Multiple Angles

The twice of angle A, thrice of angle A, fourth of angle A etc. are denoted by 2A, 3A, 4A etc. are called the multiple angles of the angle A.





































































































































































































































































9.            Prove the following:

        a) (cos 2A - cos 2B)2 + (sin 2A + sin 2B)2 = 4 sin2 (A + B)

Solution

LHS = (cos 2A - cos 2B)2 + (sin 2A + sin 2B)2

        = cos2 2A - 2cos 2A cos 2B + cos2 2B + sin2 2A + 2sin 2A sin 2B + sin2 2B

        = cos2 2A +  sin2 2A  + cos2 2B + sin2 2B - 2cos 2A cos 2B + 2sin 2A sin 2B

        = 1 + 1 - 2cos 2A cos 2B + 2sin 2A sin 2B

        = 2 - 2cos 2A cos 2B + 2sin 2A sin 2B

        = 2 - 2(cos 2A cos 2B - sin 2A sin 2B)

        = 2[1 - cos (2A + 2B)]

        = 2[1 - cos 2(A + B)]

        = 2[1 - {1 - 2sin2(A + B)}]                                      [Q cos 2A = 1 - 2sin2 A]

= 2[1 - 1 + 2sin2(A + B)]        

= 2[2sin2(A + B)]      

= 4sin2(A + B) = RHS proved

 b) (sin 2A - sin 2B)2 + (cos 2A + cos 2B)2 = 4 cos2 (A + B)

Solution

LHS = (sin 2A - sin 2B)2 + (cos 2A + cos 2B)2

        = sin2 2A - 2sin 2A sin 2B + sin2 2B + cos2 2A + 2cos 2A cos 2B + cos2 2B

        = sin2 2A +  cos2 2A  + sin2 2B + cos2 2B - 2sin 2A sin 2B + 2cos 2A cos2B

        = 1 + 1 - 2sin 2A sin 2B + 2cos 2A cos2B

        = 2 + 2cos 2A cos2B - 2sin 2A sin 2B

        = 2 + 2(cos 2A cos2B - sin 2A sin 2B)

        = 2[1 + cos (2A + 2B)]

        = 2[1 + cos 2(A + B)]

        = 2 [ 1 + 2 cos2(A + B) - 1]

        = 2 [ 2 cos2(A + B)]

        = 4 cos2(A + B) = RHS proved.

 10.          Prove the following:

a)       cos2q + sin2q. cos2a = cos2a + sin2a. cos2q                             

Soln:               

L.H.S.      = cos2q + sin2q. cos2a

                = 1 – sin2q + sin2q (1 – 2sin2a)

                = 1 – sin2q + sin2q – 2 sin2q. sin2a

                =  = 1 – 2sin2q. sin2a

R.H.S.     = cos2a + sin2a. cos2q

                = 1 – sin2a  + sin2a(1 – 2 sin2q)

                = 1 – sin2a  + sin2a – 2 sin2a  sin2q

                =  = 1 – 2sin2q. sin2a

L.H.S. = R.H.S. proved.

 b)        (1 + cos 2q + sin 2q)2 = 4 cos2 q (1 + sin 2q) 

LHS = (1 + cos 2q + sin 2q)2

= {1 + (cos 2q + sin 2q)}2

= 12 + 2.1.(cos 2q + sin 2q) + (cos 2q + sin 2q)2

= 1 + 2(cos 2q + sin 2q) + cos2 2q + 2sin 2q cos 2q + sin2 2q

= 1 + 2 cos 2q + 2 sin 2q + (cos2 2q + sin2 2q) + 2sin 2q cos 2q

= 1 + 2 cos 2q + 2 sin 2q + 1 + 2sin 2q cos 2q

= 2 + 2 cos 2q + 2 sin 2q + 2sin 2q cos 2q

= 2(1 + cos 2q + sin 2q + sin 2q cos 2q)

= 2{1(1 + cos 2q) + sin 2q(1 + cos 2q)}

= 2(1 + cos 2q) (1 + sin 2q)

= 2(1 + 2cos2 q - 1) (1 + sin 2q)

= 2´ 2cos2 q (1 + sin 2q)

= 4cos2 q (1 + sin 2q) = RHS proved.

 c)      (2 cos A + 1) (2 cos A - 1) (2 cos 2A - 1) = 1 + 2 cos 4A

LHS = (2 cos A + 1) (2 cos A - 1) (2 cos 2A - 1)

         = [(2 cos A)2 - 12] (2 cos 2A - 1)

         = (4 cos2 A - 1) (2 cos 2A - 1)

         = (4 cos2 A - 2 + 1) (2 cos 2A - 1)

         = [2 (2cos2 A - 1) + 1)] (2 cos 2A - 1)

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

         = (2 cos 2A)2 - 12

         = 4 cos2 2A - 1

         = 4 cos2 2A - 2 + 1

         = 2(2 cos2 2A - 1) + 1

         = 2 cos 2.2A + 1

         = 2 cos 4A  + 1 = RHS proved.

d)     1 + 2 cos 8q =  (2 cos 4q - 1) (2 cos 2q - 1) (2 cos q - 1)(2 cos q + 1)

LHS = (2 cos 4q - 1) (2 cos 2q - 1) (2 cos q - 1)(2 cos q + 1)

= (2 cos 4q - 1) (2 cos 2q - 1) {(2 cos q}2 - 12}

= (2 cos 4q - 1) (2 cos 2q - 1) (4 cos2 q - 1)

= (2 cos 4q - 1) (2 cos 2q - 1) (4 cos2 q - 2 + 1)

= (2 cos 4q - 1) (2 cos 2q - 1) {2 (2cos2 q - 1) + 1}

= (2 cos 4q - 1) (2 cos 2q - 1) (2 cos 2q  + 1)

= (2 cos 4q - 1) {(2cos 2q)2 - 12}

= (2 cos 4q - 1) (4cos2 2q - 1)

= (2 cos 4q - 1) (4cos2 2q - 2 + 1)

= (2 cos 4q - 1) {2(2cos2 2q - 1) + 1}

= (2 cos 4q - 1) (2cos 2.2q + 1)

= (2 cos 4q - 1) (2cos 4q + 1)

= {(2 cos 4q)2 - 12}

= 4 cos2 4q - 1

= 4 cos2 4q - 2 + 1

= 2 (2cos2 4q - 1) + 1

= 2 cos 2.4q  + 1

= 2 cos 8q  + 1 = RHS proved

 














































































12.          Prove the following:


































































15. (c) Prove that sin5θ =16sin5θ − 20sin3θ + 5sinθ

Here,

 sin5θ = sin(3θ+2θ)

=sin3θ.cos2θ+cos3θ.sin2θ

=(3sinθ − 4sin3θ)(1− 2sin2θ) + (4cos3θ − 3cosθ) 2sinθ.cosθ

=3sinθ  − 6sin3θ −4sin3θ + 8sin5θ + 8sinθ.cos4θ − 6sinθ.cos2θ

=3sinθ  − 6sin3θ −4sin3θ + 8sin5θ + 8sinθ.(cos2θ)2 − 6sinθ.cos2θ

=3sinθ −10sin3θ + 8sin5θ + 8sinθ(1− sin2θ)2  −  6sinθ(1−sin2θ)

=3sinθ −10sin3θ + 8sin5θ + 8sinθ (1−2sin2θ + sin4θ) − 6sinθ + 6sin3θ

=3sinθ −10sin3θ + 8sin5θ + 8sinθ − 16sin3θ + 8sin5θ − 6sin θ + 6sin3θ

=16sin5θ − 20sin3θ + 5sinθ


















































12.  c)  cos3A cos 3 A + sin3 A sin 3 A = cos3 2 A                   

Soln: L.H.S.    = cos3A cos 3A + sin3A sin 3 A

                                = cos3A (4 cos3A – 3 cos A) + sin3A (3 sin A – 4 sin3A)

                                = 4 cos6A– 3 cos4A+ 3 sin4A – 4sin6A

                                = 4(cos6A – sin6A) – 3 (cos4A – sin4A)

                                = 4[(cos2A)3 – (sin2A)3] – 3(cos2A – sin2A) (cos2A + sin2A)

                                = 4[(cos2A –sin2A)3+3 cos2A sin2A (cos2A – sin2A)] – 3(cos2A–sin2A)

                                = 4[cos32A+ 3cos2A sin2A. cos 2A] – 3 cos 2 A

                                = 4cos32A + 3 . 4cos2A sin2 A . cos 2A – 3 cos 2A

                                = 4 cos3 2A + 3(2 cos A sin A)2 . cos 2A – 3 cos 2A

                                = 4cos3 2A + 3 sin2 2A . cos2A – 3 cos 2A

                                = 4 cos3 2A + 3(1 – cos2 2A) . cos 2A – 3 cos 2A

                                = 4 cos2 2A + 3 cos 2A – 3 cos3 2A – 3 cos 2A = cos3 2A

                                = R.H.S. Proved.



15.    With the help of multiple angles relation of Sine and Cosine, find the values of sin18°, sin36° and sin54°. By using these values, find the values of cos18°, cos36° and cos54°. Also, find the value of tan18°, tan36° and tan54°. Share your result to your friend and prepare combine report.

Soln:

To find the value of sin 18o

     Let q = 18o

     or,    5q = 5 ´ 18o = 90o

     or,    3q + 2q  =  90o

     or,    2q  =  90o- 3q

     or,    sin 2q  =  sin(90o- 3q)

     or,    sin 2q  =  cos 3q

     or,    2 sin q cos q =  4 cos3 q - 3 cos q

     or,    2 sin q cos q -  4 cos3 q + 3 cos q = 0

     or,    cos q (2 sin q -  4 cos2 q + 3) = 0

     Since cos q ¹ 0, so

     2 sin q - 4 cos2 q + 3 = 0

or,     2 sin q -  4 (1 - sin2 q) + 3 = 0

or,     2 sin q -  4  + 4sin2 q + 3 = 0

or,     4sin2 q +  2 sin q -  1 = 0

Let sin q = x, the above equation becomes,

4x2 + 2x - 1 = 0