Question

SINQ-SIN3Q+SIN5Q-SIN7Q/COSQ-COS3Q-COS5Q+COS7Q=COT2Q

Answer 1

we have to prove that,

(sinQ - sin3Q + sin5Q - sin7Q)/(cosQ - cos3Q - cos5Q + cos7Q) = cot2Q

proof : we know,

sinC + sinD = 2sin(C + D)/2 cos(C - D)/2

cosC - cosD = 2sin(C + D)/2 sin(D - C)/2

sinQ + sin5Q = 2sin(Q + 5Q)/2 cos(5Q - Q)/2

= 2sin3Q cos2Q

sin3Q + sin7Q = 2sin(3Q + 7Q)/2 cos(7Q - 3Q)/2

= 2sin5Q cos2Q

cosQ - cos5Q = 2sin(Q + 5Q)/2 sin(5Q - Q)/2

= 2sin3Q sin2Q

cos7Q - cos3Q = 2sin(7Q + 3Q)/2 sin(3Q - 7Q)/2

= -2sin5Q sin2Q

now LHS = (sinQ - sin3Q + sin5Q - sin7Q)/(cosQ - cos3Q - cos5Q + cos7Q)

= (2sin3Q cos2Q - 2sin5Q cos2Q)/(2sin3Q sin2Q - 2sin5Q sin2Q)

= 2cos2Q(sin3Q - sin5Q)/2sin2Q(sin3Q - sin5Q)

= cos2Q/sin2Q

= cot2Q = RHS

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