Trigonometric Functions

Trigonometric Functions: Definitions, Formulas, and Applications

Introduction

📐 Trigonometric functions are the backbone of mathematics, forming the foundation of geometry, calculus, and physics. These functions help us understand the relationship between angles and the ratios of triangle sides. By studying trigonometric functions, we can solve problems involving distances, heights, and angles in real-world applications.

🔺 There are six primary trigonometric functions: sine, cosine, tangent, cosecant, secant, and cotangent. Sine (sin) is the ratio of the side opposite an angle to the hypotenuse of a right triangle. Cosine (cos) is the ratio of the adjacent side to the hypotenuse, while tangent (tan) is the ratio of the opposite side to the adjacent side.

💡 The reciprocal functions—cosecant (csc), secant (sec), and cotangent (cot)—are essential in advanced calculations in calculus and physics. Trigonometric functions also describe periodic phenomena like waves, vibrations, and circular motion, making them vital in mechanics and oscillations.

🌐 Beyond triangles, trigonometric functions are crucial in coordinate geometry. They help define points on a circle and relate linear motion to angular motion. They are also widely used to model repeating patterns such as sound waves, light waves, and other periodic phenomena.

✨ Overall, trigonometric functions provide a powerful set of tools for analyzing angles, triangles, and periodic phenomena. They link geometry, algebra, and physics together, enabling us to understand and model the world around us.

A B C

Basic Trigonometric Functions

Function Definition Formula
Sine (sin)Opposite / Hypotenusesin θ = a/c
Cosine (cos)Adjacent / Hypotenusecos θ = b/c
Tangent (tan)Opposite / Adjacenttan θ = a/b
Cosecant (csc)Hypotenuse / Oppositecsc θ = c/a
Secant (sec)Hypotenuse / Adjacentsec θ = c/b
Cotangent (cot)Adjacent / Oppositecot θ = b/a

Explore the six fundamental trigonometric functions — sine, cosine, tangent, cotangent, secant, and cosecant — through 10 interactive Chart.js diagrams with dynamic visual explanations.

1. Unit Circle : Relationship of Sin and Cos

The unit circle represents all angles θ where sin(θ) and cos(θ) correspond to the y and x coordinates respectively.

2. Sine Function of Trigonometry (sin θ)

3. Cosine Function (cos θ)

4. Tangent Function (tan θ)

5. Secant Function (sec θ)

6. Cosecant Function (csc θ)

7. Cotangent Function (cot θ)

8. Identity: sin²θ + cos²θ = 1

9. Combined sin, cos, tan Comparison

10. Arcsin (Inverse of Sine)

Important Identities & Formulas

  • sin²θ + cos²θ = 1
  • 1 + tan²θ = sec²θ
  • 1 + cot²θ = csc²θ

To understand identities and formulas, check the Trigonometric identities and their formulas section below.

📐 Trigonometric Identities Interactive Learning

Trigonometry is a branch of mathematics that studies the relationship between angles and sides of triangles. It is fundamental in physics, engineering, and mathematics. To make learning trigonometric identities interactive and intuitive, we provide visual diagrams for each key identity along with step-by-step explanations. You can navigate through each identity using the Previous and Next buttons below.

Derivatives of Trigonometric Functions

Function Derivative
sin xcos x
cos x-sin x
tan xsec² x
cot x-csc² x
sec xsec x · tan x
csc x-csc x · cot x

1. Understanding Trigonometric Derivatives

The derivative of a function measures how the function value changes as its input changes. In trigonometry, derivatives are used to study the rate of change of angles and oscillatory behavior, which is crucial in physics, engineering, and applied mathematics.

2. Derivatives of Basic Trigonometric Functions

  • sin x → cos x: The rate of change of the sine function is the cosine function. This means as sine increases or decreases, its slope at any point x is equal to cos x.
  • cos x → -sin x: The derivative of cosine is negative sine, indicating that the slope of the cosine curve decreases where sine increases.
  • tan x → sec² x: Tangent’s derivative is sec² x. This shows that the slope of tan x grows very quickly near π/2 + nπ, where n is an integer, because sec x → ∞ at these points.
  • cot x → -csc² x: Cotangent decreases as its input increases, reflected in the negative sign.
  • sec x → sec x · tan x: The derivative of secant is a product of secant and tangent, indicating a steep increase where tan x is large.
  • csc x → -csc x · cot x: Cosecant decreases as input increases, and the negative sign shows this decreasing nature.

3. How to Derive These Formulas

The derivatives of trigonometric functions can be derived using limits, chain rule, and reciprocal identities:

  • sin x: Using the limit definition of derivative, lim (h→0) [sin(x+h)-sin(x)]/h = cos x
  • cos x: Similarly, lim (h→0) [cos(x+h)-cos(x)]/h = -sin x
  • tan x: Write tan x as sin x / cos x and apply the quotient rule: d/dx [tan x] = (cos² x + sin² x)/cos² x = sec² x
  • cot x, sec x, csc x: Use reciprocal or quotient rule along with basic derivatives above.

4. Applications in Real Life

Trigonometric derivatives are applied in various fields:

  1. Physics: Study of oscillatory motion, like pendulums, springs, and wave motion.
  2. Engineering: Signal processing, alternating current circuits, and mechanical vibrations.
  3. Mathematics: Solving differential equations, curve sketching, and optimization problems.
  4. Computer Graphics: Rotations and animations often rely on sine and cosine derivatives.

5. Related Study Topics

After understanding the basic derivatives, students should explore:

  • Higher-order derivatives (second derivative, third derivative, etc.) of trigonometric functions.
  • Derivatives of inverse trigonometric functions: arcsin, arccos, arctan, etc.
  • Derivatives using the chain rule for composite functions like sin(2x), cos(3x), etc.
  • Implicit differentiation involving trigonometric functions.
  • Integration of trigonometric functions (inverse process of differentiation).

6. Visualization Tips

To fully grasp derivatives of trigonometric functions, it’s useful to visualize:

  • Graph sin x and cos x together to see how the slope of sine corresponds to cosine.
  • Plot tan x and sec² x to understand rapid slope changes near vertical asymptotes.
  • Use software like GeoGebra or Desmos to dynamically change x and see the derivative graph in real-time.

7. Practice Problems

1. Find the derivative of y = sin(3x)
2. Compute d/dx [sec²(2x)]
3. Determine the slope of y = cot(x) at x = π/4
4. If y = cos(5x) + tan(x), find dy/dx

Regular practice of these derivatives enhances problem-solving skills and prepares students for higher-level calculus.

Conclusion

Understanding the derivatives of trigonometric functions is essential for analyzing periodic behavior in mathematics and real-world applications. Mastery of these concepts, along with their applications in physics and engineering, forms a strong foundation for advanced calculus, differential equations, and mathematical modeling.

Inverse Trigonometric Functions

Inverse trigonometric functions, denoted as sin⁻¹(x), cos⁻¹(x), tan⁻¹(x), cot⁻¹(x), sec⁻¹(x), and csc⁻¹(x), are used to determine the angle when a trigonometric ratio is known. These functions are crucial in solving equations where the unknown is an angle rather than a ratio. Understanding their derivatives also builds on your knowledge of trigonometric derivatives.

Derivatives of Inverse Trigonometric Functions

Function Derivative
y = sin⁻¹ xdy/dx = 1 / √(1 – x²)
y = cos⁻¹ xdy/dx = -1 / √(1 – x²)
y = tan⁻¹ xdy/dx = 1 / (1 + x²)
y = cot⁻¹ xdy/dx = -1 / (1 + x²)
y = sec⁻¹ xdy/dx = 1 / (|x|√(x² – 1))
y = csc⁻¹ xdy/dx = -1 / (|x|√(x² – 1))

1. Understanding Inverse Functions

The inverse trigonometric function answers the question: “For a given trigonometric ratio, what is the angle?” For example, if sin θ = 1/2, then θ = sin⁻¹(1/2) = π/6. These functions are restricted to principal values to make them single-valued.

2. Domain and Range

  • sin⁻¹ x: Domain: [-1, 1], Range: [-π/2, π/2]
  • cos⁻¹ x: Domain: [-1, 1], Range: [0, π]
  • tan⁻¹ x: Domain: (-∞, ∞), Range: (-π/2, π/2)
  • cot⁻¹ x: Domain: (-∞, ∞), Range: (0, π)
  • sec⁻¹ x: Domain: (-∞, -1] ∪ [1, ∞), Range: [0, π], y ≠ π/2
  • csc⁻¹ x: Domain: (-∞, -1] ∪ [1, ∞), Range: [-π/2, π/2], y ≠ 0

3. Relationship with Trigonometric Derivatives

The derivatives of inverse trigonometric functions are closely related to trigonometric derivatives. For example, consider y = sin⁻¹ x:
If x = sin y, then differentiating both sides gives dx/dy = cos y. Using cos y = √(1 – sin² y), we get:
dy/dx = 1 / √(1 - x²).
This method can be applied to all inverse functions using implicit differentiation.

4. Applications

  • Solving triangles in geometry and trigonometry.
  • Engineering applications: calculating angles in circuits, structures, and mechanical linkages.
  • Physics: determining angular positions from known ratios in oscillatory motion.
  • Calculus: evaluating integrals of the form ∫ dx / √(1-x²), ∫ dx / (1+x²), etc.

5. Advanced Study Topics

  • Composition with other functions: sin⁻¹(f(x)), tan⁻¹(g(x)) and using chain rule.
  • Integration techniques involving inverse trigonometric functions.
  • Graphing inverse trigonometric functions and understanding symmetry.
  • Connection to hyperbolic functions: e.g., tanh⁻¹ x, coth⁻¹ x.
  • Implicit differentiation in more complex trigonometric equations.

6. Practice Problems

1. Find d/dx [sin⁻¹(2x)]
2. Compute d/dx [tan⁻¹(x²)]
3. Determine d/dx [sec⁻¹(3x)]
4. Solve for θ: cos⁻¹(θ/2) = π/3

Practicing these problems will strengthen your understanding of inverse functions and their derivatives.

Conclusion

Mastery of inverse trigonometric functions enhances your ability to solve advanced problems in calculus, physics, and engineering. They build on your knowledge of trigonometric derivatives and open doors to higher-level mathematics involving integrals, differential equations, and analytic geometry.

20 Key Concepts of Trigonometry

  1. Pythagorean Identity
  2. Graphs of sin, cos, tan
  3. Co-function Identities
  4. Angle Sum & Difference
  5. Double Angle Formulas
  6. Half Angle Formulas
  7. Product-to-Sum Identities
  8. Sum-to-Product Identities
  9. Law of Sines
  10. Law of Cosines
  11. Trigonometric Ratios in Special Angles
  12. Polar Coordinates
  13. Amplitude & Period of Functions
  14. Inverse Trig Ranges
  15. Hyperbolic Functions
  16. Euler’s Formula
  17. Trigonometry in Calculus
  18. Applications in Physics
  19. Applications in Astronomy
  20. Applications in Engineering

Applications of Trigonometry

Trigonometry is used in architecture, navigation, sound engineering, astronomy, and many real-world calculations.

Frequently Asked: Trigonometric Functions — Step-by-step Solutions

35 curated Q&A items with step-by-step solutions. Click any question to expand the detailed solution.

1 What are the six trigonometric functions?
▾
  1. Start with a right triangle: label Opposite (opp), Adjacent (adj), Hypotenuse (hyp).
  2. Define sine: sinθ = opp / hyp.
  3. Define cosine: cosθ = adj / hyp.
  4. Define tangent: tanθ = opp / adj.
  5. Reciprocals: cscθ = 1/sinθ = hyp/opp and secθ = 1/cosθ = hyp/adj.
  6. Reciprocal of tangent: cotθ = 1/tanθ = adj/opp.
  7. Remember identities: e.g., sin²θ + cos²θ = 1.
2 If sin θ = 3/5, find cos θ.
▾
  1. Given sinθ = opp/hyp = 3/5, set opp = 3 and hyp = 5.
  2. Use Pythagoras: adj² = hyp² − opp² = 25 − 9 = 16.
  3. Take square root: adj = 4.
  4. Compute cosine: cosθ = adj/hyp = 4/5.
  5. Consider quadrant: for acute θ, cos is positive.
  6. Verify: (3/5)² + (4/5)² = 9/25 + 16/25 = 1.
  7. Final: cosθ = 4/5 (assuming θ acute).
3 If sin²θ + cos²θ = 1, find cosθ when sinθ = 3/5.
▾
  1. Start with identity: sin²θ + cos²θ = 1.
  2. Substitute sinθ = 3/5 ⇒ sin²θ = 9/25.
  3. Thus cos²θ = 1 − 9/25 = 16/25.
  4. Take square root: cosθ = ±4/5.
  5. Choose sign by quadrant (positive if θ is acute).
  6. For principal acute θ: cosθ = 4/5.
  7. Answer: cosθ = ±4/5 (usually +4/5).
4 If cosθ = 12/13, find sinθ.
▾
  1. Use identity: sin²θ = 1 − cos²θ.
  2. cos²θ = (12/13)² = 144/169.
  3. So sin²θ = 1 − 144/169 = 25/169.
  4. Take square root: sinθ = ±5/13.
  5. Sign depends on quadrant: positive in QI, negative in QIII.
  6. If acute, sinθ = 5/13.
  7. Final: sinθ = ±5/13.
5 Find tanθ if sinθ = 8/17 and cosθ = 15/17.
▾
  1. Use tanθ = sinθ / cosθ.
  2. Substitute: (8/17) / (15/17).
  3. Cancel 17 factors: tanθ = 8/15.
  4. Reduce if necessary (already reduced).
  5. Check signs: both positive ⇒ tan positive.
  6. Numeric approx ≈ 0.533.
  7. Final: tanθ = 8/15.
6 If tanθ = 3/4, find sinθ and cosθ.
▾
  1. Let tanθ = opp/adj = 3/4.
  2. Choose opp = 3, adj = 4 (Pythagorean triple).
  3. Compute hypotenuse: √(3²+4²) = 5.
  4. So sinθ = opp/hyp = 3/5.
  5. And cosθ = adj/hyp = 4/5.
  6. For principal acute θ both positive.
  7. Answer: sinθ = 3/5, cosθ = 4/5.
7 If sinθ = 5/13, find secθ.
▾
  1. Given sinθ = 5/13, opp = 5, hyp = 13.
  2. Compute adj: √(13² − 5²) = √144 = 12.
  3. So cosθ = 12/13.
  4. Then secθ = 1/cosθ = 13/12.
  5. Sign of sec follows sign of cos.
  6. If acute, secθ = +13/12.
  7. Final: secθ = ±13/12 (usually +13/12).
8 Find cscθ if tanθ = 5/12.
▾
  1. Let tan = opp/adj = 5/12 so opp = 5, adj = 12.
  2. Compute hypotenuse: √(5²+12²) = 13.
  3. So sinθ = opp/hyp = 5/13.
  4. Therefore cscθ = 1/sinθ = 13/5.
  5. Sign depends on quadrant.
  6. For principal acute θ, cscθ = 13/5.
  7. Answer: cscθ = ±13/5 (usually +13/5).
9 If cotθ = 7/24, find sinθ and cosθ.
▾
  1. cot = adj/opp = 7/24 ⇒ tan = 24/7, set opp=24, adj=7.
  2. Hypotenuse: √(24²+7²) = 25.
  3. So sinθ = 24/25.
  4. And cosθ = 7/25.
  5. Signs depend on quadrant where cot is positive or negative.
  6. For QI both positive.
  7. Final: sinθ = 24/25, cosθ = 7/25.
10 Find tan²θ + 1 if tanθ = 3/4.
▾
  1. Compute tan²θ = (3/4)² = 9/16.
  2. Add 1: 9/16 + 1 = 9/16 + 16/16.
  3. Sum = 25/16.
  4. Use identity: tan²θ + 1 = sec²θ.
  5. So sec²θ = 25/16.
  6. Therefore secθ = ±5/4 (sign by quadrant).
  7. Final: tan²θ + 1 = 25/16.
11 Find 1 + cot²θ if cotθ = 5/12.
▾
  1. Compute cot²θ = (5/12)² = 25/144.
  2. Add 1: 1 + 25/144 = 144/144 + 25/144.
  3. Sum = 169/144.
  4. Identity: 1 + cot²θ = csc²θ.
  5. So csc²θ = 169/144.
  6. Thus cscθ = ±13/12 (sign by quadrant).
  7. Final: 1 + cot²θ = 169/144.
12 If sinθ = 4/5, find tanθ.
▾
  1. Given sinθ = opp/hyp = 4/5, so opp=4, hyp=5.
  2. Find adj: √(5² − 4²) = 3.
  3. Compute cosθ = 3/5.
  4. Then tanθ = sinθ / cosθ = (4/5)/(3/5).
  5. Cancel 5: tanθ = 4/3.
  6. Sign by quadrant; for QI positive.
  7. Final: tanθ = 4/3.
13 If cosθ = 3/5, find cotθ.
▾
  1. cosθ = adj/hyp = 3/5 ⇒ adj=3, hyp=5.
  2. Compute opp: √(5² − 3²) = 4.
  3. So sinθ = 4/5.
  4. cotθ = cosθ / sinθ = (3/5)/(4/5).
  5. Cancel 5: cotθ = 3/4.
  6. Sign depends on quadrant.
  7. Final: cotθ = 3/4.
14 If sinA = 0.6, find cosA.
▾
  1. sinA = 0.6 ⇒ sin²A = 0.36.
  2. cos²A = 1 − 0.36 = 0.64.
  3. cosA = ±√0.64 = ±0.8.
  4. Choose sign by quadrant (positive if acute).
  5. If A acute, cosA = 0.8.
  6. Check: 0.6² + 0.8² = 0.36 + 0.64 = 1.
  7. Final: cosA = ±0.8 (usually 0.8).
15 Find sin²30° + cos²30°.
▾
  1. sin30° = 1/2 ⇒ sin²30° = 1/4.
  2. cos30° = √3/2 ⇒ cos²30° = 3/4.
  3. Sum = 1/4 + 3/4 = 4/4.
  4. 4/4 = 1.
  5. This confirms the identity sin² + cos² = 1.
  6. Valid for any angle; here for 30° it’s exact.
  7. Final: 1.
16 Find tan45° + cot45°.
▾
  1. tan45° = 1.
  2. cot45° = 1 (reciprocal of 1).
  3. Sum = 1 + 1.
  4. Sum = 2.
  5. No quadrant ambiguity for standard 45°.
  6. Useful check for symmetry of tan/cot.
  7. Final: 2.
17 Find sec60° − csc30°.
▾
  1. cos60° = 1/2 ⇒ sec60° = 1/(1/2) = 2.
  2. sin30° = 1/2 ⇒ csc30° = 1/(1/2) = 2.
  3. Difference = 2 − 2.
  4. Difference = 0.
  5. Both are exact values for those angles.
  6. No sign ambiguity here.
  7. Final: 0.
18 Find the value of sin²60° + cos²60°.
▾
  1. sin60° = √3/2 ⇒ sin²60° = 3/4.
  2. cos60° = 1/2 ⇒ cos²60° = 1/4.
  3. Sum = 3/4 + 1/4 = 4/4.
  4. 4/4 = 1.
  5. This verifies the identity sin² + cos² = 1.
  6. True for any θ, here illustrated at 60°.
  7. Final: 1.
19 If tanθ = 1, what is θ?
▾
  1. tanθ = 1 when opposite = adjacent.
  2. Principal solution: θ = 45° (in [0°,360°) ).
  3. Tangent period = 180°, so add 180° increments.
  4. General solution: θ = 45° + n·180°, n ∈ ℤ.
  5. Examples: 45°, 225°, …
  6. In radians: θ = π/4 + nπ.
  7. Final: θ = 45° + k·180°.
20 Find cos(90° − θ) when sinθ = 4/5.
▾
  1. Use co-function identity: cos(90° − θ) = sinθ.
  2. Given sinθ = 4/5.
  3. Therefore cos(90° − θ) = 4/5.
  4. No Pythagorean calculation required.
  5. Identity holds in degrees (same in radians).
  6. Sign follows sinθ.
  7. Final: 4/5.
21 Find sin(90° − θ) when cosθ = 12/13.
▾
  1. Use co-function identity: sin(90° − θ) = cosθ.
  2. Given cosθ = 12/13.
  3. Therefore sin(90° − θ) = 12/13.
  4. Direct substitution — no Pythagoras needed.
  5. Sign follows cosθ’s sign.
  6. Valid in degrees/radians analogs.
  7. Final: 12/13.
22 If sinθ = 3/5, find cscθ.
▾
  1. Definition: cscθ = 1/sinθ.
  2. Given sinθ = 3/5.
  3. Compute cscθ = 1 / (3/5) = 5/3.
  4. No Pythagorean step needed.
  5. Sign depends on sign of sinθ.
  6. For positive sin, csc positive.
  7. Final: cscθ = 5/3.
23 Find sin²45° + cos²45°.
▾
  1. sin45° = cos45° = 1/√2.
  2. Square each: (1/√2)² = 1/2.
  3. Sum = 1/2 + 1/2.
  4. Result = 1.
  5. This verifies the Pythagorean identity for 45°.
  6. Works for any angle as an identity.
  7. Final: 1.
24 Find sin²θ + cos²θ if sinθ = 0.8.
▾
  1. sinθ = 0.8 ⇒ sin²θ = 0.64.
  2. Identity: sin²θ + cos²θ = 1.
  3. So cos²θ = 1 − 0.64 = 0.36.
  4. Then cosθ = ±0.6 depending on quadrant.
  5. Sum = 0.64 + 0.36 = 1.
  6. Identity holds for all θ.
  7. Final: 1.
25 If tanθ = 5/12, find secθ.
▾
  1. Use identity: sec²θ = 1 + tan²θ.
  2. tan²θ = (5/12)² = 25/144.
  3. 1 + tan²θ = 1 + 25/144 = 169/144.
  4. sec²θ = 169/144 ⇒ secθ = ±13/12.
  5. Sign follows cosθ (sec positive if cos positive).
  6. Magnitude is 13/12.
  7. Final: secθ = ±13/12.
26 Find cosθ if tanθ = 4/3.
▾
  1. tan = opp/adj = 4/3 → choose opp=4, adj=3.
  2. Hypotenuse = √(4²+3²) = 5.
  3. cosθ = adj/hyp = 3/5.
  4. Alternatively use sec² = 1 + tan² = 25/9 → sec = 5/3 → cos = 3/5.
  5. Sign depends on quadrant.
  6. For acute θ, cosθ = 3/5.
  7. Final: 3/5.
27 Find sin30° and cos30°.
▾
  1. Standard angle values: sin30° = 1/2.
  2. cos30° = √3/2.
  3. tan30° = sin30°/cos30° = 1/√3.
  4. sin²30° = 1/4, cos²30° = 3/4.
  5. Sum = 1/4 + 3/4 = 1.
  6. Useful for many trigonometric simplifications.
  7. Final: sin30° = 1/2, cos30° = √3/2.
28 If sinθ = -3/5 and θ is in QIII, find cosθ.
▾
  1. In QIII, both sin and cos are negative.
  2. sinθ = -3/5 implies opp = -3, hyp = 5.
  3. Compute adj magnitude: √(25 − 9) = 4.
  4. cosθ = adj/hyp = 4/5 but negative in QIII → -4/5.
  5. tanθ = sin/cos = (-3/5)/(-4/5) = 3/4 (positive in QIII).
  6. cscθ = -5/3.
  7. Final: cosθ = -4/5.
29 Prove sin²θ + cos²θ = 1.
▾
  1. Start with sin = opp/hyp and cos = adj/hyp (right triangle).
  2. Square both: sin² = opp²/hyp², cos² = adj²/hyp².
  3. Add: (opp² + adj²)/hyp².
  4. By Pythagoras, opp² + adj² = hyp².
  5. Thus sum = hyp²/hyp² = 1.
  6. Equivalent proof: unit circle x² + y² = 1 with x=cos, y=sin.
  7. Conclusion: sin²θ + cos²θ = 1.
30 If secθ = 5/3, find cosθ and tanθ (θ acute).
▾
  1. secθ = 5/3 ⇒ cosθ = 1/secθ = 3/5.
  2. Set adj=3, hyp=5.
  3. opp = √(5² − 3²) = 4.
  4. sinθ = opp/hyp = 4/5.
  5. tanθ = sinθ/cosθ = (4/5)/(3/5) = 4/3.
  6. All values positive for acute θ.
  7. Final: cosθ = 3/5, tanθ = 4/3.
31 If cosθ = -4/5 and θ in QII, find sinθ.
▾
  1. In QII cos is negative, sin is positive.
  2. Given cosθ = -4/5 ⇒ adj = -4, hyp = 5.
  3. Compute opp magnitude: √(25 − 16) = 3.
  4. sinθ = opp/hyp = 3/5 (positive in QII).
  5. tanθ = sin/cos = (3/5)/(-4/5) = -3/4.
  6. cscθ = 5/3.
  7. Final: sinθ = 3/5.
32 Find value of tan30°.
▾
  1. Use sin30° = 1/2 and cos30° = √3/2.
  2. tan30° = (1/2) / (√3/2).
  3. Cancel 1/2: tan30° = 1/√3.
  4. Rationalize: tan30° = √3/3.
  5. Numeric approx ≈ 0.577.
  6. Exact form preferred: √3/3.
  7. Final: tan30° = 1/√3 = √3/3.
33 If sin(2θ) = 1, find θ in [0°,360°).
▾
  1. sin(2θ) = 1 when 2θ = 90° + 360°·n.
  2. So 2θ = 90° + 360°·n ⇒ θ = 45° + 180°·n.
  3. Check values in [0°,360°): n=0 → θ=45°.
  4. n=1 → θ=225°.
  5. n=2 gives θ>360°, outside range.
  6. Hence solutions: 45°, 225° in the interval.
  7. Final: 45° and 225°.
34 Express tanθ in terms of sinθ and cosθ.
▾
  1. Start with sinθ = opp/hyp and cosθ = adj/hyp.
  2. Compute sinθ / cosθ = (opp/hyp) / (adj/hyp).
  3. Hypotenuse cancels: = opp/adj.
  4. opp/adj is the definition of tanθ.
  5. Therefore tanθ = sinθ / cosθ.
  6. Note undefined when cosθ = 0.
  7. Final: tanθ = sinθ/cosθ.
35 If cosA = 0.8 and A acute, find sinA and tanA.
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  1. cosA = 0.8 ⇒ cos² = 0.64.
  2. sin² = 1 − 0.64 = 0.36.
  3. sinA = √0.36 = 0.6 (positive since acute).
  4. tanA = sinA / cosA = 0.6 / 0.8.
  5. Simplify: tanA = 3/4 (0.75).
  6. Triangle interpretation: scaled 3-4-5 triangle.
  7. Final: sinA = 0.6, tanA = 3/4.
   
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