Question
Question: If we have the expression as \(y=\cos \theta +i\sin \theta \) then the value of \(y+\dfrac{1}{y}\) i...
If we have the expression as y=cosθ+isinθ then the value of y+y1 is
a). 2cosθ
b). 2sinθ
c). 2cosecθ
d). 2tanθ
Solution
So we need to find the value of y+y1 in which the value of ‘y’ is given in the trigonometric form with an imaginary number. So we will try to eliminate the imaginary function from the denominator, by using the method of rationalisation, and then we will simply solve it using trigonometric identities.
Complete step-by-step solution:
Moving ahead with the question in the step wise manner, we have y=cosθ+isinθ and we need to find out the value of y+y1. So let us first put the value of ‘y’ in y+y1, so we will get;
=cosθ+isinθ+cosθ+isinθ1
Since the value of ‘y’ is in the trigonometric function additionally with the imaginary function, since we cannot solve the imaginary number in the denominator, let us rationalise it to remove the imaginary part from the denominator.
So let us rationalise y1 by simultaneously multiplying and dividing the term with =cosθ−isinθ, so we will get;
=cosθ+isinθ+cosθ+isinθ1(cosθ−isinθcosθ−isinθ)
On simplifying it further we will get;
=cosθ+isinθ+cos2θ+sin2θcosθ−isinθ
As by the trigonometric identity we know that cos2θ+sin2θ is equal to 1, so by replacing cos2θ+sin2θ with 1 we will get;
=cosθ+isinθ+1cosθ−isinθ
On simplifying we will get;
⇒cosθ+isinθ+cosθ−isinθ⇒2cosθ
So we got 2cosθ.
Hence the answer is 2cosθ i.e. option ‘a’ is correct.
Note: Whenever the imaginary number comes in the denominator then our first step should be to rationalise it. Rationalisation means we should multiply and divide the number by term such that the denominator value will get reduced in the simpler form.