Question

if a cos^3O + 3 a cosO sin^2O = m, a sin^3O +3 a cos^2OsinO, prove that
(m + n)^2/3 + (m – n)^2/3 = 2a^⅔​

Answer 1

||✪✪ CORRECT QUESTION ✪✪||

if acos³x + 3acosxsin²x = m, and asin³x +3acos²xsinx = n , prove that (m + n)^2/3 + (m – n)^2/3 = 2a^⅔ .?

|| ✰✰ ANSWER ✰✰ ||

Let :-

→ acos³x + 3acosxsin²x = m ---------- Equation (1)

→ asin³x +3acos²xsinx = n ------------ Equation (2)

Adding Equation (1) and (2) , we get,

→ (m + n) = acos³x + 3acosx sin²x + asin³x + 3acos²xsinx

→ (m + n) = a(sinx + cosx)³

So,

→ (m + n)^(2/3) = a^(2/3)(sinx + cosx)² -------- Equation (3)

________________________

Now, Subtracting Equation (2) From Equation (1),

→ (m - n) = acos³x + 3acosx sin²x - asin³x - 3acos²xsinx

→ (m - n) = a(cosx - sinx)³

So,

→ (m - n)^(2/3) = a^(2/3)(cosx - sinx)² -------- Equation (4)

________________________

Now adding Equation (3) and (4) we get,

→ (m + n)^(2/3) + (m - n)^(2/3) = a^(2/3)(sinx + cosx)² + a^(2/3)(cosx - sinx)²

→ (m + n)^(2/3) + (m - n)^(2/3) = a^(2/3)[(sinx + cosx)² + (cosx – sinx)²]

→ (m + n)^(2/3) + (m - n)^(2/3) = a^(2/3)[ sin²x + cos²x + 2sinx*cosx + sin²x + cos²x - 2sinx*cosx ]

→ (m + n)^(2/3) + (m - n)^(2/3) = a^(2/3) [ 2sin²x + 2cos²x ]

→ (m + n)^(2/3) + (m - n)^(2/3) = a^(2/3) * 2 [ sin²x + cos²x]

Now, since Sin²A + cos²A = 1 ,

So,

→ (m + n)^(2/3) + (m - n)^(2/3) = 2a^(2/3).

✪✪ Hence Proved ✪✪

Answer 2

$\bullet{\underline{\sf{\color{grey}{Given:-}}}}$

• acos³x+3acosxsin²x= m

• asin³x+3acos²xsinx= n

$\bullet{\underline{\sf{\color{grey}{To\:Prove:-}}}}$

•(m+n)⅔+(m-n)⅔= 2a⅔

$\bullet{\underline{\sf{\color{grey}{Proof:-}}}}$

•acos³x + 3acosxsin²x = m ----------(1)

•asin³x +3acos²xsinx = n ------------ (2)

Adding Equation (1) and (2) , we get,

• (m + n) = acos³x + 3acosx sin²x + asin³x + 3acos²xsinx

•(m + n) = a(sinx + cosx)³

So,

•(m + n)^(2/3) = a^(2/3)(sinx + cosx)²-----(3)

Now, Subtracting Equation (2) From Equation (1),

•(m - n) = acos³x + 3acosx sin²x - asin³x - 3acos²xsinx

• (m - n) = a(cosx - sinx)³

So,

→ (m - n)^(2/3) = a^(2/3)(cosx - sinx)² ----(4)

Now adding Equation (3) and (4) we get,

→ (m + n)⅔ + (m - n)⅔= a⅔(sinx + cosx)² + a⅔(cosx - sinx)²

→ (m + n)⅔+ (m - n)⅔ = a^⅔[(sinx + cosx)² + (cosx – sinx)²]

→ (m + n)⅔ + (m - n)⅔ = a⅔[ sin²x + cos²x + 2sinxcosx + sin²x + cos²x - 2sinx*cosx ]

→ (m + n)⅔+ (m - n)⅔ = a⅔ [ 2sin²x + 2cos²x ]

→ (m + n)⅔ + (m - n)⅔= a⅔* 2 [ sin²x + cos²x]

As we know Sin²A + cos²A = 1 ,

So,

→ (m + n)⅔ + (m - n)⅔ = 2a⅔