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
\(\large\sin(A+B)\equiv\sin A\cos B+ \cos A \sin B\\ \large\sin(A-B)\equiv\sin A\cos B- \cos A \sin B\)
\(\large\cos(A+B)\equiv\cos A\cos B- \sin A \sin B\\ \large\cos(A-B)\equiv\cos A\cos B+ \sin A \sin B\\\)
\(\large \tan(A+B)≡\frac{\tan A +\tan B}{1-\tan A\tan B}\\ \large \tan(A-B)≡\frac{\tan A -\tan B}{1+\tan A\tan B}\)
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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 \(\large \sin B=\frac{5}{13}\\ \large \cos B=\frac{12}{13}\)
\(\large\sin(A+B)\equiv\sin A\cos B+ \cos A \sin B\)
\(\large\sin(A+B)=\frac{3}{5}\times\frac{12}{13}+\frac{4}{5}\times\frac{5}{13}=\frac{36+20}{65}=\frac{56}{65}\)
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
a =
\(\large \tan(A+B)≡\frac{\tan A +\tan B}{1-\tan A\tan B}\)
\(\large \tan(A+B)=\frac{\frac{1}{4} +\frac{3}{4}}{1-\frac{1}{4} \times\frac{3}{4}}=\frac{1}{1-\frac{3}{16}}=\frac{1}{\frac{13}{16}}={\frac{16}{13}}\)
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 \(\large \sin B=\frac{7}{25}\\ \large \cos B=\frac{24}{25}\)
\(\large\cos(A-B)\equiv\cos A\cos B+ \sin A \sin B\\\)
\(\large\cos(A-B)=-\frac{3}{5}\times\frac{24}{25}+(-\frac{4}{5})\times\frac{7}{25}=\frac{-72-28}{125}=-\frac{100}{125}=-\frac{4}{5}\)
Find \(\large \tan (A+B)\) if A and B are angles such that
\(\large \tan A=-\frac{3}{4}\\ \large \tan B=\frac{2}{3}\)
\(\large \tan(A+B)≡\frac{\tan A +\tan B}{1-\tan A\tan B}\)
\(\large \tan(A+B)=\frac{-\frac{3}{4} +\frac{2}{3}}{1-(-\frac{3}{4}) \times\frac{2}{3}}=\frac{-\frac{1}{12}}{\frac{3}{2}}=-\frac{1}{18}\)
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 \(\large \sin B=\frac{2}{3}\\ \large \cos B=\frac{\sqrt{5}}{3}\)
\(\large\sin(A+B)\equiv\sin A\cos B+ \cos A \sin B\)
\(\large\sin(A+B)=\frac{\sqrt{7}}{4}\times\frac{\sqrt{5}}{3}+\frac{3}{4}\times\frac{2}{3}=\frac{\sqrt{35}+6}{12}\)
Find \(\large \sin (A-B)\) if A and B are angles represented as follows
\(\large\sin(A-B)\equiv\sin A\cos B- \cos A \sin B\)
\(\large\sin(A-B)=\frac{4}{5}\times(-\frac{5}{13})-\frac{3}{5}\times(-\frac{12}{13})=\frac{-20+36}{65}=\frac{16}{65}\)
Find \(\large \sin (A+B)\) if A and B are angles represented as follows
\(\large\sin(A+B)\equiv\sin A\cos B+ \cos A \sin B\)
\(\large\sin(A+B)=\frac{2}{3}\times(\frac{2\sqrt{6}}{7})+(-\frac{\sqrt{5}}{3})\times(-\frac{5}{7})=\frac{4\sqrt{6}+5\sqrt{5}}{21}\)
Find \(\large \cos(A+B)\) if A and B are angles represented as follows
\(\large\cos(A+B)\equiv\cos A\cos B- \sin A \sin B\\\)
\(\large\cos(A+B)=-\frac{4}{5}\times\frac{12}{13}-(-\frac{3}{5})\times\frac{5}{13}=\frac{-48+15}{65}=-\frac{33}{65}\)
If A and B are acute angles such that
\(\large \tan A=0.2\\ \large \tan B=0.5\)
Find a if \(\large\tan(A+B)=\frac{a}{9}\)
a =
\(\large \tan(A+B)≡\frac{\tan A +\tan B}{1-\tan A\tan B}\)
\(\large \tan(A+B)=\frac{0.2 +0.5}{1-0.2\times 0.5}={\frac{0.7}{0.9}}={\frac{7}{9}}\)
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
\(\large\cos(A-B)\equiv\cos A\cos B+ \sin A \sin B\\\)
\(\large\cos(A-B)=-\frac{4}{5}\times\frac{24}{25}+\frac{3}{5}\times\frac{7}{25}=\frac{-96+21}{125}=-\frac{75}{125}=-\frac{3}{5}\)
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
Full Solution
If \(\large \sin(x+30°)=2\cos(x+60°)\), then show that \(\large \tan x=\frac{\sqrt{3}}{9}\)
Hint
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
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
Full Solution
How much of Compound Angle Formulae have you understood?
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