Question

If sinA=1/2(a+1/a) Show that Sin3A=1/2(a^3+1/a^3​

Answer 1

Answer:

Your answer attached in the photo

Answer 2

EXPLANATION.

⇒ sinα = 1/2(a + 1/a).

Show that :

⇒ sin3α = -1/2(a³ + 1/a³).

As we know that,

Formula of :

⇒ sin3θ = 3sinθ - 4sin³θ.

⇒ (a + b)³ = a³ + 3a²b + 3ab² + b³.

Using this formula in this question, we get.

⇒ sin3α = 3sinα - 4sin³α.

Put the value in the equation, we get.

⇒ sin3α = 3[1/2(a + 1/a)] - 4[1/2(a + 1/a)]³.

⇒ sin3α = 3/2(a + 1/a) - (4/8)(a + 1/a)³.

⇒ sin3α = 3/2(a + 1/a) - 1/2[a³ + 3(a²)(1/a) + 3(a)(1/a²) + (1/a)³].

⇒ sin3α = 3/2(a + 1/a) - 1/2[a³ + 1/a³ +  3a + 3/a].

⇒ sin3α = 3/2(a + 1/a) - 1/2[a³ + 1/a³ + 3(a + 1/a)].

⇒ sin3α = 3/2(a + 1/a) - 1/2(a³ + 1/a³) - 3/2(a + 1/a).

⇒ sin3α = - 1/2(a³ + 1/a³).

Hence proved.