Question

let θ represent the radian measure of the angle below: by dragging the terminal point in the applet, adjust the given angle so that cos(θ) ≈ - 0.74 and sin(θ) ≈ 0.68 hint

• try again remember that the cosine and sine values represent displacements measured in radius lengths. the displacements in the applet are measured in cm how can we determine the location of the terminal point

measured in cm when cos(θ) = - 0.74 and sin(θ) = 0.68 given the circles radius length?

Explanation:

Step1: Recall the definitions of cosine and sine in the unit - circle (generalized)

In a circle of radius \(r\), if the terminal point of an angle \(\theta\) has coordinates \((x,y)\), then \(\cos\theta=\frac{x}{r}\) and \(\sin\theta=\frac{y}{r}\). We can rewrite these formulas as \(x = r\cos\theta\) and \(y = r\sin\theta\).

Step2: Use the given values of \(x\) and \(y\) (displacements)

We are given that \(x=- 1.11\) (approximate value related to \(\cos\theta=-0.74\)) and \(y = 1.72\) (approximate value related to \(\sin\theta = 0.68\)). From \(x = r\cos\theta\) and \(y = r\sin\theta\), we can also use the identity \(r=\sqrt{x^{2}+y^{2}}\) (derived from the Pythagorean theorem \(x^{2}+y^{2}=r^{2}\cos^{2}\theta + r^{2}\sin^{2}\theta=r^{2}(\cos^{2}\theta+\sin^{2}\theta)\) and since \(\cos^{2}\theta+\sin^{2}\theta = 1\)).

Substitute \(x=-1.11\) and \(y = 1.72\) into the formula \(r=\sqrt{x^{2}+y^{2}}\):

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Another way: Using \(r=\frac{x}{\cos\theta}\) (since \(\cos\theta=\frac{x}{r}\)), substituting \(x=-1.11\) and \(\cos\theta=-0.74\)

\(r=\frac{-1.11}{-0.74}\approx1.5\)

Using \(r = \frac{y}{\sin\theta}\) (since \(\sin\theta=\frac{y}{r}\)), substituting \(y = 1.72\) and \(\sin\theta=0.68\)

\(r=\frac{1.72}{0.68}\approx2.53\)

The more accurate way is using the Pythagorean - based formula \(r=\sqrt{x^{2}+y^{2}}\). But if we assume the displacements \(x\) and \(y\) are directly proportional to \(\cos\theta\) and \(\sin\theta\) (in the sense of the unit - circle generalization \(x = r\cos\theta,y = r\sin\theta\)), we can also note that if we consider the ratios.

Let's use the formula \(r=\frac{x}{\cos\theta}\) (assuming the relationship \(x = r\cos\theta\)). Given \(x\) (displacement along the \(x\) - axis) and \(\cos\theta\)

\(r=\frac{- 2.5}{\cos\theta}\) (if we take \(x=-2.5\) (from the green line, assuming it's the \(x\) - coordinate in the formula \(x = r\cos\theta\)) and \(\cos\theta=-0.74\))

\(r=\frac{-2.5}{-0.74}\approx3.38\)

Using \(r=\frac{y}{\sin\theta}\) (where \(y = 1.72\) and \(\sin\theta=0.68\))

\(r=\frac{1.72}{0.68}\approx2.53\)

The most consistent value (since in the unit - circle \(x = r\cos\theta,y = r\sin\theta\)):

If we use \(x=-2.5\) (assuming the green line is \(x\) in \(x = r\cos\theta\))

\(r=\frac{-2.5}{-0.74}\approx3.38\)

If we use \(y = 1.72\) (assuming the blue line is \(y\) in \(y = r\sin\theta\))

\(r=\frac{1.72}{0.68}\approx2.53\)

The correct formula is \(r=\sqrt{x^{2}+y^{2}}\). If \(x=-2.5\) and \(y = 1.72\)

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Answer:

The radius \(r\) of the circle is approximately \(3\space cm\)