Inverse Operations for Trigonometric Ratios
Recall from a previous lesson that calculators have inverse operations to compute the angle corresponding to a given trigonometric ratio to a very high precision.
- inverse sine operation is denoted sin-1
- inverse cosine operation is denoted cos-1
- inverse tangent operation is denoted tan-1
For example, the angle \(\theta\) in \(\triangle ABC\) can be found by making an equation using any of the primary trigonometric ratios and solving for \(\theta\).
Sine Ratio
\[\begin{align*} \sin \left(\theta\right)&=\dfrac{3}{5}\\ \sin^{-1}\left(\sin \left(\theta\right)\right)&=\sin^{-1}\left(\dfrac{3}{5}\right)\\ \theta &=\sin ^{-1}\left(\dfrac{3}{5}\right)\\ \theta &\approx 36.87^\circ \end{align*}\]
Cosine Ratio
\[\begin{align*} \cos \left(\theta\right)&=\dfrac{4}{5}\\ \theta &=\cos ^{-1}\left(\dfrac{4}{5}\right)\\ \theta &\approx 36.87^\circ \end{align*}\]
Tangent Ratio
\[\begin{align*} \tan \left(\theta\right)&=\dfrac{3}{4}\\ \theta&=\tan ^{-1}\left(\dfrac{3}{4}\right)\\ \theta &\approx 36.87^\circ \end{align*}\]
Solving for \(\theta\) in each of the three equations gives the same answer of approximately \(36.87^\circ\).
In the next example, when angles are not acute, our calculators may not directly provide the answer we are looking for.
Example 1
Use the fact that \(\sin \left(\alpha\right)=\dfrac{5}{13}, 0^\circ \lt \alpha\lt 90^\circ\), to answer the following questions.
- Draw a diagram showing \(\alpha\) and another angle, \(A\), between \(0^\circ\) and \(360^\circ\) so that \(\sin\left(A\right)=\sin\left(\alpha\right)\) and \(A \neq \alpha\). How are these two angles related?
- Find the measures of \(A\) and \(\alpha\), to the nearest degree.
For any angle \(\theta\) in standard position whose terminal arm passes through the point \(P\left(x,y\right)\), the primary trigonometric ratios are
- \(\sin\left(\theta\right)=\dfrac{y}{r}\)
- \(\cos\left(\theta\right)=\dfrac{x}{r}\)
- \(\tan\left(\theta\right)=\dfrac{y}{x}\)
where \(r=\sqrt{x^2+y^2}\) is the distance from the origin to the point \(P\) (or the radius of the circle that \(P\) lies on).
Solution — Part A
Since \(\sin \left(\alpha\right)=\dfrac{5}{13}\) and \(\alpha\) lies in the first quadrant (because we were told that \(0^\circ \lt \alpha\lt 90^\circ\)), a point \(P\left(x,y\right)\) on its terminal arm could have \(y=5\) and \(r=13\). We can determine the value of \(x.\)
\[\begin{align*} r&=\sqrt{x^2+y^2} \\ r^2&=x^2+y^2 \\ 13^2&=x^2 +5^2\\ x^2&=169-25\\ x&=\pm \sqrt{144}\\ x&= \pm 12 \end{align*}\]
Since \(\alpha\) is in the first quadrant, the \(x\)-coordinate of \(P\) is \(12\).
The sine ratio is positive in Quadrant Ⅰ and Quadrant Ⅱ, so the terminal arm of \(A\) must be in Quadrant Ⅱ, with \(\sin \left(A\right)=\dfrac{5}{13}\), \(y=5\), \(r=13\), and \(x=-12\).
Since the point \(\left(-12,5\right)\) is a reflection of the point \(\left(12,5\right)\) in the \(y\)-axis, \(A\) and \(\alpha\) are supplementary angles. That is, \( A=180^\circ - \alpha\).
Solution — Part B
First solve for \(\alpha\) using our calculator.
\[\begin{align*} \sin \left(\alpha\right)&=\dfrac{5}{13}\\ \alpha &=\sin ^{-1}\left(\dfrac{5}{13}\right)\\ \alpha &\approx 22^\circ \end{align*}\]
Next, calculate the obtuse angle \(A\) given what we learned in part a).
\[\begin{align*} A&=180^\circ- \alpha\\ &\approx 180^\circ -22^\circ\\ &\approx 158^\circ \end{align*}\]
Note: Calculators are programmed to give one answer for inverse trig calculations. We were able to determine \(\alpha\) directly, but we had to use the value of \(\alpha\) along with our knowledge of angles in standard position to determine the value for \(A\), since its terminal arm was in Quadrant II. Since \(\alpha\) is a critical part of this calculation, we call it the reference angle or the related acute angle of \(A\).
Example 2
Use the fact that \(\sin \left(\alpha\right)=\dfrac{5}{13}, 0^\circ \lt \alpha\lt 90^\circ\), to answer the following questions.
- Determine another angle, \(B\), between \(0^\circ\) and \(360^\circ\) so that \(\cos\left(B\right)=\cos\left(\alpha\right)\) and \(B \neq \alpha\). How are these two angles related?
- Determine another angle, \(C\), between \(0^\circ\) and \(360^\circ\) so that \(\tan\left(C\right)=\tan\left(\alpha\right)\) and \(C \neq \alpha\). How are these two angles related?
Solution — Part A
Since \(\sin \left(\alpha\right)=\dfrac{5}{13}\) and \(\alpha\) lies in the first quadrant, a point \(P\left(x,y\right)\) on its terminal arm could have \(y=5\), \(r=13\), and \(x=12\). Previously, we found that \( \alpha \approx 22^\circ\).
The cosine ratio is positive in Quadrant Ⅰ and Quadrant Ⅳ. Since \(B \neq \alpha\), the terminal arm of \(B\) must be in Quadrant Ⅳ and the point \(\left(12,-5\right)\) is a point on its terminal arm.
From the diagram, the relationship between \( B\) and \(\alpha\) is \( B=360^\circ- \alpha\), so \( B\approx360^\circ-22^\circ=338^\circ\).
Solution — Part B
The tangent ratio is positive in Quadrant Ⅰ and Quadrant Ⅲ and \(C \neq \alpha\) so the terminal arm of \( C\) must be in Quadrant Ⅲ and the point \(\left(-12,-5\right)\) is a point on its terminal arm.
From the diagram, the relationship between \( C\) and \(\alpha\) is \( C=180^\circ+ \alpha\), so \( C\approx180^\circ+22^\circ=202^\circ\).