Multiple Angles

 

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 18^o

     Let q = 18^o

     or,    5q = 5 ´ 18^o = 90^o

     or,    3q + 2q  =  90^o

     or,    2q  =  90^o- 3q

     or,    sin 2q  =  sin(90^o- 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