Question Number 66517 by mhmd last updated on 16/Aug/19
$${find}\:{the}\:{area}\:{cos}\left(\mathrm{2}\theta\right) \\ $$
Commented by MJS last updated on 16/Aug/19
$$\mathrm{you}\:\mathrm{must}\:\mathrm{give}\:\mathrm{borders}… \\ $$
Answered by Smail last updated on 18/Aug/19
![S=8∫_0 ^(π/4) (1/2)(r^2 dθ)=4∫_0 ^(π/4) cos^2 (2θ)dθ =2∫_0 ^(π/4) (1+cos(4θ))dθ =2[θ+((sin(4θ))/4)]_0 ^(π/4) =(π/2)](https://www.tinkutara.com/question/Q66645.png)
$${S}=\mathrm{8}\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \frac{\mathrm{1}}{\mathrm{2}}\left({r}^{\mathrm{2}} {d}\theta\right)=\mathrm{4}\int_{\mathrm{0}} ^{\pi/\mathrm{4}} {cos}^{\mathrm{2}} \left(\mathrm{2}\theta\right){d}\theta \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \left(\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)\right){d}\theta \\ $$$$=\mathrm{2}\left[\theta+\frac{{sin}\left(\mathrm{4}\theta\right)}{\mathrm{4}}\right]_{\mathrm{0}} ^{\pi/\mathrm{4}} =\frac{\pi}{\mathrm{2}} \\ $$