Use the information given below to find $\cos(\alpha-\beta)$ . $\newline$ $\cos \alpha=\frac{3}{5}$ , with $\alpha$ in quadrant I $\newline$ $\tan \beta=\frac{3}{4}$ , with $\beta$ in quadrant III $\newline$ Give the exact answer, not a decimal approximation. $\newline$ $\cos(\alpha-\beta)=$

Full solution

Q. Use the information given below to find

\cos(\alpha-\beta)

\newline

\cos \alpha=\frac{3}{5}

, with

\alpha

in quadrant I

\newline

\tan \beta=\frac{3}{4}

, with

\beta

in quadrant III

\newline

Give the exact answer, not a decimal approximation.

\newline

\cos(\alpha-\beta)=

Apply cosine difference identity: Use the cosine difference identity: $\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$ .
Find $\sin(\alpha)$ : Given $\cos(\alpha) = \frac{3}{5}$ . Since $\alpha$ is in quadrant I, $\sin(\alpha)$ is positive. Use the Pythagorean identity $\sin^2(\alpha) + \cos^2(\alpha) = 1$ to find $\sin(\alpha)$ .
Calculate $\sin(\alpha)$ : Calculate $\sin(\alpha)$ : $\sin(\alpha) = \sqrt{1 - \cos^2(\alpha)} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ .
Use $\tan(\beta)$ to find $\sin(\beta)$ and $\cos(\beta)$ : Given $\tan(\beta) = \frac{3}{4}$ and $\beta$ is in quadrant III, both sine and cosine are negative. Use $\tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)}$ to find $\sin(\beta)$ and $\cos(\beta)$ .
Find $\sin(\beta)$ and $\cos(\beta)$ : Find $\sin(\beta)$ and $\cos(\beta)$ : $\sin(\beta) = -\frac{3}{5}$ and $\cos(\beta) = -\frac{4}{5}$ , since the hypotenuse of the right triangle formed is $5$ ( $3^2 + 4^2 = 5^2$ ).