Signs of Trig Functions

Section 10.2 Signs of Trig Functions

As we vary the quadrant of angle, the coordinates of a point on its terminal ray change sign. As a result, the quadrant of an angle determines the sign of each trigonometric function.

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Example 10.2.1. Determining the Sign.

Suppose \(\theta\) is a Quadrant III angle. Determine which trigonometric function values of \(\theta\) are positive and which are negative.

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Solution.

Points \((x,y)\) in Quadrant III satisfy \(x \lt 0\) and \(y \lt 0\text{.}\)

Note that \(r = \sqrt{x^2+y^2} \gt 0\) is always positive. Thus, by Definition 10.1.1

\begin{gather*} \sin(\theta) = \frac{y}{r} \quad \text{is negative}\\ \cos(\theta) = \frac{x}{r} \quad \text{is negative}\\ \tan(\theta) = \frac{y}{x} \quad \text{is positive}\\ \csc(\theta) = \frac{r}{y} \quad \text{is negative}\\ \sec(\theta) = \frac{r}{x} \quad \text{is negative}\\ \cot(\theta) = \frac{x}{y} \quad \text{is positive} \end{gather*}

So in Quadrant III only the tangent and its reciprocal are positive.

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Theorem 10.2.2. Sign of Trigonometric Functions.

Summarizing for all four quadrants, we have the following.

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Figure 10.2.3. QII: \(x \lt 0\text{,}\) \(y \gt 0\text{.}\) Only *sine* and its reciprocal *cosecant* are positive.
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Figure 10.2.4. QI: \(x \gt 0\text{,}\) \(y \gt 0\text{.}\) *All* trig function values are positive.
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Figure 10.2.5. QIII: \(x \lt 0\text{,}\) \(y \lt 0\text{.}\) Only *tangent* and its reciprocal *cotangent* are positive.
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Figure 10.2.6. QIV: \(x \gt 0\text{,}\) \(y \lt 0\text{.}\) Only *cosine* and its reciprocal *secant* are positive.
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Example 10.2.7. Determining the Quadrant.

Suppose \(\csc(\theta) = -4\text{.}\) What quadrant could \(\theta\) be in?

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Solution.

Since \(\csc(\theta) = r/y \lt 0\) and \(r \gt 0\text{,}\) we conclude that \(y \lt 0\text{.}\) This occurs in Quadrants III and IV.

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Example 10.2.8. Determining the Quadrant.

Suppose \(\tan(\theta)\) is positive. What quadrant could \(\theta\) be in?

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Solution.

Since \(\tan(\theta) = y/x \gt 0\text{,}\) we conclude that \(x\) and \(y\) have the same sign. This occurs in Quadrants I and Quadrant III.

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While there are six trigonometric functions, their values are all related. In fact, one trigonometric function value and the quadrant is sufficient to determine the rest.

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Principle 10.2.9. Fundamental Principle of Trigonometry.

If you know one trigonometric function value of a non-quadrantal angle and its quadrant, then the remaining trigonometric function values are completely determined by this information.

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Let’s illustrate this principle.

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Example 10.2.10. Finding Trigonometric Function Values.

Suppose \(\cot(\theta) = -5/4\) and \(\theta\) is a Quadrant IV angle. We should be able to find the remaining trigonometric function values of \(\theta\text{.}\)

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Now \(\cot(\theta)= x/y = -5/4\text{.}\) Since \(\theta\) is in Quadrant \(IV\text{,}\) \(x \gt \) and \(y \lt 0\text{.}\) So we can take \((x,y) = (+5,-4)\) as a point on the terminal ray of \(\theta\text{.}\) Then

\begin{equation*} r = \sqrt{5^2+(-4)^2} = \sqrt{25+16} = \sqrt{41}. \end{equation*}

We can now compute:

\begin{gather*} \sin(\theta) = -\frac{4}{\sqrt{41}}\\ \cos(\theta) = \frac{5}{\sqrt{41}}\\ \tan(\theta) = -\frac{4}{5} \end{gather*}

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\begin{gather*} \csc(\theta) = \frac{\sqrt{41}}{5}\\ \sec(\theta) = -\frac{\sqrt{41}}{4}\\ \cot(\theta) = -\frac{5}{4} \end{gather*}

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Example 10.2.11. Finding Trigonometric Function Values.

Suppose \(\csc(\theta) = 2.25\) and \(\cos(\theta) \lt 0\text{.}\) Find the sine, cosine, and tangent of \(\theta\text{.}\)

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Solution.

Immediately,

\begin{equation*} \sin(\theta) = \frac{1}{\csc(\theta)} = \frac{1}{2.25} \approx 0.44. \end{equation*}

Sine (\(y/r\)) is positive and cosine (\(x/r\)) is negative only in Quadrant II. Since

\begin{equation*} \csc(\theta) = \frac{2.25}{1} = \frac{r}{y} \end{equation*}

so that we may take \(r=2.25\) and \(y=1\text{.}\) Then

\begin{equation*} r^2 = x^2 + y^2 \Rightarrow x = \pm \sqrt{(2.25)^2 -1^2} \approx \pm 2.02. \end{equation*}

Since \(x\lt 0\) in Quadrant II, we must take \(x \approx - 2.02\text{.}\) Then,

\begin{equation*} \cos(\theta) \approx - \frac{2.02}{2.25} \approx - 0.90 \end{equation*}

and

\begin{equation*} \tan(\theta) \approx \frac{1}{-2.02} \approx -0.50. \end{equation*}

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