In this page, we will will learn about the Compound Angle Formulae used in Trigonometry. These are really important because they open up so many other formulae in trigonometry. In particular, we can derive the double angle formula. There are 6 formula which are written in a shortened form in the IB formula booklet. You need to be very careful with positive/negative signs.
On this page, you should learn about the double angle identities for sine and cosine
\(\large\sin(A+B)\equiv\sin A\cos B+ \cos A \sin B\\ \large\sin(A-B)\equiv\sin A\cos B- \cos A \sin B\)
\(\large\cos(A+B)\equiv\cos A\cos B- \sin A \sin B\\ \large\cos(A-B)\equiv\cos A\cos B+ \sin A \sin B\\\)
\(\large \tan(A+B)≡\frac{\tan A +\tan B}{1-\tan A\tan B}\\ \large \tan(A-B)≡\frac{\tan A -\tan B}{1+\tan A\tan B}\)
This quiz is about the Compound Angle formulae for sin(A+B) , cos(A+B) and tan(A+B)
START QUIZ! Find \(\large \sin (A+B)\) if A and B are acute angles such that
\(\large \sin A=\frac{3}{5}\\ \large \cos A=\frac{4}{5}\) and \(\large \sin B=\frac{5}{13}\\ \large \cos B=\frac{12}{13}\)
\(\large\sin(A+B)\equiv\sin A\cos B+ \cos A \sin B\)
\(\large\sin(A+B)=\frac{3}{5}\times\frac{12}{13}+\frac{4}{5}\times\frac{5}{13}=\frac{36+20}{65}=\frac{56}{65}\)
Check
A and B are acute angles such that
\(\large \tan A=\frac{1}{4}\\ \large \tan B=\frac{3}{4}\)
If \(\large \tan(A+B)=\frac{a}{13}\) , find a
\(\large \tan(A+B)≡\frac{\tan A +\tan B}{1-\tan A\tan B}\)
\(\large \tan(A+B)=\frac{\frac{1}{4} +\frac{3}{4}}{1-\frac{1}{4} \times\frac{3}{4}}=\frac{1}{1-\frac{3}{16}}=\frac{1}{\frac{13}{16}}={\frac{16}{13}}\)
Check
Find \(\large \cos (A-B)\) if A and B are angles such that
\(\large \sin A=-\frac{4}{5}\\ \large \cos A=-\frac{3}{5}\) and \(\large \sin B=\frac{7}{25}\\ \large \cos B=\frac{24}{25}\)
\(\large\cos(A-B)\equiv\cos A\cos B+ \sin A \sin B\\\)
\(\large\cos(A-B)=-\frac{3}{5}\times\frac{24}{25}+(-\frac{4}{5})\times\frac{7}{25}=\frac{-72-28}{125}=-\frac{100}{125}=-\frac{4}{5}\)
Check
Find \(\large \tan (A+B)\) if A and B are angles such that
\(\large \tan A=-\frac{3}{4}\\ \large \tan B=\frac{2}{3}\)
\(\large \tan(A+B)≡\frac{\tan A +\tan B}{1-\tan A\tan B}\)
\(\large \tan(A+B)=\frac{-\frac{3}{4} +\frac{2}{3}}{1-(-\frac{3}{4}) \times\frac{2}{3}}=\frac{-\frac{1}{12}}{\frac{3}{2}}=-\frac{1}{18}\)
Check
Find \(\large \sin (A+B)\) if A and B are acute angles such that
\(\large \sin A=\frac{\sqrt{7}}{4}\\ \large \cos A=\frac{3}{4}\) and \(\large \sin B=\frac{2}{3}\\ \large \cos B=\frac{\sqrt{5}}{3}\)
\(\large\sin(A+B)\equiv\sin A\cos B+ \cos A \sin B\)
\(\large\sin(A+B)=\frac{\sqrt{7}}{4}\times\frac{\sqrt{5}}{3}+\frac{3}{4}\times\frac{2}{3}=\frac{\sqrt{35}+6}{12}\)
Check
Find \(\large \sin (A-B)\) if A and B are angles represented as follows
\(\large\sin(A-B)\equiv\sin A\cos B- \cos A \sin B\)
\(\large\sin(A-B)=\frac{4}{5}\times(-\frac{5}{13})-\frac{3}{5}\times(-\frac{12}{13})=\frac{-20+36}{65}=\frac{16}{65}\)
Check
Find \(\large \sin (A+B)\) if A and B are angles represented as follows
\(\large\sin(A+B)\equiv\sin A\cos B+ \cos A \sin B\)
\(\large\sin(A+B)=\frac{2}{3}\times(\frac{2\sqrt{6}}{7})+(-\frac{\sqrt{5}}{3})\times(-\frac{5}{7})=\frac{4\sqrt{6}+5\sqrt{5}}{21}\)
Check
Find \(\large \cos(A+B)\) if A and B are angles represented as follows
\(\large\cos(A+B)\equiv\cos A\cos B- \sin A \sin B\\\)
\(\large\cos(A+B)=-\frac{4}{5}\times\frac{12}{13}-(-\frac{3}{5})\times\frac{5}{13}=\frac{-48+15}{65}=-\frac{33}{65}\)
Check
If A and B are acute angles such that
\(\large \tan A=0.2\\ \large \tan B=0.5\)
Find a if \(\large\tan(A+B)=\frac{a}{9}\)
\(\large \tan(A+B)≡\frac{\tan A +\tan B}{1-\tan A\tan B}\)
\(\large \tan(A+B)=\frac{0.2 +0.5}{1-0.2\times 0.5}={\frac{0.7}{0.9}}={\frac{7}{9}}\)
Check
Find \(\large \cos (A-B)\) , given that
\(\large \cos A=-\frac{4}{5}\) , \(\large \frac{\pi}{2}\le A\le \pi\)
\(\large \tan B=\frac{7}{24}\) , \(0\le B\le \frac{\pi}{2}\)
Let's put the information into triangles and find the missing lengths using Pythagoras' Theorem
\(\large\cos(A-B)\equiv\cos A\cos B+ \sin A \sin B\\\)
\(\large\cos(A-B)=-\frac{4}{5}\times\frac{24}{25}+\frac{3}{5}\times\frac{7}{25}=\frac{-96+21}{125}=-\frac{75}{125}=-\frac{3}{5}\)
Check
If \(\large \sin A=\frac{4}{5}\) , where \(\large 0\le A\le\frac{\pi}{2}\)
and \(\large \cos B=-\frac{12}{13}\) , where \(\large \pi \le B\le\frac{3\pi}{2}\)
work out \(\large \cos(B-A)\)
Hint You will need to work out cosA and sinB
You will need to use the compound formula \(\large\cos(B-A)\equiv\cos B\cos A+ \sin B \sin A\)
Full Solution
If \(\large \sin(x+30°)=2\cos(x+60°)\) , then show that \(\large \tan x=\frac{\sqrt{3}}{9}\)
Hint You will need to remember the exact values for sin30°, cos30°, sin60° and cos60° Full Solution
a) By writing 15° as 45° - 30° , find the value of sin15°
b) Hence , show that the area of this triangle \(\large =4(\sqrt{3}-1)\)
Hint a) We need to use the compound angle formula to evaluate sin(45°-30°)
b) We need to use the area of the triangle formula. \(\large A= \frac{1}{2}ab\sin C\)
Full Solution
Prove that
\(\large \frac{\sin(A+B)+\sin(A-B)}{\cos(A+B)+\cos(A-B)}=\tan A\)
Hint Full Solution
Prove that \(\large \tan 3x\equiv \frac{3\tan x-\tan^3x}{1-3\tan^2x}\)
Hint Use the compound angle formula to work out \(\large \tan(2x + x)\)
Very careful manipulation is required in this question. Take your time to make sure that you do not make a mistake.
Full Solution
a) Write \(\large \cos4x\) in terms of \(\large\cos x\)
b) Hence , solve \(\large 8\cos^4x-8\cos^2x+1=\sin4x\) , for \(\large 0\le x\le\pi\)
Hint a) Write \(\large\cos4x\equiv \cos(2x+2x)\) and use the double angle formula for \(\large \cos2x\) and \(\large \sin2x\)
b) Remember that \(\large \tan x\equiv\frac{\sin x}{\cos x}\)
Full Solution MY PROGRESS
Self-assessment How much of Compound Angle Formulae have you understood?
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