Elimination of Trigonometric Ratios

Here we will learn about the elimination of trigonometric ratios with the help of different types of problems.

In order to eliminate the T-ratios from the given relations, we make use of the fundamental trigonometrical identities, in the following examples.

Worked-out examples on elimination of trigonometric ratios:

1. If sin θ + sin² θ = 1, prove that cos² θ + cos⁴ θ = 1

Solution:

sin θ + sin² θ = 1

⇒ sin θ = 1 - sin² θ, [subtract sin² θ from both the sides]

⇒ sin θ = cos² θ, [since, 1 – sin² θ = cos² θ]



⇒ sin² θ = cos⁴ θ, [squaring both the sides]

⇒ 1 - cos² θ = cos⁴ θ, [since sin² θ = 1 – cos² θ]

⇒ 1 = cos⁴ θ + cos² θ, [adding cos² θ on both the sides]

⇒ cos⁴ θ + cos² θ = 1

Therefore, cos² θ + cos⁴ θ = 1


2. If (cos θ + sin θ) = √2 cos θ, shown that (cos θ - sin θ) = √2 sin θ

Solution:

(cos θ + sin θ) = √2 cos θ ………… (A)

⇒ (cos θ + sin θ) ² = 2 cos² θ, [squaring both the sides]

⇒ cos² θ + sin² θ + 2 sin θ cos θ = 2 cos² θ

⇒ 2 sin θ cos θ = 2 cos² θ - cos² θ - sin² θ

⇒ 2 sin θ cos θ = cos² θ - sin² θ

⇒ cos² θ - sin² θ = 2 sin θ cos θ

⇒ (cos θ + sin θ) (cos θ - sin θ) = 2 sin θ cos θ

⇒ (√2 cos θ) (cos θ - sin θ) = 2 sin θ cos θ ………… using (A)

⇒ (cos θ - sin θ) = (2 sin θ cos θ)/(√2 cos θ)

⇒ (cos θ - sin θ) = √2 sin θ

Therefore, (cos θ - sin θ) = √2 sin θ


3. If 3 sin θ + 5 cos θ = 5, prove that (5 sin θ - 3 cos θ) = ± 3.

Solution:

(3 sin θ + 5 cos θ)² + (5 sin θ - 3 cos θ)²

                                = (9 sin² θ + 25 cos² θ + 30 sin θ cos θ) + (25 sin² θ                                   + 9 cos² θ - 30 sin θ cos θ)

                               = 34 sin² θ + 34 cos² θ

                               = 34 (sin² θ + cos² θ)

                               = 34 (1)

                               = 34

⇒ (3 sin θ + 5 cos θ)² + (5 sin θ - 3 cos θ)² = 34

⇒ (5)² + (5 sin θ - 3 cos θ)² = 34, [since, (3 sin θ + 5 cos θ) = 5]

⇒ 25 + (5 sin θ - 3 cos θ)² = 34

⇒ (5 sin θ - 3 cos θ)² = 9 [subtract 25 from both the sides]

⇒ (5 sin θ - 3 cos θ) = ± 3

Therefore, (5 sin θ - 3 cos θ) = ± 3.

The above problems on elimination of trigonometric ratios are explained step-by-step so, that students get the clear concept how to make use of the fundamental trigonometrical identities.

● Trigonometric Functions

• Basic Trigonometric Ratios and Their Names
• Restrictions of Trigonometrical Ratios
• Reciprocal Relations of Trigonometric Ratios
• Quotient Relations of Trigonometric Ratios
  
• Limit of Trigonometric Ratios
  
• Trigonometrical Identity
• Problems on Trigonometric Identities
• Elimination of Trigonometric Ratios
• Eliminate Theta between the equations
• Problems on Eliminate Theta
• Trig Ratio Problems
• Proving Trigonometric Ratios
• Trig Ratios Proving Problems
• Verify Trigonometric Identities
• Trigonometrical Ratios of 0°
• Trigonometrical Ratios of 30°
• Trigonometrical Ratios of 45°
• Trigonometrical Ratios of 60°
• Trigonometrical Ratios of 90°
• Trigonometrical Ratios Table
• Problems on Trigonometric Ratio of Standard Angle
  
• Trigonometrical Ratios of Complementary Angles
• Rules of Trigonometric Signs
• Signs of Trigonometrical Ratios
  
• All Sin Tan Cos Rule
• Trigonometrical Ratios of (- θ)
• Trigonometrical Ratios of (90° + θ)
• Trigonometrical Ratios of (90° - θ)
• Trigonometrical Ratios of (180° + θ)
• Trigonometrical Ratios of (180° - θ)
• Trigonometrical Ratios of (270° + θ)
• Trigonometrical Ratios of (270° - θ)
• Trigonometrical Ratios of (360° + θ)
• Trigonometrical Ratios of (360° - θ)
• Trigonometrical Ratios of any Angle
• Trigonometrical Ratios of some Particular Angles
• Trigonometric Ratios of an Angle
• Trigonometric Functions of any Angles
• Problems on Trigonometric Ratios of an Angle
• Problems on Signs of Trigonometrical Ratios

10th Grade Math

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