frac{16}{65}

No approximate angle was needed. The answer lies between −1 and 1, as a sine value must.

Your turnPractice

Using the same $A,B$ , find $\cos(A-B)$ exactly.

Show worked answer

\begin{gathered}\cos(A-B)\\=-\frac{12}{13}\frac45+\frac5{13}\frac35\end{gathered}

Answer

-\frac{33}{65}