Question

find sin θ and cos θ, 17. 18. 19. 20.

Explanation:

1. For problem 17:

• Step 1: Calculate the radius \(r\) using the distance - formula \(r=\sqrt{x^{2}+y^{2}}\)
• Given the point \((x = 3,y = 4)\), then \(r=\sqrt{3^{2}+4^{2}}=\sqrt{9 + 16}=\sqrt{25}=5\).
• Step 2: Use the definitions of sine and cosine in the unit - circle context (\(\sin\theta=\frac{y}{r}\) and \(\cos\theta=\frac{x}{r}\))
• \(\sin\theta=\frac{y}{r}=\frac{4}{5}\), \(\cos\theta=\frac{x}{r}=\frac{3}{5}\).

1. For problem 18:

• Step 1: Calculate the radius \(r\) using the distance - formula \(r=\sqrt{x^{2}+y^{2}}\)
• Given the point \((x=-3,y = - 1)\), then \(r=\sqrt{(-3)^{2}+(-1)^{2}}=\sqrt{9 + 1}=\sqrt{10}\).
• Step 2: Use the definitions of sine and cosine in the unit - circle context (\(\sin\theta=\frac{y}{r}\) and \(\cos\theta=\frac{x}{r}\))
• \(\sin\theta=\frac{y}{r}=\frac{-1}{\sqrt{10}}=-\frac{\sqrt{10}}{10}\), \(\cos\theta=\frac{x}{r}=\frac{-3}{\sqrt{10}}=-\frac{3\sqrt{10}}{10}\).

1. For problem 19:

• Step 1: Calculate the radius \(r\) using the Pythagorean theorem \(r=\sqrt{x^{2}+y^{2}}\)
• Given \(x = 12\) and \(y = 5\), then \(r=\sqrt{12^{2}+5^{2}}=\sqrt{144 + 25}=\sqrt{169}=13\).
• Step 2: Use the definitions of sine and cosine in the unit - circle context (\(\sin\theta=\frac{y}{r}\) and \(\cos\theta=\frac{x}{r}\))
• \(\sin\theta=\frac{y}{r}=\frac{5}{13}\), \(\cos\theta=\frac{x}{r}=\frac{12}{13}\).

1. For problem 20:

• Step 1: Calculate the radius \(r\) using the Pythagorean theorem \(r=\sqrt{x^{2}+y^{2}}\)
• Given \(x = 2\) and \(y=-\sqrt{5}\), then \(r=\sqrt{2^{2}+(-\sqrt{5})^{2}}=\sqrt{4 + 5}=\sqrt{9}=3\).
• Step 2: Use the definitions of sine and cosine in the unit - circle context (\(\sin\theta=\frac{y}{r}\) and \(\cos\theta=\frac{x}{r}\))
• \(\sin\theta=\frac{y}{r}=\frac{-\sqrt{5}}{3}\), \(\cos\theta=\frac{x}{r}=\frac{2}{3}\).

Answer:

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• Problem 17: \(\sin\theta=\frac{4}{5},\cos\theta=\frac{3}{5}\)
• Problem 18: \(\sin\theta=-\frac{\sqrt{10}}{10},\cos\theta=-\frac{3\sqrt{10}}{10}\)
• Problem 19: \(\sin\theta=\frac{5}{13},\cos\theta=\frac{12}{13}\)
• Problem 20: \(\sin\theta=-\frac{\sqrt{5}}{3},\cos\theta=\frac{2}{3}\)