If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ – y cos θ = 0, then prove that x2 + y2 = 1, (where, sin θ ≠ 0 and cos θ ≠ 0).

2
by Deleted account

2016-03-31T09:52:28+05:30

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Solutions

x sin θ - y cos θ = 0, (Given)

⇒ x sin θ = y cos θ

⇒ y cos θ = x sin θ

Now dividing both sides by cos θ we get,

y = x ∙ (sin θ/cos θ)

Again, x sin3 θ + y cos3 θ = sin θ cos θ

⇒ x sin3 θ + x ∙ (sin θ /cos θ) ∙ cos3 θ = sin θ cos θ [Since, y = x ∙ (sin θ/cos θ)]

⇒ x sin θ ( sin2 θ + cos2 θ) = sin θ cos θ, [since, cos θ ≠ 0]

⇒ x sin θ (1) = sin θ cos θ,[since, sin2 θ + cos2 θ = 0]

⇒ x sin θ = sin θ cos θ

Now dividing both sides by sin θ we get,

⇒ x = cos θ, [since, sin θ ≠ 0]

Therefore, y = x ∙ (sin θ/cos θ)

⇒ y = cos θ ∙ (sin θ/cos θ), [Putting x = cos θ]

⇒ y = sin θ

Now, x2 + y

= cos2 θ + sin2 θ

= 1.

Therefore, x2 + y2 = 1.

2016-03-31T18:32:21+05:30

This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Given that,

x sin θ - y cos θ = 0,

⇒ x sin θ = y cos θ

⇒ y cos θ = x sin θ

Dividing both sides by cos θ,

y = x ∙ (sin θ/cos θ)

Now, x sin3 θ + y cos3 θ = sin θ cos θ

⇒ x sin3 θ + x ∙ (sin θ /cos θ) ∙ cos3 θ = sin θ cos θ (∵ y = x ∙ (sin θ/cos θ))

⇒ x sin θ ( sin2 θ + cos2 θ) = sin θ cos θ, (∵ cos θ ≠ 0)

⇒ x sin θ (1) = sin θ cos θ,(∵ sin2 θ + cos2 θ = 0)

⇒ x sin θ = sin θ cos θ

Dividing both sides by sin θ,

⇒ x = cos θ, (∵ sin θ ≠ 0)

∴y = x ∙ (sin θ/cos θ)

⇒ y = cos θ ∙ (sin θ/cos θ), (∵x = cos θ)

⇒ y = sin θ

Also, x2 + y2

= cos2 θ + sin2 θ

= 1

∴ x2 + y2 = 1.