The Trig Detective
Finding other trig ratios from a given ratio
Follow the steps below to find all six trig ratios from just one ratio and a quadrant.
Step 1 of 6
We know \(\sin\theta = \frac{3}{5}\) and \(\theta\) is in Quadrant II. You are not allowed to find \(\theta\). Sine is opposite over hypotenuse. What are the opposite side and the hypotenuse?
Enter positive whole numbers.
Step 2 of 6
Now find the missing side using the Pythagorean theorem. If \(\text{opp} = 3\) and \(\text{hyp} = 5\), what is the length of the adjacent side?
Enter a positive number (the length).
Step 3 of 6
We are in Quadrant II. Is the adjacent side \(4\) or \(-4\)?
Step 4 of 6
With opposite = 3, adjacent = -4, and hypotenuse = 5, compute all six trig ratios.
Select the correct value for each ratio (tolerance within 0.01 for decimals).
Step 5 of 6
Watch where the point lands on the unit circle. The simulation places the point at \((-4/5,\, 3/5)\). Every ratio reads directly from the triangle.
Step 6 of 6
New problem: \(\cos\theta = -\frac{5}{13}\) and \(\theta\) is in Quadrant III. Find all six ratios. This time with less scaffolding.
Adjacent = -5, Hypotenuse = 13. Use the Pythagorean theorem to find the opposite side. In Q3, the opposite (y) is also negative.
Explore: Build Any Triangle
Choose a trig ratio, enter its value, pick the quadrant, and watch the simulation build the reference triangle and compute all six ratios step by step.
Apply: Real Problems
\(\cos\theta = -\frac{5}{13}\) and \(\theta\) is in Quadrant III. Find \(\sin\theta\) and \(\tan\theta\).
What is \(\sin\theta\)?
\(\tan\theta = -2\) and \(\cos\theta\) is positive. The quadrant is not given. Determine the quadrant first from the two sign conditions, then find \(\sin\theta\).
Which quadrant is \(\theta\) in?
\(\sec\theta = -\sqrt{5}\) and \(\sin\theta\) is positive. Find the remaining five ratios and explain how you determined the quadrant.
Which quadrant is \(\theta\) in?
- Given one trigonometric ratio and a quadrant, find all six ratios without finding the angle.
- Use the Pythagorean theorem to find the missing side of a reference triangle.
- Apply quadrant sign rules to assign the correct sign to each side.
- Recognize that one ratio plus one quadrant uniquely determines a point on the unit circle.
Quick Check
If \(\sin\theta = \frac{7}{25}\) and \(\theta\) is in Quadrant I, what is \(\cos\theta\)?
Instructor Notes
Teaching Notes
This simulation targets a specific misconception: students often believe they must find the angle (using inverse trig) before computing the other ratios. The Discover sequence deliberately forbids finding the angle to force the Pythagorean-plus-signs approach. Emphasize that the angle is never needed, only the three sides with correct signs.
The "trap" in Step 3 is designed around the most common error: taking the Pythagorean result as always positive. Roughly 60-70% of students will say "4" on their first attempt. Let the surprise land before explaining.
Common Student Errors
- Forgetting to apply the quadrant sign after using the Pythagorean theorem. The theorem gives a length (always positive); the quadrant determines the sign of x and y.
- Confusing which coordinate is which: in Q II, x is negative and y is positive, not the reverse.
- Attempting to use inverse trig to find the angle first, which introduces rounding and loses exact values.
- Mixing up reciprocal ratios: csc is 1/sin, not 1/cos.
Discussion Questions
- Why does one ratio plus one quadrant pin down the point on the unit circle? Could two different points share the same sine value in the same quadrant?
- When would you actually need to find the angle, and when is the ratio approach sufficient?
- The Stretch problem gives tan and the sign of cos but no quadrant. How do sign conditions narrow the quadrant?
Exam Connection
This is a standard exam item: "Given that [ratio] = [value] and [angle] is in Q[n], find all six trig ratios." The Pythagorean identity approach (which this simulation builds) is faster and more reliable on exams than the angle-finding approach, and it preserves exact values.