How do you use the reference angles to find sin210cos330-tan 135sin210cos330−tan135?

1 Answer
Zor Shekhtman
Feb 3, 2015

sin(210)=sin(180+30)=-sin(30)=-1/2sin(210)=sin(180+30)=−sin(30)=−12
cos(330)=cos(360-30)=cos(-30)=cos(30)=sqrt(3)/2cos(330)=cos(360−30)=cos(−30)=cos(30)=√32
tan(135)=sin(180-45)/cos(180-45)=sin(45)/-cos(45)=(sqrt(2)/2)/(-sqrt(2)/2)=-1tan(135)=sin(180−45)cos(180−45)=sin(45)−cos(45)=√22−√22=−1
The answer, therefore, is
(-1/2)*(sqrt(3)/2)-(-1)=1-sqrt(3)/4=0.567(−12)⋅(√32)−(−1)=1−√34=0.567 (approximately)

The simple trigonometric identities used in the above calculations are:
sin(x+π) = −sin(x)
sin(x−π) = −sin(x)
sin(π−x) = sin(x)
cos(x+π) = −cos(x)
cos(x−π) = −cos(x)
cos(π−x) = −cos(x)
These any many other useful properties of trigonometric functions are explained in details in the chapter on Trigonometry on a Web site Unizor - a free Web site dedicated to advanced math for high school students.

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