Simplify the expression: $\frac{(\cos 2\theta + i\sin 2\theta)^7}{(\cos 4\theta + i\sin 4\theta)^3}$
- A$\cos 2\theta + i\sin 2\theta$
- B$\cos 2\theta - i\sin 2\theta$
- C$\cos 4\theta + i\sin 4\theta$
- D$\cos 4\theta - i\sin 4\theta$
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If $1, \omega, \omega^2$ are the cube roots of unity,then
$1(2+\frac{1}{\omega})(2+\frac{1}{\omega^2})+2(3+\frac{1}{\omega})(3+\frac{1}{\omega^2})+3(4+\frac{1}{\omega})(4+\frac{1}{\omega^2})+\ldots 10 \text{ terms} =$
$1(2+\frac{1}{\omega})(2+\frac{1}{\omega^2})+2(3+\frac{1}{\omega})(3+\frac{1}{\omega^2})+3(4+\frac{1}{\omega})(4+\frac{1}{\omega^2})+\ldots 10 \text{ terms} =$
$\left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^{12} = $
If $n$ is an integer which leaves a remainder of $1$ when divided by $3$,then $(1+\sqrt{3}i)^n + (1-\sqrt{3}i)^n$ equals
DifficultTS EAMCET 2009
View SolutionIf $x$ is a cube root of unity other than $1$,then $\left(x+\frac{1}{x}\right)^2+\left(x^2+\frac{1}{x^2}\right)^2+\ldots+\left(x^{12}+\frac{1}{x^{12}}\right)^2=$
$(1+\sqrt{5}+i \sqrt{10-2 \sqrt{5}})^5=$
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