Question

prove that
$\frac{1}{3} {cos}^{3} \alpha \sin(3 \alpha ) + {sin}^{3} \alpha \cos(3 \alpha ) = \frac{1}{4} \sin(4 \alpha )$
​

Answer 1

Correct Question :-

Prove That :-

→ (1/3) [ cos³A*sin3A + sin³A*cos3A ] = (1/4)*sin4A

Formula used :-

• cos³A = (cos3A + 3cosA)/4
• sin³A = (3sinA - sin3A)/4
• sinA*cosB + cosA*sinB = sin(A+B)

Solution :-

Taking LHS, we get,

→ (1/3) [ cos³A*sin3A + sin³A*cos3A ]

Putting Both Formula of cos³A & sin³A we get,

→ (1/3)[[{(cos3A + 3cosA)/4} * sin3A] + [{(3SinA - Sin3A)/4} * cos3A ]]

→ (1/3*4) [(cos3A + 3cosA)*sin3A + (3SinA - Sin3A)*cos3A]

→ (1/3*4) [ (cos3A*sin3A + 3cosA*sin3A) + (3sinA*cos3A - sin3A*cos3A) ]

→ (1/3*4) [ cos3A*sin3A - sin3A*cos3A + 3(cosA*sin3A + sinA*cos3A) ]

→ (1/3*4) [ 3(cosA*sin3A + sinA*cos3A) ]

→ (1/4)[ (cosA*sin3A + sinA*cos3A) ]

using sinA*cosB + cosA*sinB = sin(A+B) now, we get,

→ (1/4)[ sin(A + 3A) ]

→ 1(/4) * sin4A . = RHS = Hence , Proved.

Answer 2

_______________________

$\huge\tt{CORRECT~QUESTION:}$

→ (1/3) [ cos³A*sin3A + sin³A*cos3A ] = (1/4)*sin4A

_______________________

$\huge\tt{FORMULAS~USED:}$

• cos³A = (cos3A + 3cosA)/4
• sin³A = (3sinA - sin3A)/4
• sinA*cosB + cosA*sinB = sin(A+B)

_______________________

$\huge\tt{SOLUTION:}$

↪ (1/3) [ cos³A×sin3A + sin³A×cos3A ]

↪ (1/3)[[{(cos3A + 3cosA)/4} ×sin3A] + [{(3SinA - Sin3A)/4} ×cos3A ]]

↪ (1/3×4) [(cos3A + 3cosA)×sin3A + (3SinA - Sin3A)×cos3A]

↪ (1/3×4) [ (cos3A×in3A + 3cosA×sin3A) + (3sinA×cos3A - sin3A×cos3A) ]

↪(1/3×4) [ cos3A×sin3A - sin3A×cos3A + 3(cosA×sin3A + sinA×cos3A) ]

↪ (1/3×4) [ 3(cosA×sin3A + sinA×cos3A) ]

↪ (1/4)[ (cosA×sin3A + sinA×cos3A) ]

↪ (1/4)[ sin(A + 3A) ]

↪ 1(/4) × sin4A

_______________________

$\huge\tt{L.H.S=R.H.S}$

$\tt{Hence,~Proved}$

_______________________