How do you use the sum or difference identity to find the exact value of sin 255 degrees?

2 Answers
Mar 27, 2018

The result is #(-sqrt6-sqrt2)/4#.

Explanation:

You can use the #sin# angle sum formula:

#sin(color(red)A+color(blue)B)=sincolor(red)Acoscolor(blue)B+sincolor(blue)Bcoscolor(red)A#

Since #255^@# is the sum of #225^@# and #30^@#, we can write:

#color(white)=sin(255^@)#

#=sin(color(red)(225^@)+color(blue)(30^@))#

#=sin(color(red)(225^@))cos(color(blue)(30^@))+sin(color(blue)(30^@))cos(color(red)(225^@))#

Here's a unit circle to remind us of some #sin# and #cos# values:

#color(white)=sin(color(red)(225^@))cos(color(blue)(30^@))+sin(color(blue)(30^@))cos(color(red)(225^@))#

#=color(red)(color(red)-sqrt2/2)*cos(color(blue)(30^@))+sin(color(blue)(30^@))cos(color(red)(225^@))#

#=color(red)(color(red)-sqrt2/2)*color(blue)(sqrt3/2)+sin(color(blue)(30^@))cos(color(red)(225^@))#

#=color(red)(color(red)-sqrt2/2)*color(blue)(sqrt3/2)+color(blue)(1/2)*color(red)cos(color(red)(225^@))#

#=color(red)-color(red)(sqrt2/2)*color(blue)(sqrt3/2)+color(blue)(1/2)*color(red)-color(red)(sqrt2/2)#

#=color(red)-color(red)(sqrt2/2)*color(blue)(sqrt3/2)color(purple)-(color(blue)1*color(red)sqrt2)/(color(blue)2*color(red)2)#

#=color(red)-color(red)(sqrt2/2)*color(blue)(sqrt3/2)color(purple)-color(purple)sqrt2/color(purple)4#

#=color(purple)-(color(red)sqrt2*color(blue)sqrt3)/(color(red)2*color(blue)2)color(purple)-color(purple)sqrt2/color(purple)4#

#=color(purple)-color(purple)sqrt6/color(purple)4color(purple)-color(purple)sqrt2/color(purple)4#

#=color(purple)(-sqrt6)/color(purple)4+color(purple)(-sqrt2)/color(purple)4#

#=color(purple)(-sqrt6-sqrt2)/color(purple)4#

This is the result. You can use a calculator to check your work:

https://www.desmos.com/calculator

Hope this helps!

Mar 27, 2018

#- (sqrt2 + sqrt6)/4#

Explanation:

sin 255 = sin (75 + 180) = - sin 75.
Find sin 75 by using the trig identity:
sin (a + b) = sin a.cos b + sin b.cos a
In this case:
sin 75 = sin (30 + 45) = sin 30.cos 45 + sin 45.cos 30
#sin 75 = (1/2)(sqrt2/2) + (sqrt2/2)(sqrt3/2) = #
#= (sqrt2/2)((1 + sqrt3)/2) = (sqrt2 + sqrt6)/4#
Finally,
#sin 255 = - sin 75 = - (sqrt2 + sqrt6)/4#