Question

honors pre - calculus
hw #7 (lt 6) triangles
day 1 required set #1 - 5 day 2 required set: #6 - 9 day 3 required set #10 - 18
find the values of the six trigonometric functions for an angle in standard position if a point with the given coordinates lies on its terminal side. (must show triangle and work!)

1. (-4, -8)
2. (-8, 15)

sin -\frac{2\sqrt{5}}{5} cos -\frac{\sqrt{5}}{5} tan 2
csc -\frac{\sqrt{5}}{2} sec -\sqrt{5} cot \frac{1}{2}
sin \frac{15}{17} cos -\frac{8}{17} tan -\frac{15}{8}
csc \frac{17}{15} sec -\frac{17}{8} cot -\frac{8}{15}
mixed answers on next page --->
suppose theta is an angle in standard position whose terminal side lies in the given quadrant. find the values of the remaining five trigonometric ratios for theta. (must show triangle & work!)

1. tan \theta=-1 ; quadrant 4
2. cot \theta = \frac{1}{2} ; quadrant 1

Explanation:

Step1: Recall trigonometric - ratio definitions

For a point \((x,y)\) on the terminal side of an angle \(\theta\) in standard - position, \(r=\sqrt{x^{2}+y^{2}}\), \(\sin\theta=\frac{y}{r}\), \(\cos\theta=\frac{x}{r}\), \(\tan\theta=\frac{y}{x}(x
eq0)\), \(\csc\theta=\frac{r}{y}(y
eq0)\), \(\sec\theta=\frac{r}{x}(x
eq0)\), \(\cot\theta=\frac{x}{y}(y
eq0)\)

Step2: Solve for problem 8

Given \(\tan\theta=- 1=\frac{y}{x}\) and the angle is in Quadrant 4, so \(x > 0\) and \(y<0\). Let \(y=-1\) and \(x = 1\). Then \(r=\sqrt{x^{2}+y^{2}}=\sqrt{1 + 1}=\sqrt{2}\)

• \(\sin\theta=\frac{y}{r}=\frac{-1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\)
• \(\cos\theta=\frac{x}{r}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
• \(\tan\theta=-1\)
• \(\csc\theta=\frac{r}{y}=-\sqrt{2}\)
• \(\sec\theta=\frac{r}{x}=\sqrt{2}\)
• \(\cot\theta=-1\)

Step3: Solve for problem 9

Given \(\cot\theta=\frac{1}{2}=\frac{x}{y}\) and the angle is in Quadrant 1, so \(x>0\) and \(y > 0\). Let \(x = 1\) and \(y = 2\). Then \(r=\sqrt{x^{2}+y^{2}}=\sqrt{1 + 4}=\sqrt{5}\)

• \(\sin\theta=\frac{y}{r}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}\)
• \(\cos\theta=\frac{x}{r}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}\)
• \(\tan\theta = 2\)
• \(\csc\theta=\frac{r}{y}=\frac{\sqrt{5}}{2}\)
• \(\sec\theta=\frac{r}{x}=\sqrt{5}\)
• \(\cot\theta=\frac{1}{2}\)

Answer:

Problem 8	Value	Problem 9	Value
\(\cos\)	\(\frac{\sqrt{2}}{2}\)	\(\cos\)	\(\frac{\sqrt{5}}{5}\)
\(\tan\)	\(-1\)	\(\tan\)	\(2\)
\(\csc\)	\(-\sqrt{2}\)	\(\csc\)	\(\frac{\sqrt{5}}{2}\)
\(\sec\)	\(\sqrt{2}\)	\(\sec\)	\(\sqrt{5}\)
\(\cot\)	\(-1\)	\(\cot\)	\(\frac{1}{2}\)