How do you write the complex number in standard form #3.75(cos((3pi)/4))+isin((3pi)/4))#? Precalculus Complex Numbers in Trigonometric Form Trigonometric Form of Complex Numbers 1 Answer Shwetank Mauria Aug 31, 2016 #3.75(cos((3pi)/4)+isin((3pi)/4))=-2.652+i2.652# Explanation: As #cos((3pi)/4)=-1/sqrt2# and #sin((3pi)/4)=1/sqrt2# #3.75(cos((3pi)/4)+isin((3pi)/4))# = #3.75(-1/sqrt2+i×1/sqrt2)# Now as #1/sqrt2=sqrt2/2# above is equal to #3.75(-sqrt2/2+isqrt2/2)# = #-(3.75×sqrt2)/2+i(3.75×sqrt2)/2# = #-(3.75×1.4142)/2+i(3.75×1.4142)/2# = #-2.652+i2.652# Answer link Related questions How do I find the trigonometric form of the complex number #-1-isqrt3#? How do I find the trigonometric form of the complex number #3i#? How do I find the trigonometric form of the complex number #3-3sqrt3 i#? How do I find the trigonometric form of the complex number #sqrt3 -i#? How do I find the trigonometric form of the complex number #3-4i#? How do I convert the polar coordinates #3(cos 210^circ +i\ sin 210^circ)# into rectangular form? What is the modulus of the complex number #z=3+3i#? What is DeMoivre's theorem? How do you find a trigonometric form of a complex number? Why do you need to find the trigonometric form of a complex number? See all questions in Trigonometric Form of Complex Numbers Impact of this question 3622 views around the world You can reuse this answer Creative Commons License