Question

If sin A/3= 1/2(a+1/a) prove that SinA= -1/2(a^3+1/a^3)​

Answer 1

Answer:

Step-by-step explanation:

Correct Question :-

If sinA = 1/2 (a + 1/a ) prove Sin3A = -1/2 ( a³+ 1/ a³)

Given,

sinA = 1/2(a + 1a)

To Prove :-

sin3A = -1/2(a³ + 1/a³)

How To Do :-

We need to take the L.H.S(sin3A) and we need to use the formula of sin3θ and we need to expand it after we can substitute the given value in that and we need to prove that L.H.S = R.H.S.

Formula Required :-

sin3θ = 3sinθ - 4sin³θ

(ab)³ = a³.b³

(a + b)³ = a³ + b³ + 3ab(a + b)

Solution :-

Taking L.H.S :-

= sin3A

= 3sinA - 4sin³A

[ ∴ sin3θ = 3sinθ - 4sin³θ]

$=3\left(\dfrac{1}{2}\left(a+\dfrac{1}{a}\right)\right)-4\left(\dfrac{1}{2}\left(a+\dfrac{1}{a}\right)\right)^3$

$=\dfrac{3}{2}\left(a+\dfrac{1}{a}\right)-4\left(\dfrac{1^3}{2^3}\left(a+\dfrac{1}{a}\right)^3\right)$

[ ∴ (ab)³ = a³.b³]

$=\dfrac{3}{2}\left(a+\dfrac{1}{a}\right)-4\left(\dfrac{1}{8}\left(a+\dfrac{1}{a}\right)^3\right)$

$=\dfrac{3}{2}\left(a+\dfrac{1}{a}\right)-\dfrac{4}{8}\left(a+\dfrac{1}{a}\right)^3$

$=\dfrac{3}{2}\left(a+\dfrac{1}{a}\right)-\dfrac{1}{2}\left(a+\dfrac{1}{a}\right)^3$

[∴ (a + b)³ = a³ + b³ + 3ab(a + b)]

$=\dfrac{3}{2}\left(a+\dfrac{1}{a}\right)-\dfrac{1}{2}\left(a^3+\left(\dfrac{1}{a}\right)^3+3a\dfrac{1}{a}\left(a+\dfrac{1}{a}\right)\right)$

$=\dfrac{3}{2}\left(a+\dfrac{1}{a}\right)-\dfrac{1}{2}\left(a^3+\dfrac{1^3}{a^3}+3\left(a+\dfrac{1}{a}\right)\right)$

$=\dfrac{3}{2}\left(a+\dfrac{1}{a}\right)-\dfrac{1}{2}\left(\left(a^3+\dfrac{1}{a^3}\right)+3\left(a+\dfrac{1}{a}\right)\right)$

$=\dfrac{3}{2}\left(a+\dfrac{1}{a}\right)-\dfrac{1}{2}\left(a^3+\dfrac{1}{a^3}\right)-\dfrac{3}{2}\left(a+\dfrac{1}{a}\right)$

$=-\dfrac{1}{2}\left(a^3+\dfrac{1}{a^3}\right)$

= R.H.S

Hence Proved