How do you find the exact values of the sine, cosine, and tangent of the angle #285^circ#?

1 Answer
May 15, 2018

#cos 285 ^\circ = cos (330 ^\circ - 45 ^\circ )= 1/4(\sqrt{6}-sqrt{2})#

#sin 285 ^\circ = sin(330 ^\circ -45) = -1/4 (sqrt{6} + sqrt{2}) #

#tan285 ^\circ = {sin 285 ^\circ }/{cos 285 ^\circ } = -2-sqrt{3}#

Explanation:

Why don't all questions have that comment saying when they we're created? I prefer to answer the recent ones.

#285^circ# is coterminal with (has the same trig functions as) #285 ^circ -360^circ =-75^circ.# So we see this is again the Two Tired Triangle of Trig, 30/60/90 plus 45/45/90.

So we just make #285^circ# the sum or difference of two angles we know, say

#285^\circ = 330^circ - 45^circ#

and use the difference angle formulas.

I'll drop the degree signs; they're too hard to type. We note before starting:

#cos 330 = cos(-30)= cos(30) =\sqrt{3}/2#

#sin 330 = -1/2#

#cos 45 = sin 45 = \sqrt{2}/2#

Now the difference angle formulas:

# cos (a-b) = cos a cos b + sin a sin b#

#sin(a-b)= sin a cos b + cos a sin b#

#cos 285 = cos (330 - 45) = cos 330 cos 45 + sin 330 sin 45 = \sqrt{2}/2 ( sqrt{3}/2 - 1/2) = 1/4(\sqrt{6}-sqrt{2})#

#sin 285 = sin(330-45) = sin 330 sin 45 - cos 330 cos 45 = \sqrt{2}/2 ( -1/2 - \sqrt{3}/2) = -1/4 (sqrt{6} + sqrt{2}) #

#tan285 = {sin 285}/{cos 285} = {-1/4 (sqrt{6} + sqrt{2}) } / { 1/4(\sqrt{6}-sqrt{2})} \times {sqrt{6}+sqrt{2})/{sqrt{6)+sqrt{2}} = - {6 + 2 + 2 \sqrt{6 times 2} } /{6-2} = {8+4 sqrt{3}}/4 = -2-sqrt{3}#