Trig Ratios Proving Problems

In trig ratios proving problems we will learn how to proof the questions step-by-step using trigonometric identities.

1. If (1 + cos A)( 1 + cos B)( 1 + cos C) = (1 - cos A)( 1 - cos B)( 1 - cos C) then prove that each side = ± sin A sin B sin C.

Solution:  Let, (1 + cos A) (1 + cos B) (1 + cos C) = k         …. (i)

Therefore, according to the problem,

(1 - cos A) (1 - cos B) (1 - cos C) = k                         ….. (ii)

Now multiplying both sides of (i) and (ii) we get,

(1 + cos A)(1 + cos B)(1 + cos C)(1 - cos A)(1 - cos B)(1 - cos C) = k²

⇒ k² = (1 - cos² A) (1 - cos² B) (1 - cos² C)

⇒ k² = sin² A sin² B sin² C

 k = ± sin A sin B sin C.

Therefore, each side of the given condition

= k = ± sin A sin B  sin C 
                                           Proved.

More solved examples on trig ratios proving problems.

2. If u_n = cosⁿ θ + sinⁿ θ then prove that, 2u₆ - 3u₄ + 1 = 0.

Solution:

Since, u_n = cosⁿ θ + sinⁿ θ

Therefore, u₆ = cos⁶ θ + sin⁶ θ

⇒ u₆ = (cos² θ)³ + (sin² θ)³

⇒ u₆ = (cos² θ + sin² θ)³ - 3 cos² θ ∙ sin² θ (cos² θ + sin² θ)

⇒ u₆ = 1 - 3cos² θ sin² θ and u₄ = cos⁴ θ + sin⁴ θ

⇒ u₄ = (cos² θ)² + (sin² θ)²

⇒ u₄ = (cos² θ + sin² θ)² - 2 cos² θ sin² θ

⇒ u₄ = 1 - 2 cos² θ sin² θ

Therefore,

2u₆ - 3u₄ + 1

= 2(1 - 3cos² θ sin² θ) - 3(1 - 2 cos² θ sin² θ) + 1

= 2 - 6 cos² θ sin² θ - 3 + 6 cos² θ sin² θ + 1

= 0.

Therefore, 2u₆ - 3u₄ + 1 = 0.

                                           Proved.

3. If a sin θ - b cos θ = c then prove that, a cos θ + b sin θ = ± √(a² + b² - c²).

Solution:

Given: a sin θ - b cos θ = c

⇒ (a sin θ - b cos θ)² = c², [Squaring both sides]

⇒ a² sin² θ + b² cos² θ - 2ab sin θ cos θ = c²

⇒ - a² sin² θ - b² cos² θ + 2ab sin θ cos θ = - c²

⇒ a² - a² sin² θ + b² - b² cos² θ + 2ab sin θ cos θ = a² + b² - c²

⇒ a²(1 - sin² θ) + b²(1 - cos² θ) + 2ab sin θ cos θ = a² + b² - c²

⇒ a² cos² θ + b² sin² θ + 2 ∙ a cos θ ∙ b sin θ = a² + b² - c²

⇒ (a cos θ + b sin θ)² = a² + b² - c²

Now taking square root on both the sides we get,

⇒ a cos θ + b sin θ = ± √(a² + b² - c²).

                                                      Proved.

The above three trig ratios proving problems will help us to solve more basic problems on T-ratio.

Basic Trigonometric Ratios 

Relations Between the Trigonometric Ratios

Problems on Trigonometric Ratios

Reciprocal Relations 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 

10th Grade Math

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