\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \] \[ \sin(A - B) = \sin A \cos B - \cos A \sin B \] \[ \cos(A + B) = \cos A \cos B - \sin A \sin B \] \[ \cos(A - B) = \cos A \cos B + \sin A \sin B \] \[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] \[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]
Compound angle formulae allow us to simplify or expand trigonometric expressions involving sums or differences of two angles. These identities are particularly useful in solving trigonometric equations, proving identities, and simplifying expressions.
The sine of the sum or difference of two angles is given by: \[ \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B \] - The sign in the formula matches the sign between \(A\) and \(B\). - Example: \(\sin(60^\circ + 30^\circ) = \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ\).
The cosine of the sum or difference of two angles is given by: \[ \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B \] - Notice the opposite sign: for \(\cos(A + B)\), we use a minus; for \(\cos(A - B)\), we use a plus. - Example: \(\cos(60^\circ - 30^\circ) = \cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ\).
The tangent of the sum or difference of two angles is given by: \[ \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B} \] - The signs in the numerator and denominator are opposite. - Example: \(\displaystyle \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ}\).
Find \(\sin 75^\circ\), \(\cos 15^\circ\), and \(\tan 75^\circ\) exactly.
\(\displaystyle \sin 75^\circ=\sin(45^\circ+30^\circ)=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ\) \[ =\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot\frac12 =\frac{\sqrt6+\sqrt2}{4}. \]
\(\displaystyle \cos 15^\circ=\cos(45^\circ-30^\circ)=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ\) \[ =\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot\frac12 =\frac{\sqrt6+\sqrt2}{4}. \]
\(\displaystyle \tan 75^\circ=\tan(45^\circ+30^\circ) =\frac{\tan45^\circ+\tan30^\circ}{1-\tan45^\circ\tan30^\circ} =\frac{1+\frac{1}{\sqrt3}}{1-\frac{1}{\sqrt3}} =\frac{\sqrt3+1}{\sqrt3-1} =2+\sqrt3.\)
Solve \(\cos(\theta+30^\circ)=\tfrac12\) for \(0^\circ\le \theta<360^\circ\).
\(\cos u=\tfrac12\Rightarrow u=60^\circ\) or \(u=300^\circ\) (mod \(360^\circ\)). Let \(u=\theta+30^\circ\). Then: \[ \theta+30^\circ=60^\circ \ \Rightarrow\ \theta=30^\circ;\qquad \theta+30^\circ=300^\circ \ \Rightarrow\ \theta=270^\circ. \] Hence \(\boxed{\theta=30^\circ,\,270^\circ}\).
Evaluate \(\sin 50^\circ\cos 10^\circ+\cos 50^\circ\sin 10^\circ\).
Using \(\sin(A+B)=\sin A\cos B+\cos A\sin B\) with \(A=50^\circ,\,B=10^\circ\), \[ \sin 50^\circ\cos 10^\circ+\cos 50^\circ\sin 10^\circ=\sin(60^\circ)=\frac{\sqrt3}{2}. \]
Find \(\sin\frac{\pi}{12}\) exactly.
\(\displaystyle \frac{\pi}{12}=15^\circ=45^\circ-30^\circ\). \[ \sin\frac{\pi}{12}=\sin(45^\circ-30^\circ) =\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ =\frac{\sqrt6-\sqrt2}{4}. \]
Given \(\tan\alpha=2\) and \(\tan\beta=\tfrac13\), find \(\tan(\alpha+\beta)\).
\[ \tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\,\tan\beta} =\frac{2+\frac13}{1-2\cdot\frac13} =\frac{\frac73}{\frac13}=7. \] (Valid provided \(1-\tan\alpha\tan\beta\neq0\), i.e. \(\alpha+\beta\neq 90^\circ+180^\circ k\).)