Question Number 68021 by anaplak last updated on 03/Sep/19
Commented by Tanmay chaudhury last updated on 03/Sep/19
Answered by Tanmay chaudhury last updated on 03/Sep/19
![p−1=(1/2)cos2α−(1/(2.4))cos4α+((1.3)/(2.4.6))cos6α+... q=(1/2)sin2α−(1/(2.4))sin4α+((1.3)/(2.4.6))sin6α+... (p−1)+iq=(1/2)e^(i2α) −(1/(2.4))e^(i4α) +((1.3)/(2.4.6))e^(i6α) −... p+iq=1+(1/2)e^(i2α) −(1/(2.4))e^(i4α) +((1.3)/(2.4.6))e^(i6α) −... p+iq=(1+e^(i2α) )^(1/2) =(1+cos2α+isin2α)^(1/2) p+iq=[2cos^2 α+i2sinαcosα]^(1/2) p+iq=[2cosα(e^(iα) )]^(1/2) p+iq=2^(1/2) (cosα)^(1/2) e^(i×(α/2)) p+iq=2^(1/2) (cosα)^(1/2) (cos(α/2)+isin(α/2)) so p=2^(1/2) .(cosα)^(1/2) .cos(α/2) q=2^(1/2) .(cosα)^(1/2) .sin(α/2) pls check...](https://www.tinkutara.com/question/Q68030.png)
$${p}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{2}\alpha−\frac{\mathrm{1}}{\mathrm{2}.\mathrm{4}}{cos}\mathrm{4}\alpha+\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}{cos}\mathrm{6}\alpha+… \\ $$$${q}=\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}\alpha−\frac{\mathrm{1}}{\mathrm{2}.\mathrm{4}}{sin}\mathrm{4}\alpha+\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}{sin}\mathrm{6}\alpha+… \\ $$$$\left({p}−\mathrm{1}\right)+{iq}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\mathrm{2}\alpha} −\frac{\mathrm{1}}{\mathrm{2}.\mathrm{4}}{e}^{{i}\mathrm{4}\alpha} +\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}{e}^{{i}\mathrm{6}\alpha} −… \\ $$$${p}+{iq}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\mathrm{2}\alpha} −\frac{\mathrm{1}}{\mathrm{2}.\mathrm{4}}{e}^{{i}\mathrm{4}\alpha} +\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}{e}^{{i}\mathrm{6}\alpha} −… \\ $$$${p}+{iq}=\left(\mathrm{1}+{e}^{{i}\mathrm{2}\alpha} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\left(\mathrm{1}+{cos}\mathrm{2}\alpha+{isin}\mathrm{2}\alpha\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${p}+{iq}=\left[\mathrm{2}{cos}^{\mathrm{2}} \alpha+{i}\mathrm{2}{sin}\alpha{cos}\alpha\right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${p}+{iq}=\left[\mathrm{2}{cos}\alpha\left({e}^{{i}\alpha} \right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${p}+{iq}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \left({cos}\alpha\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {e}^{{i}×\frac{\alpha}{\mathrm{2}}} \\ $$$${p}+{iq}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \left({cos}\alpha\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left({cos}\frac{\alpha}{\mathrm{2}}+{isin}\frac{\alpha}{\mathrm{2}}\right) \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{p}}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} .\left(\boldsymbol{{cos}}\alpha\right)^{\frac{\mathrm{1}}{\mathrm{2}}} .{cos}\frac{\alpha}{\mathrm{2}} \\ $$$${q}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} .\left({cos}\alpha\right)^{\frac{\mathrm{1}}{\mathrm{2}}} .{sin}\frac{\alpha}{\mathrm{2}} \\ $$$${pls}\:{check}… \\ $$
Commented by mathmax by abdo last updated on 03/Sep/19
$${sir}\:{Tanmay}\:{you}\:{have}\:{used}\:\:{relation}\:\mathrm{20}.\mathrm{12}\:{you}\:{answer}\:\:{is}\:{correct}. \\ $$
Commented by Tanmay chaudhury last updated on 03/Sep/19
$${thank}\:{you}\:{sir}… \\ $$