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Chapter 5: Problem 45

Suppose \(\theta\) is not an odd multiple of \(\frac{\pi}{2}\). Explain why the point \((\tan \theta, 1)\) is on the line containing the point \((\sin \theta, \cos \theta)\) and the origin.

Short Answer

Expert verified
Since the slopes of the lines passing through the points \((0, 0)\) and \((\sin \theta, \cos \theta)\) and through the points \((0, 0)\) and \((\tan \theta, 1)\) are both \(\frac{\cos \theta}{\sin \theta}\), the point \((\tan \theta, 1)\) is collinear with the origin and the point \((\sin \theta, \cos \theta)\), thus it lies on the line containing the point \((\sin \theta, \cos \theta)\) and the origin.

Step by step solution

01

Find the slope of the line passing through the points \((0, 0)\) and \((\sin \theta, \cos \theta)\)

To find the slope of the line, we can use the formula: Slope = \(\frac{\text{change in } y}{\text{change in } x}\) So, the slope will be: Slope = \(\frac{\cos \theta - 0}{\sin \theta - 0}\) Slope = \(\frac{\cos \theta}{\sin \theta}\)
02

Find the slope of the line passing through the points \((0, 0)\) and \((\tan \theta, 1)\)

Similarly, we can find the slope of the line passing through the points \((0, 0)\) and \((\tan \theta, 1)\), following the same formula. The slope will be: Slope = \(\frac{1 - 0}{\tan \theta - 0}\) Slope = \(\frac{1}{\tan \theta}\) Recall that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) when \(\theta\) is not an odd multiple of \(\frac{\pi}{2}\). Therefore, the slope will be: Slope = \(\frac{1}{\frac{\sin \theta}{\cos \theta}}\) Slope = \(\frac{\cos \theta}{\sin \theta}\)
03

Compare the slopes

Now, we can see that both lines have the same slope: \(\frac{\cos \theta}{\sin \theta}\) This implies that the point \((\tan \theta, 1)\) is collinear with the origin and the point \((\sin \theta, \cos \theta)\). Thus, the point \((\tan \theta, 1)\) is on the line containing the point \((\sin \theta, \cos \theta)\) and the origin.

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Most popular questions from this chapter

Show that $$ \sin \left(t+\frac{\pi}{2}\right)=\cos t $$ for every number \(t\).

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