Inverse Trigonometric

Grade 11 · algebra · 101 practice problems · read aloud

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Inverse Trigonometric Functions

What Are They? 🤔

Inverse trigonometric functions (like sin⁻¹, cos⁻¹, tan⁻¹) reverse the work of sine, cosine, and tangent. If sin(θ) = 0.5, then sin⁻¹(0.5) = θ. They are crucial for finding an angle when you know the ratio of the sides of a right triangle.

How to Solve Problems

  1. Identify the Ratio: Determine which trigonometric ratio (sin, cos, tan) you are given.
  2. Apply the Inverse: Use the corresponding inverse function (sin⁻¹, cos⁻¹, tan⁻¹) on your calculator.
  3. Find the Angle: The calculator's output is the principal angle value. Remember the range restrictions.
  4. Consider Other Angles: Depending on the problem, there may be more than one correct angle in a 0° to 360° range.

Worked Examples

Example 1: Basic Angle Finding

Find θ if cos(θ) = -√3/2, for 0° ≤ θ ≤ 180°.

  1. We are given a cosine value: cos(θ) = -√3/2.
  2. Apply the inverse: θ = cos⁻¹(-√3/2).
  3. The calculator gives 150°. This is within our range.
  4. Answer: θ = 150°.

Example 2: Composition

Find the exact value of sin(tan⁻¹(1)).

  1. First, find the angle: tan⁻¹(1) = 45° (or π/4 radians).
  2. Now, find the sine of that angle: sin(45°) = √2/2.
  3. Answer: √2/2.

Common Mistakes to Avoid 🚫

Forgetting Range Restrictions: sin⁻¹(x) outputs angles between -90° and 90°. cos⁻¹(x) outputs angles between 0° and 180°. tan⁻¹(x) outputs angles between -90° and 90°. Your answer must be in this range.

Confusing with Reciprocals: sin⁻¹(x) is NOT the same as 1/sin(x). The "-1" means the inverse function, not the reciprocal.

Angle Mode: Always check if your calculator is in DEGREE or RADIAN mode based on the problem.

Tips & Tricks

Memory Aid: Remember the ranges by thinking of the unit circle. cos⁻¹ gives angles from the top half, sin⁻¹ from the right half, and tan⁻¹ is "trapped" between the two vertical asymptotes.

Strategy: For problems like Example 2, drawing a right triangle can help you visualize the relationship between the sides and find the exact value.

How to Practice

  • Start with basic "find the angle" problems for all three functions.
  • Practice composition problems, like cos(sin⁻¹(x)).
  • Solve real-world problems involving angles of elevation/depression where you have to find the angle using an inverse function.
  • Always check your answers by plugging the angle back into the original trigonometric function.

Practice problems

6 of the 101, worked through step by step — try them before opening the answer.

1 sin⁻¹(1/2) = ?

Hint: Consider the unit circle and the angles where the sine function produces common values. Think about which angle in the principal range gives this sine value.

Show the answer

Answer: π/6

  1. Understand the meaning of sin⁻¹(x). The expression sin⁻¹(1/2) means: "Find the angle θ such that sin(θ) = 1/2, and θ is in the range [-π/2, π/2] for the principal branch of inverse sine."
  2. Recall common sine values from the unit circle. We know that: sin(π/6) = 1/2 sin(5π/6) = 1/2
  3. Check which of these angles lies in the correct range for sin⁻¹. The range for sin⁻¹ is [-π/2, π/2] (or [-90°, 90°] in degrees). - π/6 is about 0.523, which is in [-π/2, π/2]. - 5π/6 is about 2.618, which is not in [-π/2, π/2]. So only π/6 is valid for the principal value.
  4. Conclude the answer. Since sin(π/6) = 1/2 and π/6 is within the principal range, we have: sin⁻¹(1/2) = π/6. Final answer: π/6

We are asked to find sin⁻¹(1/2).

2 arcsin(1/2) = ?

Hint: Consider the unit circle and which standard angle has a sine value that matches the given input. Remember that inverse trigonometric functions return angle measures.

Show the answer

Answer: π/6

  1. Understand what arcsin means arcsin(1/2) means "find the angle whose sine is 1/2"
  2. Recall the range of arcsin The arcsin function returns angles between -pi/2 and pi/2 (or -90° to 90°)
  3. Think about the unit circle On the unit circle, sin(theta) = y-coordinate We need an angle where the y-coordinate is 1/2
  4. Recall common sine values From special triangles, we know: sin(30°) = 1/2 sin(pi/6) = 1/2
  5. Verify this is in the correct range pi/6 radians = 30°, which is between -pi/2 and pi/2 This is in the first quadrant, so it's within the valid range for arcsin
  6. Check if there are other angles with sin = 1/2 While 150° (5pi/6) also has sin = 1/2, it's not in the range of arcsin The arcsin function only returns the principal value between -pi/2 and pi/2
  7. Conclusion Therefore, arcsin(1/2) = pi/6 Final answer: pi/6

Step-by-step solution:

3 sin(2arcsin(1/2)) = ?

Hint: Use the double-angle identity for sine and consider the range of the inverse sine function.

Show the answer

Answer: √3/2

  1. Understand arcsin(1/2)** arcsin(1/2) means "the angle whose sine is 1/2". We know sin(pi/6) = 1/2, and arcsin returns values in the range [-pi/2, pi/2]. So arcsin(1/2) = pi/6. Thus: 2 * arcsin(1/2) = 2 * (pi/6) = pi/3. --- **
  2. Rewrite the problem** sin(2 * arcsin(1/2)) = sin(pi/3). --- **
  3. Evaluate sin(pi/3)** From basic trigonometry: sin(pi/3) = sqrt(3)/2. --- **
  4. Final answer** sin(2 * arcsin(1/2)) = sqrt(3)/2. --- ANSWER: sqrt(3)/2

Let's solve step by step. We are given: sin(2 * arcsin(1/2)) --- **

4 sin(2arcsin(3/5)) = ?

Hint: Use the double-angle identity for sine and consider the relationship between trigonometric functions and their inverses.

Show the answer

Answer: 24/25

Let θ = arcsin(3/5). That means sin(θ) = 3/5, and θ is in the range [-π/2, π/2] (where cosine is nonnegative). We want sin(2θ). Using the double-angle identity: sin(2θ) = 2 sin(θ) cos(θ) We know sin(θ) = 3/5. We need cos(θ). Using the Pythagorean identity: sin²(θ) + cos²(θ) = 1 (3/5)² + cos²(θ) = 1 9/25 + cos²(θ) = 1 cos²(θ) = 1 - 9/25 cos²(θ) = 25/25 - 9/25 cos²(θ) = 16/25 Since θ is in [-π/2, π/2], cos(θ) ≥ 0, so: cos(θ) = 4/5 Now substitute into the double-angle formula: sin(2θ) = 2 * (3/5) * (4/5) sin(2θ) = (2 * 3 * 4) / (5 * 5) sin(2θ) = 24/25 Therefore, sin(2arcsin(3/5)) = 24/25.

5 sin(2arcsin(4/5)) = ?

Hint: Consider the double-angle identity for sine and the relationship between sine and arcsine.

Show the answer

Answer: 24/25

  1. Let θ = arcsin(4/5). This means sin(θ) = 4/5.
  2. We need to find sin(2θ). Use the double-angle identity: sin(2θ) = 2 sin(θ) cos(θ).
  3. We know sin(θ) = 4/5. Find cos(θ) using the Pythagorean identity: cos²(θ) = 1 - sin²(θ) = 1 - (4/5)² = 1 - 16/25 = 9/25.
  4. Since θ = arcsin(4/5) and the range of arcsin is [-π/2, π/2], cos(θ) is positive. So cos(θ) = √(9/25) = 3/5.
  5. Substitute into the double-angle formula: sin(2θ) = 2 × (4/5) × (3/5) = 24/25.

The answer is 24/25.

6 sin(2arctan(3/4)) = ?

Hint: Consider using a right triangle to represent the arctangent relationship and then apply the double-angle identity for sine.

Show the answer

Answer: 24/25

  1. Let θ = arctan(3/4). This means tan(θ) = 3/4.
  2. Construct a right triangle where the opposite side is 3 and the adjacent side is 4. The hypotenuse is sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.
  3. From the triangle, sin(θ) = opposite/hypotenuse = 3/5 and cos(θ) = adjacent/hypotenuse = 4/5.
  4. Use the double-angle identity: sin(2θ) = 2 sin(θ) cos(θ).
  5. Substitute the values: sin(2θ) = 2 * (3/5) * (4/5) = 2 * (12/25) = 24/25.

The answer is 24/25.

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