Question #09e0d Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer A. S. Adikesavan Jul 24, 2016 0 Explanation: #-60^o# is in Quadrant IV (Q IV)i; Cosine is positive. So #cos (-60^o)=cos(60^o)=1/2# #135^0=(180-45)^o# is in Q II; tan is negative. So, #tan(135^o)=tan(180-45)^o=-tan 45^o=-1# #315^o=(360-45)^o# is in Q IV; tangent is negative. So, #tan(315^o)=tan(360-45)^o=-tan45^o=-1# #660^0=2X320-60)^0# is in Q IV; cos is positive. So, #cos660^0=cos(2X360-60)^o=cos60^o=1/2#. Now, the given expression# = 1/2-(-1)(-1)+1/2=0#.. Answer link Related questions How do you find the trigonometric functions of any angle? What is the reference angle? How do you use the ordered pairs on a unit circle to evaluate a trigonometric function of any angle? What is the reference angle for #140^\circ#? How do you find the value of #cot 300^@#? What is the value of #sin -45^@#? How do you find the trigonometric functions of values that are greater than #360^@#? How do you use the reference angles to find #sin210cos330-tan 135#? How do you know if #sin 30 = sin 150#? How do you show that #(costheta)(sectheta) = 1# if #theta=pi/4#? See all questions in Trigonometric Functions of Any Angle Impact of this question 1401 views around the world You can reuse this answer Creative Commons License