Question

Sin θ. Sin (∝-θ) Sin (∝+θ) = *


​

Answer 1

Answer:

Solution:

We have,

sin θ = sin ∝

⇒ sin θ - sin ∝ = 0

⇒ 2 cos θ+∝2θ+∝2 sin θ−∝2θ−∝2 = 0

Therefore either cos θ+∝2θ+∝2 = 0 or, sin θ−∝2θ−∝2 = 0

Now, from cos θ+∝2θ+∝2 = 0 we get, θ+∝2θ+∝2 = (2m + 1)π2π2, m ∈ Z

⇒ θ = (2m + 1)π - ∝, m ∈ Z i.e., (any odd multiple of π) - ∝ ……………….(i)

And from sin θ−∝2θ−∝2 = 0 we get,

θ−∝2θ−∝2 = mπ, m ∈ Z

⇒ θ = 2mπ + ∝, m ∈ Z i.e., (any even multiple of π) + ∝ …………………….(ii)

Now combining the solutions (i) and (ii) we get,

θ = nπ + (-1)nn ∝, where n ∈ Z.

Hence, the general solution of sin θ = sin ∝ is θ = nπ + (-1)nn ∝, where n ∈ Z.

Note: The equation csc θ = csc ∝ is equivalent to sin θ = sin ∝ (since, csc θ = 1sinθ1sinθ and csc ∝ = 1sin∝1sin∝). Thus, csc θ = csc ∝ and sin θ = sin ∝ have the same general solution.

Hence, the general solution of csc θ = csc ∝ is θ = nπ + (-1)nn ∝, where n ∈ Z.

1. Find the general values of x which satisfy the equation sin 2x = -1212

solution:

sin 2x = -1212

sin 2x = - sin π6π6

⇒ sin 2x = sin (π + π6π6)

⇒ sin 2x = sin 7π67π6

⇒ 2x = nπ + (-1)nn 7π67π6, n ∈ Z

⇒ x = nπ2nπ2 + (-1)nn 7π127π12, n ∈ Z

Therefore the general solution of sin 2x = -1212 is x = nπ2nπ2 + (-1)nn 7π127π12, n ∈ Z

2. Find the general solution of the trigonometric equation sin 3θ = √32√32.

Solution:

sin 3θ = √32√32

⇒ sin 3θ = sin π3π3

⇒ 3θ = = nπ + (-1)nn π3π3, where, n = 0, ± 1, ± 2, ± 3, ± 4 .....

⇒ θ = nπ3nπ3 + (-1)nn π9π9,where, n = 0, ± 1, ± 2, ± 3, ± 4 .....

Therefore the general solution of sin 3θ = √32√32 is θ = nπ3nπ3 + (-1)nn π9π9, where, n = 0, ± 1, ± 2, ± 3, ± 4 .....

3. Find the general solution of the equation csc θ = 2

Solution:

csc θ = 2

⇒ sin θ = 1212

⇒ sin θ = sin π6π6

⇒ θ = nπ + (-1)nn π6π6, where, n ∈ Z, [Since, we know that the general solution of the equation sin θ = sin ∝ is θ = 2nπ + (-1)nn ∝, where n = 0, ± 1, ± 2, ± 3, ……. ]

Therefore the general solution of csc θ = 2 is θ = nπ + (-1)nn π6π6, where, n ∈ Z

4. Find the general solution of the trigonometric equation sin22 θ = 3434.

Solution:

sin22 θ = 3434.

⇒ sin θ = ± √32√32

⇒ sin θ = sin (± π3π3)

⇒ θ = nπ + (-1)nn ∙ (±π3π3), where, n ∈ Z

⇒ θ = nπ ±π3π3, where, n ∈ Z

Therefore the general solution of sin22 θ = 3434 is θ = nπ ±π3π3, where, n ∈ Z