prove that 1/3 (cos³a sin3a+ sin³acos3a) = 1/4 sin4a
solve it!! and get points..!!
Answers
FORMULAS USED:
cos³A = (cos3A + 3cosA)/4
sin³A = (3sinA - sin3A)/4
sinA*cosB + cosA*sinB = sin(A+B)
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↪ (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
Given :- prove that 1/3 (cos³a sin3a+ sin³acos3a) = 1/4 sin4a ?
Solution :-
solving LHS,
→ 1/3 (cos³a * sin 3a+ sin³a * cos 3a)
putting :-
- cos³a = (cos 3a + 3cos a)/4
- sin³a = (3sin a - sin 3a)/4
→ (1/3)[(cos 3a + 3cos a)/4 * sin 3a + ((3sin a - sin 3a)/4 * cos 3a)]
taking (1/4) common,
→ (1/12)[cos 3a * sin 3a + 3cos a * sin 3a + 3sin a * cos 3a - sin 3a * cos 3a]
sin 3a * cos 3a will be cancel ,
→ (1/12)[3cos a * sin 3a + 3sin a * cos 3a]
taking 3 common now,
→ (1/4)[cos a * sin 3a + sin a * cos 3a]
using :-
- sin(A + B) = sinA * cosB + cosA * sinB
→ (1/4) * sin(3a + a)
→ (1/4)sin 4a = RHS (Proved)
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