Express the following expression in the form | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given expression: $E = (\cos 2	heta + i\sin 2	heta )^{ - 5} (\cos 3	heta - i\sin 3	heta )^6 (\sin 	heta - i\cos 	heta )^3$. First,simplify t

Vedclass · Accepted Answer

$i(\cos 25	heta - i\sin 25	heta )$

Vedclass · Answer

(C) Given expression: $E = (\cos 2	heta + i\sin 2	heta )^{ - 5} (\cos 3	heta - i\sin 3	heta )^6 (\sin 	heta - i\cos 	heta )^3$. First,simplify the terms using De Moivre's Theorem: $(\cos 2	heta + i\sin 2	heta )^{ - 5} = \cos(-10	heta) + i\sin(-10	heta)$. $(\cos 3	heta - i\sin 3	heta )^6 = (\cos(-3	heta) + i\sin(-3	heta))^6 = \cos(-18	heta) + i\sin(-18	heta)$. $(\sin 	heta - i\cos 	heta )^3 = [-i(\cos 	heta + i\sin 	heta )]^3 = (-i)^3 (\cos 	heta + i\sin 	heta )^3 = i(\cos 3	heta + i\sin 3	heta)$. Multiplying these: $E = [\cos(-10	heta) + i\sin(-10	heta)] [\cos(-18	heta) + i\sin(-18	heta)] [i(\cos 3	heta + i\sin 3	heta)]$. $E = i [\cos(-10	heta - 18	heta + 3	heta) + i\sin(-10	heta - 18	heta + 3	heta)]$. $E = i [\cos(-25	heta) + i\sin(-25	heta)] = i(\cos 25	heta - i\sin 25	heta)$.

Express the following expression in the form $A + iB$: $(\cos 2\theta + i\sin 2\theta )^{ - 5} (\cos 3\theta - i\sin 3\theta )^6 (\sin \theta - i\cos \theta )^3$.

Similar Questions

If $\omega$ is an imaginary cube root of unity,then the value of $\sin \left[ (\omega^{10} + \omega^{23})\pi - \frac{\pi}{4} \right]$ is

If $\alpha, \beta$ are the roots of the equation $x^2-4x+8=0$,then for any $n \in N$,$\alpha^{2n}+\beta^{2n}$ equals

$(1+\sqrt{3}i)^6-(\sqrt{3}+i)^6=$

The value of $i^{1/3}$ is

$\left(\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4} i\right)^{2020} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module