How do you write the complex number in standard form #8(cos(pi/12)+isin(pi/12))#?

1 Answer
Jan 24, 2017

Answer:

#8(cos(pi/12)+isin(pi/12))=(2sqrt6+2sqrt2)+i(2sqrt6-2sqrt2)#

Explanation:

To write this complex number in standard form, what is needed is value of #cos(pi/12)=cos15^@# and #sin(pi/12)=sin15^@#

#cos15^@=cos(45^@-30^@)=cos45^@cos30^@+sin45^@sin30^@#

= #1/sqrt2xxsqrt3/2+1/sqrt2xx1/2=(sqrt3+1)/(2sqrt2)#

Similarly #sin15^@=sin(45^@-30^@)=sin45^@cos30^@-cos45^@sin30^@#

= #1/sqrt2xxsqrt3/2-1/sqrt2xx1/2=(sqrt3-1)/(2sqrt2)#

Hence, #8(cos(pi/12)+isin(pi/12))#

= #8(sqrt3+1)/(2sqrt2)+i8xx(sqrt3-1)/(2sqrt2)#

= #2sqrt2(sqrt3+1)+i2sqrt2(sqrt3-1)#

= #(2sqrt6+2sqrt2)+i(2sqrt6-2sqrt2)#