7.2 Sum and difference identities  (Page 3/6)

We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas.

1. To find $\sin \left(\alpha +\beta \right),$ we begin with $\sin \alpha =\frac{3}{5}$ and $0<\alpha <\frac{\pi }{2}.$ The side opposite $\alpha$ has length 3, the hypotenuse has length 5, and $\alpha$ is in the first quadrant. See [link] . Using the Pythagorean Theorem, we can find the length of side $a:$
   $$\begin{array}{l}{a}^2+3^2=5^2\hfill \\ \text{}{a}^2=16\hfill \\ a=4\hfill \end{array}$$

   

   Since $\cos \beta =-\frac{5}{13}$ and $\pi <\beta <\frac{3\pi }{2},$ the side adjacent to $\beta$ is $\mathrm{-5},$ the hypotenuse is 13, and $\beta$ is in the third quadrant. See [link] . Again, using the Pythagorean Theorem, we have

   $\begin{array}{l}{\left(-5\right)}^2+a^2={13}^2\hfill \\ 25+a^2=169\hfill \\ a^2=144\hfill \\ a=\pm 12\hfill \end{array}$

   Since $\beta$ is in the third quadrant, $a=\mathrm{–12.}$

   

   The next step is finding the cosine of $\alpha$ and the sine of $\beta .$ The cosine of $\alpha$ is the adjacent side over the hypotenuse. We can find it from the triangle in [link] : $\cos \alpha =\frac{4}{5}.$ We can also find the sine of $\beta$ from the triangle in [link] , as opposite side over the hypotenuse: $\sin \beta =-\frac{12}{13}.$ Now we are ready to evaluate $\sin \left(\alpha +\beta \right).$

   $$\begin{array}{l}\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta \hfill \\ =\left(\frac{3}{5}\right)\left(-\frac{5}{13}\right)+\left(\frac{4}{5}\right)\left(-\frac{12}{13}\right)\hfill \\ =-\frac{15}{65}-\frac{48}{65}\hfill \\ =-\frac{63}{65}\hfill \end{array}$$
2. We can find $\cos \left(\alpha +\beta \right)$ in a similar manner. We substitute the values according to the formula.
   $$\begin{array}{l}\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta \hfill \\ =\left(\frac{4}{5}\right)\left(-\frac{5}{13}\right)-\left(\frac{3}{5}\right)\left(-\frac{12}{13}\right)\hfill \\ =-\frac{20}{65}+\frac{36}{65}\hfill \\ =\frac{16}{65}\hfill \end{array}$$
3. For $\tan \left(\alpha +\beta \right),$ if $\sin \alpha =\frac{3}{5}$ and $\cos \alpha =\frac{4}{5},$ then
   $$\tan \alpha =\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$$

   If $\sin \beta =-\frac{12}{13}$ and $\cos \beta =-\frac{5}{13},$ then

   $$\tan \beta =\frac{\frac{-12}{13}}{\frac{-5}{13}}=\frac{12}{5}$$

   Then,

   $$\begin{array}{l}\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\hfill \\ =\frac{\frac{3}{4}+\frac{12}{5}}{1-\frac{3}{4}\left(\frac{12}{5}\right)}\hfill \\ =\frac{\frac{63}{20}}{-\frac{16}{20}}\hfill \\ \text{}=-\frac{63}{16}\hfill \end{array}$$
4. To find $\tan \left(\alpha -\beta \right),$ we have the values we need. We can substitute them in and evaluate.
   $\begin{array}{l}\tan \left(\alpha -\beta \right)=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }\hfill \\ =\frac{\frac{3}{4}-\frac{12}{5}}{1+\frac{3}{4}\left(\frac{12}{5}\right)}\hfill \\ =\frac{-\frac{33}{20}}{\frac{56}{20}}\hfill \\ =-\frac{33}{56}\hfill \end{array}$