Signs of Trig Functions

Section 9.2 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.

Example 9.6. Determining the Sign.

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

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 9.1

\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.

Theorem 9.7. Sign of Trigonometric Functions.

Summarizing for all four quadrants, we have the following.

Figure 9.8. QII: \(x \lt 0\text{,}\) \(y \gt 0\text{.}\) Only *sine* and its reciprocal *cosecant* are positive.

Figure 9.9. QI: \(x \gt 0\text{,}\) \(y \gt 0\text{.}\) *All* trig function values are positive.

Figure 9.10. QIII: \(x \lt 0\text{,}\) \(y \lt 0\text{.}\) Only *tangent* and its reciprocal *cotangent* are positive.

Figure 9.11. QIV: \(x \gt 0\text{,}\) \(y \lt 0\text{.}\) Only *cosine* and its reciprocal *secant* are positive.

Example 9.12. Determining the Quadrant.

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

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.

Example 9.13. Determining the Quadrant.

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

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.

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.

Principle 9.14. 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.

Let’s illustrate this principle.

Example 9.15. 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 \(\theta\text{.}\)

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*}

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

Example 9.16. Finding Trigonometric Function Values.

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

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*}

Prev Top Next