Problems on Eliminate Theta
Here we will solve various types of problems on eliminate theta from the given equations.
We know, “eliminate theta from the equations” means that the equations are combined in such a way into one equation that it remains valid without the theta (θ) appearing in this new equation.
Workedout problems on eliminate theta (θ) between the equations:
1. Eliminate theta between the equations:
x = a sin θ + b cos θ and y = a cos θ – b sin θ
OR,
If x = a sin θ + b cos θ and y = a cos θ –b sin θ, prove that
x
^{2} + y
^{2} = a
^{2} + b
^{2}.
Solution:
We have x
^{2} + y
^{2} = (a sin θ + b cos θ)
^{2} + (a cos θ – b sin θ)
^{2}
= (a
^{2} sin
^{2} θ + b
^{2} cos
^{2} θ + 2ab sin θ cos θ) + (a
^{2} cos
^{2} θ + b
^{2} sin
^{2} θ  2ab sin θ cos θ)
= a
^{2} sin
^{2} θ + b
^{2} cos
^{2} θ + 2ab sin θ cos θ + a
^{2} cos
^{2} θ + b
^{2} sin
^{2} θ  2ab sin θ cos θ
= a
^{2} sin
^{2} θ + b
^{2} cos
^{2} θ + a
^{2} cos
^{2} θ + b
^{2} sin
^{2} θ
= a
^{2} sin
^{2} θ + a
^{2} cos
^{2} θ + b
^{2} sin
^{2} θ + b
^{2} cos
^{2} θ
= a
^{2} (sin
^{2} θ + cos
^{2} θ) + b
^{2} (sin
^{2} θ + cos
^{2} θ)
= a
^{2} (1) + b
^{2} (1); [since, sin
^{2} θ + cos
^{2} θ = 1]
= a
^{2} + b
^{2}
Therefore, x
^{2} + y
^{2} = a
^{2} + b
^{2}
which is the required θeliminate.
2. Using the trigidentity we will solve the problems on eliminate theta (θ) between the equations:
tan θ  cot θ = a and cos θ + sin θ = b.
Solution:
tan θ – cot θ = a ………. (A)
cos θ + sin θ = b ………. (B)
Squaring both sides of (B) we get,
cos
^{2} θ + sin
^{2} θ + 2cos θ sin θ = b
^{2}
or, 1 + 2 cos θ sin θ = b
^{2}
or, 2 cos θ sin θ = b
^{2}  1 ………. (C)
Again, from (A) we get, (sin θ/cos θ) – (cos θ/sin θ) = a
or, (sin
^{2} θ  cos
^{2} θ)/(cos θ sin θ) = a
or, sin
^{2}θ  cos
^{2}θ = a sin θ cos θ
or, (sin θ + cos θ) (sin θ  cos θ) = a ∙ (b
^{2}  1)/2 ………. [by (C)]
or, b(sin θ  cos θ)= (½) a (b
^{2}  1) [by (B)]
or, b
^{2} (sin θ  cos θ)
^{2} = (1/4) a
^{2} (b
^{2}  1)
^{2}, [Squaring both the sides]
or, b
^{2} [(sin θ + cos θ)
^{2}  4 sinθ cos θ] = (1/4) a
^{2} (b
^{2}  1)
^{2}
or, b
^{2} [b
^{2}  2 ∙ (b
^{2}  1)] = (1/4) a
^{2} (b
^{2}  1)
^{2} [from (B) and (C)]
or, 4b
^{2} (2  b
^{2}) = a
^{2} (b
^{2}  1)
^{2}
which is the required θeliminate.
Show how to use the trigonometric identities to solve the problems on eliminate theta form the given two equations.
3. x sin θ  y cos θ = √(x
^{2} + y
^{2}) and cos
^{2} θ/a
^{2} + sin
^{2} θ/b
^{2} = 1/(x
^{2} + y
^{2})
Solution:
x sin θ  y cos θ = √(x
^{2} + y
^{2}) ..........…. (A)
cos
^{2} θ/a
^{2} + sin
^{2} θ/b
^{2} = 1/(x
^{2} + y
^{2}) ..........…. (B)
Squaring both sides of (A) we get,
x
^{2} sin
^{2} θ + y
^{2} cos
^{2} θ  2xy sin θ cos θ = x
^{2} + y
^{2}
or, x
^{2} (1  sin
^{2} θ) + y
^{2} (1  cos
^{2} θ) + 2xy sin θ cos θ = 0
or, x
^{2} cos
^{2} θ + y
^{2} sin
^{2} θ + 2 ∙ x cos θ ∙ y sin θ = 0
or, (x cos θ + y sin θ)
^{2} = 0
or, x cos θ + y sin θ = 0
or, x cos θ =  y sin θ
or, cos θ/(y) = sin θ/x
or, cos
^{2} θ/y
^{2} = sin
^{2} θ/x
^{2} = (cos
^{2} θ + sin
^{2} θ)/(y
^{2} + x
^{2}) = 1/(x
^{2} + y
^{2})
Therefore, cos
^{2} θ = y
^{2}/(x
^{2} + y
^{2}) and sin
^{2} θ = x
^{2}/(x
^{2} + y
^{2} )
Putting the values of cos
^{2} θ and sin
^{2} θ in (B) we get,
(1/a
^{2}) ∙ {y
^{2}/(x
^{2}} + y
^{2}) + (1/b
^{2}) ∙ {x
^{2}/(x
^{2} + y
^{2})} = 1/(x
^{2} + y
^{2})
Or, y
^{2}/a
^{2} + x
^{2}/b
^{2} = 1 (Since, x
^{2} + y
^{2} ≠0)
which is the required θeliminate.
The explanation will help us to understand how the steps are used technically to workout the problems on eliminate theta form the given equations.
● Trigonometric Functions
10th Grade Math
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