Trig Ratios Proving Problems
In trig ratios proving problems we will learn how to proof the questions
stepbystep 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) = k
^{2}
⇒ k
^{2} = (1  cos
^{2} A) (1  cos
^{2} B) (1  cos
^{2} C)
⇒ k
^{2} = sin
^{2} A sin
^{2} B sin
^{2} 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} = cos
^{n} θ + sin
^{n} θ then prove that, 2u
_{6}  3u
_{4} + 1 = 0.
Solution:
Since, u
_{n} = cos
^{n} θ + sin
^{n} θ
Therefore, u
_{6} = cos
^{6} θ + sin
^{6} θ
⇒ u
_{6} = (cos
^{2} θ)
^{3} + (sin
^{2} θ)
^{3}
⇒ u
_{6} = (cos
^{2} θ + sin
^{2} θ)
^{3}  3 cos
^{2} θ ∙ sin
^{2} θ (cos
^{2} θ + sin
^{2} θ)
⇒ u
_{6} = 1  3cos
^{2} θ sin
^{2} θ
and u
_{4} = cos
^{4} θ + sin
^{4} θ
⇒ u
_{4} = (cos
^{2} θ)
^{2} + (sin
^{2} θ)
^{2}
⇒ u
_{4} = (cos
^{2} θ + sin
^{2} θ)
^{2}  2 cos
^{2} θ sin
^{2} θ
⇒ u
_{4} = 1  2 cos
^{2} θ sin
^{2} θ
Therefore,
2u
_{6}  3u
_{4} + 1
= 2(1  3cos
^{2} θ sin
^{2} θ)  3(1  2 cos
^{2} θ sin
^{2} θ) + 1
= 2  6 cos
^{2} θ sin
^{2} θ  3 + 6 cos
^{2} θ sin
^{2} θ + 1
= 0.
Therefore, 2u
_{6}  3u
_{4} + 1 = 0.
Proved.
3. If a sin θ  b cos θ = c then prove that,
a cos θ + b sin θ = ± √(a
^{2} + b
^{2}  c
^{2}).
Solution:
Given: a sin θ  b cos θ = c
⇒ (a sin θ  b cos θ)
^{2} = c
^{2}, [Squaring both sides]
⇒ a
^{2} sin
^{2} θ + b
^{2} cos
^{2} θ  2ab sin θ cos θ = c
^{2}
⇒  a
^{2} sin
^{2} θ  b
^{2} cos
^{2} θ + 2ab sin θ cos θ =  c
^{2}
⇒ a
^{2}  a
^{2} sin
^{2} θ + b
^{2}  b
^{2} cos
^{2} θ + 2ab sin θ cos θ = a
^{2} + b
^{2}  c
^{2}
⇒ a
^{2}(1  sin
^{2} θ) + b
^{2}(1  cos
^{2} θ) + 2ab sin θ cos θ = a
^{2} + b
^{2}  c
^{2}
⇒ a
^{2} cos
^{2} θ + b
^{2} sin
^{2} θ + 2 ∙ a cos θ ∙ b sin θ = a
^{2} + b
^{2}  c
^{2}
⇒ (a cos θ + b sin θ)
^{2} = a
^{2} + b
^{2}  c
^{2}
Now taking square root on both the sides we get,
⇒ a cos θ + b sin θ = ± √(a
^{2} + b
^{2}  c
^{2}).
Proved.
The above three trig ratios proving problems will help us to solve more basic problems on Tratio.
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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