sin C + sin D = 2 sin [(C + D)/2] cos [(C - D)/2]
sin C - sin D = 2 cos [(C + D)/2] sin [(C - D)/2]
cos C + cos D = 2 cos [(C + D)/2] cos [(C - D)/2]
cos C - cos D = -2 sin [(C + D)/2] sin [(C - D)/2]
Problem 1 :
Prove that
sin(4A-2B) + sin(4B-2A) /cos(4A-2B) + cos(4B-2A) = tan(A+B)
Solution :
sin(4A-2B) + sin(4B-2A)
= 2sin(A + B)cos(3A+3B)
cos(4A-2B) + cos(4B-2A)
= 2cos(A+B)cos(3A+3B)
Then,
sin(4A-2B) + sin(4B-2A) /cos(4A-2B) + cos(4B-2A) :
= 2sin(A+B)cos(3A+3B) /2cos(A+B)cos(3A+3B)
= tan(A+B)
Hence proved.
Problem 2 :
Prove that :
cot(A+15) - tan(A-15) = 4cos2A/(1 + 2sin2A)
Solution :
cot(A+15) = cos(A+15) / sin(A+15)
tan(A+15) = sin(A-15) / cos(A-15)
Then,
cot(A+15) - tan(A-15) :
= [cos(A+15)/sin(A-15)] - [sin(A-15)/cos(A-15)]
= [cos (A+15)cos(A-15) - sin(A-15)sin(A+15)] / sin(A+15)cos (A-15) -------(1)
cos(A+15)cos(A-15) = (1/2)[2cos(A+15) cos (A-15)]
= (1/2)[cos2A + cos30]
sin(A-15)sin(A+15) = (-1/2)[-2 sin (A-15) sin (A+15)]
= (-1/2)[cos2A - cos30]
Numerator of (1) :
cos(A+15)cos(A-15) - sin(A-15)sin(A+15) :
= (1/2)[cos2A + cos30] - (-1/2)[cos2A - cos30]
= (1/2)[2cos2A]
= cos2A
Denominator of (1) :
sin (A+15)cos(A-15) :
= (1/2)[2sin(A+15)cos(A-15)]
= (1/2)[sin2A + sin30]
= (1/2)[sin2A + (1/2)]
= (1/4)[2sin2A + 1]
Then,
(1)-----> = cos 2A/[(1/4)(2sin2A + 1)]
= 4cos2A / (2sin2A + 1)
Hence it is proved.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Oct 06, 24 05:49 AM
Oct 06, 24 05:44 AM
Oct 04, 24 09:58 AM