Here we are going to see some example problems to show how to solve problems involving trigonometric angles.

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]

## Grade 11 Trigonometry Problems Involving Angles

Question 1 :

Prove that

sin (4A − 2B) + sin(4B − 2A) /cos (4A − 2B) + cos (4B − 2A)

= tan(A + B).

Solution :

L.H.S

=  sin (4A−2B) + sin(4B−2A) /cos (4A−2B) + cos (4B−2A)

sin (4A−2B) + sin(4B−2A)

=   2 sin (A + B) cos (3A+3B) ----(1)

cos (4A−2B) + cos (4B−2A)

=  2 cos (A + B) cos (3A + 3B)  ----(2)

(1)/(2)

=  2 sin (A + B) cos (3A+3B) /2 cos (A + B) cos (3A + 3B)

=  tan (A + B)

Hence proved.

Question 2 :

Show that cot (A + 15) − tan (A − 15) = 4 cos 2A/1+2sin2A

Solution :

L.H.S

=  cot (A + 15) − tan (A − 15)

cot (A + 15)  =  cos (A + 15) / sin (A + 15) ------(1)

tan (A + 15)  =  sin (A - 15) / cos (A - 15) ------(2)

(1) - (2)

=  [cos (A+15) cos (A-15) - sin (A-15) sin (A+15)]/sin (A+15) cos (A-15)  -------(A)

cos (A+15) cos (A-15)  =  (1/2)(2 cos (A+15) cos (A-15) )

=  (1/2) [cos 2A + cos 30]

sin (A-15) sin (A+15)  =  (-1/2)(-2 sin (A-15) sin (A+15) )

=  (-1/2) [cos 2A - cos 30]

Numerator of (A) :

cos (A+15) cos (A-15) - sin (A-15) sin (A+15)

=  (1/2) [cos 2A + cos 30] - (-1/2) [cos 2A - cos 30]

=  (1/2)[2 cos 2A]

=  cos 2A

Denominator of (A) :

sin (A+15) cos (A-15)  =  (1/2) (2sin (A+15) cos (A-15))

=  (1/2) [sin 2A + sin 30]

=  (1/2) [sin 2A + (1/2)]

Numerator/Denominator

=  4 cos 2A / (2 sin 2A + 1)

Hence it is proved.

After having gone through the stuff given above, we hope that the students would have understood, how to solve problems using trigonometric identities and values of trigonometric ratios.

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