Question

use each given point on the terminal side of an angle of rotation θ to evaluate the three basic trigonometric functions of θ. 1. ( - 6, 8 ) 2. (112, - 16)

Explanation:

Problem 1

Step1: Find the value of \( r \)

For a point \((x, y)\) on the terminal side of an angle \(\theta\), \( r=\sqrt{x^{2}+y^{2}} \). Here \( x = - 6\) and \( y = 8\), so \( r=\sqrt{(-6)^{2}+8^{2}}=\sqrt{36 + 64}=\sqrt{100} = 10\)

Step2: Calculate \(\sin\theta\)

The formula for \(\sin\theta\) is \(\sin\theta=\frac{y}{r}\). Substituting \( y = 8\) and \( r = 10\), we get \(\sin\theta=\frac{8}{10}=\frac{4}{5}\)

Step3: Calculate \(\cos\theta\)

The formula for \(\cos\theta\) is \(\cos\theta=\frac{x}{r}\). Substituting \( x=-6\) and \( r = 10\), we get \(\cos\theta=\frac{-6}{10}=-\frac{3}{5}\)

Step4: Calculate \(\tan\theta\)

The formula for \(\tan\theta\) is \(\tan\theta=\frac{y}{x}\) (where \( x
eq0\)). Substituting \( x = - 6\) and \( y = 8\), we get \(\tan\theta=\frac{8}{-6}=-\frac{4}{3}\)

Step1: Find the value of \( r \)

For a point \((x, y)\) on the terminal side of an angle \(\theta\), \( r=\sqrt{x^{2}+y^{2}} \). Here \( x = 12\) and \( y=-16\), so \( r=\sqrt{12^{2}+(-16)^{2}}=\sqrt{144 + 256}=\sqrt{400}=20\)

Step2: Calculate \(\sin\theta\)

The formula for \(\sin\theta\) is \(\sin\theta=\frac{y}{r}\). Substituting \( y=-16\) and \( r = 20\), we get \(\sin\theta=\frac{-16}{20}=-\frac{4}{5}\)

Step3: Calculate \(\cos\theta\)

The formula for \(\cos\theta\) is \(\cos\theta=\frac{x}{r}\). Substituting \( x = 12\) and \( r = 20\), we get \(\cos\theta=\frac{12}{20}=\frac{3}{5}\)

Step4: Calculate \(\tan\theta\)

The formula for \(\tan\theta\) is \(\tan\theta=\frac{y}{x}\) (where \( x
eq0\)). Substituting \( x = 12\) and \( y=-16\), we get \(\tan\theta=\frac{-16}{12}=-\frac{4}{3}\)

Answer:

\(\sin\theta=\frac{4}{5}\), \(\cos\theta =-\frac{3}{5}\), \(\tan\theta=-\frac{4}{3}\)

Problem 2 (Assuming the point is \((12,-16)\) as per the image)