Question

(1 + cos theta + i sin thera ) ( 1+ cos theta - i sin thera = ?

Answer 1

Answer:

Step-by-step explanation:

(1+cosA)(1-cosA)=(1)sq. -(cosA )sq

[by (a+b)(a-b)= (a)sq-(b)sq]

=hence (1-cosAsq)

=sinAsq

Answer 2

The complete question is

(1 + cosθ + isinθ) (1 + cosθ - isinθ) = ?

Solution :

Now, (1 + cosθ + isinθ) (1 + cosθ - isinθ)

= {(1 + cosθ) + isinθ} {(1 + cosθ) - isinθ}

= (1 + cosθ)² - (isinθ)²

    { using the algebraic identity

    a² - b² = (a + b) (a - b) }

= (1 + cosθ)² - i²sin²θ

    { since (ab)² = a²b² }

= (1 + 2cosθ + cos²θ) - (- 1) sin²θ

    { using the algebraic identity

    (a + b)² = a² + 2ab + b²

    and since i² = - 1 }

= 1 + 2cosθ + cos²θ + sin²θ

= 1 + 2cosθ + (cos²θ + sin²θ)

= 1 + 2cosθ + 1

    { using trigonometric identity

    sin²θ + cos²θ = 1 }

= (1 + 1) + 2cosθ

= 2 + 2cosθ

= 2 (1 + cosθ)

    { we can complete simplification

    here or, we can proceed to next }

= 2 {1 + 2cos²(θ/2) - 1}

   { since cos2θ = 2cos²θ - 1 }

= 2 * 2cos²(θ/2)

= 4cos²(θ/2) ,

which is the required simplification.