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

Solution:

Since, un = 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.

10th Grade Math

From Trig Ratios Proving Problems to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.