Trigonometry

“Trigonometry plays a significant role in various competitive exams, especially SSC, RRB, Banking, and Defence exams. The basic concepts and formulas are essential for solving problems involving angles, heights, distances, and more.”

Type 1: Trigonometric Ratios
Example: If the angle of elevation is 30°, find the values of sin 30°, cos 30°, and tan 30°.
Solution:
sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
Type 2: Applications of Basic Trigonometric Ratios
Example: In a right triangle, if the opposite side is 4 and the hypotenuse is 5, find sin θ.
Solution:
sin θ = opposite/hypotenuse = 4/5
Type 3: Pythagorean Theorem and Trigonometry
Example: If one side of a right triangle is 3 and the hypotenuse is 5, find the third side.
Solution:
Using Pythagoras' theorem: 5² = 3² + b² ⇒ b = √16 = 4.
Type 4: Trigonometric Identities
Example: Prove that sin² θ + cos² θ = 1.
Solution:
This is a standard identity in trigonometry which holds true for any angle θ.
Type 5: Angle of Elevation and Depression
Example: A person standing on a 20 m tall building observes a car at a 30° angle of depression. How far is the car from the base of the building?
Solution:
tan 30° = 20/x ⇒ x = 20/ tan 30° = 20/√3 ≈ 11.55 meters.
Type 6: Trigonometric Ratios of Compound Angles
Example: Find the value of sin(45° + 30°).
Solution:
sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30° = (√2/2 × √3/2) + (√2/2 × 1/2) ≈ 0.9659.
Type 7: Trigonometric Ratios of Multiple Angles
Example: Find the value of cos 60° × cos 30°.
Solution:
cos 60° = 1/2, cos 30° = √3/2. Therefore, cos 60° × cos 30° = (1/2) × (√3/2) = √3/4.
Type 8: Heights and Distances
Example: A man standing 20 meters away from a tower of height 40 meters, finds the angle of elevation to be 45°. Find the height of the tower.
Solution:
tan 45° = height/distance ⇒ height = distance × tan 45° = 20 × 1 = 40 meters.
Type 9: Trigonometric Formulas for Sum and Difference of Angles
Example: Find the value of sin (60° + 45°).
Solution:
sin (60° + 45°) = sin 60° cos 45° + cos 60° sin 45° = (√3/2 × √2/2) + (1/2 × √2/2) = √2/2 + √2/4 ≈ 0.9659.
Type 10: Trigonometric Applications in Real-Life
Example: A ladder leans against a wall making a 60° angle with the ground. If the length of the ladder is 10 meters, how high is the top of the ladder from the ground?
Solution:
sin 60° = height/ladder length ⇒ height = 10 × sin 60° = 10 × √3/2 ≈ 8.66 meters.
Type 11: Multiple Angle Trigonometry
Example: Find the value of sin 2θ when θ = 30°.
Solution:
sin 2θ = 2 sin θ cos θ = 2 × sin 30° × cos 30° = 2 × 1/2 × √3/2 = √3/2 ≈ 0.866.
Type 12: Trigonometric Ratios of Complementary Angles
Example: If cos θ = 3/5, find sin (90° - θ).
Solution:
sin (90° - θ) = cos θ = 3/5.
Type 13: Trigonometric Ratios of Supplementary Angles
Example: If sin θ = 1/2, find cos (180° - θ).
Solution:
cos (180° - θ) = - cos θ, since cos θ = √3/2, cos (180° - θ) = -√3/2.
Type 14: Trigonometric Ratios of Specific Angles
Example: Find sin 30°, cos 45°, and tan 60°.
Solution:
sin 30° = 1/2, cos 45° = √2/2, tan 60° = √3.
Type 15: Trigonometric Equations
Example: Solve for θ if sin θ = 1/2 and θ is in the range 0° ≤ θ ≤ 180°.
Solution:
sin θ = 1/2 at θ = 30° and θ = 150°.
Type 16: Solve for Angle Using Trigonometric Ratio
Example: If tan θ = 1, find θ.
Solution:
tan θ = 1 at θ = 45°.
Type 17: Trigonometric Ratio of Multiple Angles
Example: Find sin 3θ if θ = 30°.
Solution:
sin 3θ = 3sin θ - 4sin³ θ = 3 × 1/2 - 4 × (1/2)³ = 3/2 - 1/2 = 1.
Type 18: Trigonometric Ratio in Terms of Other Ratios
Example: Express sin θ in terms of cos θ.
Solution:
sin² θ = 1 - cos² θ. Therefore, sin θ = √(1 - cos² θ).
Type 19: Trigonometric Functions in Quadrants
Example: In which quadrant is sin θ positive and cos θ negative?
Solution:
sin θ is positive and cos θ is negative in the second quadrant.
Type 20: Height and Distance in Real Life Scenarios
Example: A tree’s shadow is 30 meters long, and the angle of elevation of the sun is 60°. Find the height of the tree.
Solution:
tan 60° = height/30 ⇒ height = 30 × √3 ≈ 51.96 meters.

Type 21: Trigonometric Ratio of Specific Angles
Example: Find sin 90°, cos 0°, and tan 45°.
Solution:
sin 90° = 1, cos 0° = 1, tan 45° = 1.
Type 22: Reciprocal Trigonometric Ratios
Example: If cos θ = 3/5, find sec θ.
Solution:
sec θ = 1/cos θ = 1/(3/5) = 5/3.
Type 23: Trigonometric Ratios of Complementary Angles
Example: If sin θ = 1/2, find cos (90° - θ).
Solution:
cos (90° - θ) = sin θ = 1/2.
Type 24: Trigonometric Ratios of Supplementary Angles
Example: If sin θ = 1/2, find cos (180° - θ).
Solution:
cos (180° - θ) = -cos θ.
Type 25: Solving Right Angled Triangles
Example: If one side of a right-angled triangle is 4 and the hypotenuse is 5, find the third side.
Solution:
By Pythagoras theorem: 5² = 4² + b² ⇒ b = √(25 - 16) = 3.
Type 26: Trigonometric Ratios in Terms of Other Ratios
Example: Express cos θ in terms of sin θ.
Solution:
cos² θ = 1 - sin² θ ⇒ cos θ = √(1 - sin² θ).
Type 27: Finding Angle Using Inverse Trigonometric Functions
Example: Find θ if sin θ = 1/2.
Solution:
θ = sin⁻¹ (1/2) = 30°.
Type 28: Trigonometric Ratios of Angles between 0° and 90°
Example: Find tan 30°, sin 45°, and cos 60°.
Solution:
tan 30° = 1/√3, sin 45° = √2/2, cos 60° = 1/2.
Type 29: Using Trigonometric Ratios to Solve Word Problems
Example: A flagpole casts a shadow of 15 meters. The angle of elevation of the sun is 30°. Find the height of the flagpole.
Solution:
tan 30° = height/15 ⇒ height = 15 × tan 30° ≈ 15 × 1/√3 ≈ 8.66 meters.
Type 30: Trigonometric Equations Involving Multiple Angles
Example: Solve sin 2θ = 1/2 for θ in the range 0° ≤ θ ≤ 180°.
Solution:
2θ = 30° or 2θ = 150° ⇒ θ = 15° or θ = 75°.
Type 31: Trigonometric Identities and Simplification
Example: Simplify sin² θ + cos² θ.
Solution:
Using the identity sin² θ + cos² θ = 1, the result is 1.
Type 32: Solving Problems Involving Multiple Trigonometric Ratios
Example: If tan θ = 3/4, find sin θ and cos θ.
Solution:
Using tan θ = opposite/adjacent, we can form a right-angled triangle and solve for sin θ and cos θ.
Type 33: Trigonometric Ratios in Terms of Sides of a Triangle
Example: In a right triangle, the lengths of the sides are 6, 8, and 10. Find sin θ and cos θ.
Solution:
sin θ = opposite/hypotenuse = 6/10 = 3/5, cos θ = adjacent/hypotenuse = 8/10 = 4/5.
Type 34: Angle of Elevation and Depression in Word Problems
Example: A man is standing 20 meters away from a building. If the angle of elevation to the top of the building is 30°, find the height of the building.
Solution:
tan 30° = height/20 ⇒ height = 20 × tan 30° ≈ 20 × 1/√3 ≈ 11.55 meters.
Type 35: Solving for Specific Angles in Trigonometric Functions
Example: Solve for θ if cos θ = 1/2.
Solution:
θ = 60° or θ = 360° - 60° = 300°.
Type 36: Solving Equations Involving Sine and Cosine
Example: Solve cos θ = sin (90° - θ).
Solution:
This is a trigonometric identity that holds true for all angles θ.
Type 37: Trigonometric Ratios in Quadrants
Example: If sin θ = -1/2 and θ lies in the third quadrant, find cos θ.
Solution:
cos² θ = 1 - sin² θ ⇒ cos² θ = 1 - (1/2)² = 3/4 ⇒ cos θ = -√3/2 (since θ is in the third quadrant).
Type 38: Sine Rule for Solving Triangles
Example: In triangle ABC, if angle A = 30°, side a = 10, and angle B = 45°, find side b.
Solution:
Using the sine rule: (a/sin A) = (b/sin B) ⇒ b = (10 × sin 45°)/sin 30° ≈ 10 × 0.7071 / 0.5 ≈ 14.14.
Type 39: Cosine Rule for Solving Triangles
Example: In triangle ABC, if angle C = 90°, side a = 6, and side b = 8, find the length of side c.
Solution:
Using the cosine rule: c² = a² + b² ⇒ c = √(6² + 8²) = √100 = 10.
Type 40: Area of Triangle Using Trigonometry
Example: In a triangle, if sides a = 5, b = 7, and angle C = 30°, find the area of the triangle.
Solution:
Area = 1/2 × a × b × sin C = 1/2 × 5 × 7 × sin 30° = 1/2 × 5 × 7 × 1/2 = 8.75 square units.