Let z= θ+i θ. Then the value of _{m=1}¹5} I_{m}(z²m-1}) at θ=2^{} is

Correct Answer is : 14sin⁡2∘{1}{4 2^{}}4sin2∘1

Correct Answer is : 14sin⁡2∘{1}{4 2^{}}4sin2∘1

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AP EAPCET 24-Aug-2021 Shift 1 Paper

Section: Mathematics

Question : 11 of 160

Marks: +1, -0

Let

z=\cos \theta+i \sin \theta

\sum\limits_{m=1}^{15} I_{m}(z^{2m-1})

\theta=2^{\circ}

Solution:

z=\cos \theta+i \sin \theta=e^{i \theta}\;\;\;\;\;

[in Euler's form]

Using Demoivre theorem

z^{2}=\cos 2\theta+i \sin 2\theta

z^{3}=\cos 3\theta+i \sin 3\theta

Now, according to the question

\sum\limits_{m=1}^{15} I_{m}(z^{2m-1})

⇒ Imaginary part of

(z+z^{3}+z^{5}+...+z^{29})

⇒

\sin \theta+\sin 3\theta+\sin 5\theta+...\sin 29\theta)

\Rightarrow \frac{\sin 15 \frac{2\theta}{2}}{\sin \frac{2\theta}{2}}\left[\sin \left(\theta+(15-1)\frac{2\theta}{2}\right)\right]

[\because \sin \alpha+\sin (\alpha+\beta)+\dots

n \text{term}

=\frac{\sin \frac{n\beta}{2}}{\sin \frac{\beta}{2}}\left[\sin \left(\alpha+(n-1)\frac{\beta}{2}\right)\right]

Here,

\alpha=\theta, \beta=2\theta, n=15]

\Rightarrow \frac{\sin 15\theta}{\sin \theta} \cdot \sin (15\theta)

Now,

\theta=2^{\circ}

=\frac{\sin 30^{\circ} \sin 30^{\circ}}{\sin 2^{\circ}}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{\sin 2^{\circ}}

=\frac{1}{4\sin 2^{\circ}}

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