Jul 31, 2016 · sin ( (5pi)/3)=-sqrt(3)/2 sin((5pi)/3)=sin(2pi-pi/3) sin(2pi-pi/3)=-sin(pi/3) Period of sin is 2pi and 2pi-pi/3 is in the 4th quadrant. so sin is negative.

Apr 20, 2016 · Trig table and unit circle --> #sin ((-5pi)/4) = sin ((-5pi)/4 + (8pi)/4) = sin ((3pi)/4) =# #= sin (pi/4) = sqrt2/2#

Oct 25, 2016 · sqrt(2 + sqrt3)/2 Unit circle and property of complementary arcs --> sin ((-5pi)/12) = sin (pi/12 - (6pi)/12) = sin (pi/12 - pi/2) = cos (pi/12) Evaluate cos (pi/12 ...

Jul 31, 2016 · Trig table of special arcs, unit circle, property of supplementary arcs --> #sin ((5pi)/4) = sin (pi/4 + (4pi)/4) = sin (pi/4 + pi) = - sin (pi/4) =- sqrt2/2#

Aug 25, 2016 · 5/6pi. The calculator is programmed to give only the principal value for inverse trigonometric functions. 5/6pi radian = 150^o and sin 150^o is displayed as 0.5, and inversely, arc sin (0.5) is displayed as the principal value 30^o=pi/6 radian. Considering sin x as a bijective function over a short interval enclosing x = 5/6pi, the answer is 5/6pi, using that f^(-1)f(a) = a. If the answer is ...

Apr 24, 2016 · The period P of f(t) is given by f(t+P) = f(t), for the least value of P. The period of sin kt = #(2pi)/k#. Here, k = #5pi#.

Jul 17, 2016 · arc (5pi)/3 is in Quadrant IV. sin ((5pi)/3) = sin ((-pi/3 + (6pi)/3) = sin (-pi/3 + 2pi) = = sin (-pi/3) = - sin (pi/3) = - sqrt3/2 cos ((5pi)/3) = cos (-pi/3 + 2pi ...

Jun 25, 2016 · #sin((5pi)/4)=sin(pi+pi/4)# #sin(a+b)=sin a*cos b+cos a*sin b#. #sin(pi+pi/4)=sin pi*cos (pi/4)+cos pi*sin(pi/4)#

Oct 29, 2016 · How do you write the trigonometric form in complex form #6(cos((5pi)/6)+isin((5pi)/6)))#?

Also, recognize the sine addition formula: sin(A+B)=sin(A)cos(B)+cos(A)sin(B) So, we can write (5pi)/12 as ...