Question

sin2a+sin5a-sin3a/1+cosa-2sin^2a

Answer 1

(sin2a + sin5a - sin3a )/(1 + cosa - 2isn²a)

= {sin2a + 2cos(5a + 3a)/2.sin(5a - 3a)/2}/{cosa + (1 - 2sin²a)}

[ from sinC - sinD = 2cos(C + D)/2.sin(C - D)/2 ]

= {sin2a + 2cos4a.sina}/{cosa + cos2a )

[From cos2A = 1 - 2sin²A]

= {2sina.cosa + 2cos4a.sina}/{2cos(a + 2a)/2.cos(2a-a)/2}

[From sin2X = 2sinX .cosX and cosC + cosD = 2cos(C+D)/2.cos(C-D)/2]

= 2sina(cosa + cos4a)/2cos(3a/2).cos(a/2)

= 2sina{2cos(5a/2).cos(3a/2)/2}/2cos(3a)/2.cos(a/2)

= 2sina.cos(5a/2)/cos(a/2).

= 4sin(a/2).cos(a/2)cos(5a/2)/cos(a/2)

= 4sin(a/2).cos(5a/2)