Menu Close

hello-show-that-sin-5pi-18-cos-13pi-18-1-2-sin-4pi-9-




Question Number 75147 by mathocean1 last updated on 07/Dec/19
hello  show that   sin((5π)/(18))cos((13π)/(18))=−(1/2)sin((4π)/9)
$$\mathrm{hello} \\ $$$$\mathrm{show}\:\mathrm{that}\: \\ $$$$\mathrm{sin}\frac{\mathrm{5}\pi}{\mathrm{18}}\mathrm{cos}\frac{\mathrm{13}\pi}{\mathrm{18}}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\frac{\mathrm{4}\pi}{\mathrm{9}} \\ $$
Answered by MJS last updated on 07/Dec/19
sin a cos b =(1/2)(sin (a−b) +sin (a+b))  (1/2)(sin (((5π)/(18))−((13π)/(18))) +sin (((5π)/(18))+((13π)/(18))))=  =(1/2)(sin (−((4π)/9)) +sin π)=(1/2)sin (−((4π)/9)) =  =−(1/2)sin ((4π)/9)
$$\mathrm{sin}\:{a}\:\mathrm{cos}\:{b}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\:\left({a}−{b}\right)\:+\mathrm{sin}\:\left({a}+{b}\right)\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{18}}−\frac{\mathrm{13}\pi}{\mathrm{18}}\right)\:+\mathrm{sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{18}}+\frac{\mathrm{13}\pi}{\mathrm{18}}\right)\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\:\left(−\frac{\mathrm{4}\pi}{\mathrm{9}}\right)\:+\mathrm{sin}\:\pi\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\left(−\frac{\mathrm{4}\pi}{\mathrm{9}}\right)\:= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\frac{\mathrm{4}\pi}{\mathrm{9}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *