Question

If a = sin θ + cos θ, b = sin³ θ + cos³ θ, then ______.

Options

• a³ – 3a + 2b = 0

• a³ + 3a + 2b = 0

• a³ – 3a – 2b = 0

• a³ + 3a – 2b = 0

Answer 1

If a = sin θ + cos θ, b = sin³ θ + cos³ θ, then a³ – 3a + 2b = 0.

Explanation:

Given, a = sin θ + cos θ

and b = sin³ θ + cos³ θ

So, a³ = sin³ θ + cos³ θ + 3 sin θ cos θ (sin θ + cos θ)

and 3a = 3(sin θ + cos θ)

a³ – 3a = sin³ θ + cos³ θ + 3 sin θ cos θ (sin θ + cos θ) – 3 (sin θ + cos θ)

= sin³ θ + cos³ θ + 3 (sin θ + cos θ) (sin θ cos θ – 1)

= sin³ θ + cos³ θ – 3 (sin θ + cos θ) (1 – sin θ cos θ)

= sin³ θ + cos³ θ – 3(sin θ + cos θ)(sin² θ + cos² θ – sin θ . cos θ)   ...[using identity, a³ + b³ = (a + b)(a² + b² – ab)]

= sin³ θ + cos³ θ – 3 (sin³ θ + cos³ θ)

= – 2 (sin³ θ + cos³ θ)

`\implies` a³ – 3a = – 2b or a³ – 3a + 2b = 0