GRADE 11 TRIGONOMETRY PROBLEMS

Sum to Product Identities

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.

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