Precalculus (6th Edition) Blitzer

(a) We are given the 80th day of the year. So, $t=80$. Now, put the value of t in the main expression: \begin{align} & H\left( t \right)=12+2.4\sin \left[ \frac{2\pi }{365}\left( t-80 \right) \right] \\ & =12+2.4\sin \left[ \frac{2\pi }{365}\left( 80-80 \right) \right] \\ & =12+2.4sin0 \\ & =12 \end{align} Thus, the number of hours of daylight in San Diego on March 21 will be 12 hours. (b) We are given the 172nd day of the year. So, $t=172$. Now, put the value of t in the main expression: \begin{align} & H\left( t \right)=12+2.4\sin \left[ \frac{2\pi }{365}\left( t-80 \right) \right] \\ & =12+2.4\sin \left[ \frac{2\pi }{365}\left( 172-80 \right) \right] \\ & =12+2.4\sin \left[ \frac{2\pi }{365}\times 92 \right] \\ & =12+2.4\sin \left[ 1.5837069 \right] \end{align} Also, calculate the value of $\sin \left[ 1.5837069 \right]$ by the use of the calculator in radian mode, and put this value in the above expression: \begin{align} & H\left( t \right)=12+2.4\sin \left[ 1.5837069 \right] \\ & =12+2.4\times 0.9999166 \\ & =12+2.39979984 \\ & \approx 14.4\text{ hours} \end{align} Thus, the number of hours of daylight in San Diego on June 21, the 172nd day of the year, will be 14.4 hours. (c) We are given the 355th day of the year. So, $t=355$. Now, put the value of t in the main expression: \begin{align} & H\left( t \right)=12+2.4\sin \left[ \frac{2\pi }{365}\left( 355-80 \right) \right] \\ & =12+2.4\sin \left[ \frac{2\pi }{365}\left( 355-80 \right) \right] \\ & =12+2.4\sin \left[ \frac{2\pi }{365}\times 275 \right] \\ & =12+2.4\sin \left[ 4.7339067 \right] \end{align} Also, calculate the value of $\sin \left[ 4.7339067 \right]$, and put this value in the above expression: \begin{align} & H\left( t \right)=12+2.4\sin \left[ 4.7339067 \right] \\ & =12+2.4\times \left( -0.9997685 \right) \\ & =12-2.399 \\ & \approx 9.6\text{ hours} \end{align} Thus, the number of hours of daylight in San Diego on December 21, the 355th day of the year, will be 9.6 hours.