Is #cos 27^@ + cos 93^@ + cos 143^@ = 0# ?

1 Answer
Jan 16, 2018

No

Explanation:

As a quick check, note that the trigonometric functions of any angle that is a multiple of #3^@# is expressible using square roots, but the trigonometric functions of any other multiple of #1^@# are not.

See https://socratic.org/s/aMANhumV for example.

Note that #27^@# and #93^@# are both multiples of #3#, so #cos 27^@# and #cos 93^@# are expressible in terms of square roots.

Note that #143^@# is a whole number of degrees but not a multiple of #3^@#. So #cos 143^@# is not expressible in terms of square roots.

Therefore it is not possible that #cos 27^@ + cos 93^@ + cos 143^@ = 0#, since otherwise we would have:

#cos 143^@ = -cos 27^@-cos 93^@#

with the left hand expression not being expressible in terms of square roots, while the right hand expression is.

In fact, we find:

#cos 27^@ + cos 93^@ + cos 147^@ = 0#