How would I perform the operation #14(cos 63° + isin 63°)##/##20(cos 15° + isin 15°)#? Make sure to leave the result in trigonometric form.

1 Answer
Bill K.
Oct 9, 2015

The quick way to do it is to divide the moduli ("lengths") and subtract the arguments ("angles"). The answer is #7/10 (cos(48^{circ})+i sin(48^{circ}))#

Explanation:

Note that #14(cos(63^{circ})+isin(63^{circ}))# has modulus 14 and argument #63^{circ}# while #20(cos(15^{circ})+i sin(15^{circ}))# has modulus 20 and argument #15^{circ}#.

The answer follows since #14/20=7/10# and #63^{circ}-15^{circ}=48^{circ}#.

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