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Trigonometry: Addition Formulas

Sol 1

\begin{displaymath}\cos 75^\circ=\cos(45^\circ+30^\circ)\end{displaymath}


\begin{displaymath}=\cos45^\circ\cos30^\circ-\sin45^\circ\sin30^\circ\end{displaymath}


\begin{displaymath}=\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\frac{1}{2}\end{displaymath}


\begin{displaymath}=\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}=\frac{\sqrt{6}-\sqrt{2}}{4}\end{displaymath}

.

Sol 2

\begin{displaymath}\sin15^\circ=\sin(45^\circ-30^\circ)\end{displaymath}


\begin{displaymath}=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ\end{displaymath}


\begin{displaymath}\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\frac{1}{2}\end{displaymath}


\begin{displaymath}=\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}=\frac{\sqrt{6}-\sqrt{2}}{4}\end{displaymath}

.

Sol 3

\begin{displaymath}\cos105^\circ=\cos(60^\circ+45^\circ)\end{displaymath}


\begin{displaymath}\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ\end{displaymath}


\begin{displaymath}=\frac{1}{2}\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}\end{displaymath}


\begin{displaymath}=\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4}=\frac{\sqrt{2}-\sqrt{6}}{4}\end{displaymath}

.

Sol 4 $\sin(\theta+\pi/2)=\sin\theta\cos\pi/2+\cos\theta\sin\pi/2=$ $(\sin\theta)(0)+(\cos\theta)(1)=\cos\theta$.

Sol 5 $\cos(\pi-\theta)=\cos\pi\cos\theta+\sin\pi\sin\theta=$ $(-1)\cos\theta+(0)\sin\theta=-\cos\theta$.

Sol 6

\begin{displaymath}\tan15^\circ=\tan(60^\circ-45^\circ)\end{displaymath}


\begin{displaymath}=\frac{\tan60^\circ-\tan45^\circ}{1+\tan60^\circ\tan45^\circ}\end{displaymath}


\begin{displaymath}=\frac{\sqrt{3}-1}{1+\sqrt{3}(1)}=\frac{\sqrt{3}-1}{1+\sqrt{3}}\end{displaymath}

.

Sol 7 $\frac{1}{2}[\sin(A+B)+\sin(A-B)]$ $=\frac{1}{2}[(\sin A\cos B+\cos A\sin B)+(\sin A\cos B-\cos A\sin B)]$ $=\frac{1}{2}[2\sin A\cos B]=\sin A\cos B$.

Sol 8 Let $\theta_1$ be the angle between the first line and the positive x-axis, and let $\theta_2$ be the angle between the second line and the positive x-axis. Then the angle between the two lines is given by $\theta=\theta_1-\theta_2$, and $\tan\theta_1=m_1=3/2$ and $\tan\theta_2=m_2=1/5$; so

\begin{displaymath}\tan\theta=\tan(\theta_1-\theta_2)\end{displaymath}


\begin{displaymath}=\frac{\tan\theta_1-\tan\theta_2}{1+\tan\theta_1\tan\theta_2}\end{displaymath}


\begin{displaymath}=\frac{3/2-1/5}{1+\frac{3}{2}\frac{1}{5}}=\frac{13/10}{13/10}=1\end{displaymath}

, so $\theta=45^\circ$.



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Lawrence Marx 2002-07-11