Aug 4, 2015 · How do you find the exact values #sin (pi/4)# using the special triangles?

Feb 2, 2016 · sin(x+pi)=-sin(x)=sin(-x)=cos(pi/2 +x) cos (x+pi)=-cos(x) So According to the question cos(pi/2 +x)=cos(x) Using solution of cos we get (pi/2) + x = 2npi +- (x) So case I:- (pi/2) + x = 2npi + (x) But here, the value of x is cancelled, so case II:- (pi/2) + x = 2npi - (x) x=npi-pi/4 The solution of this equation is x=npi-pi/4

Aug 20, 2016 · sqrt(2 + sqrt3)/2 Trig unit circle --> sin ((7pi)/12) = sin (pi/12 + pi/2) = cos (pi/12) Find cos (pi/12) by using the trig identity: cos 2a = 2cos^2 a - 1 cos ((2pi ...

May 4, 2018 · 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#?

May 11, 2018 · #sin(pi/5 - pi/2) = sin(36^circ - 90^circ) = cos(90 - (36^circ - 90^circ)) = cos(180^circ-36^circ) = cos 144^circ # Strap in, this is where it gets interesting. First we note #cos(3 theta)=cos(2 theta)# has solutions #3 theta = pm 2 theta + 360^circ k,# integer #k# , or just using the subsuming minus sign,

Feb 17, 2017 · How do you apply the fundamental identities to values of #theta# and show that they are true

Jun 27, 2016 · How do you use a double-angle identity to find the exact value of sin 120°? How do you use double angle identities to solve equations? How do you find all solutions for #sin 2x = cos x# for the interval #[0,2pi]#?

Aug 19, 2015 · #sin (pi/12)# #=sin(pi/3 - pi/4)# #=sin(pi/3)cos(pi/4)-cos(pi/3)sin(pi/4)# #=(sqrt(3)/2)(1/sqrt(2))-(1/2)(1/sqrt(2))#

Apr 8, 2018 · #arcsin(sin(pi/9))=pi/9# The above makes sense if you think about what the inverse of a function is. If we have an angle #x# and we put this through the function: #f(x)=sinx# we get a ratio returned. When we put the ratio in the inverse function . …

Feb 2, 2017 · sin(a-pi/2)=-m We can use two formulas here. One sin(-A)=-sinA and sin(pi/2-A)=cosA Hence sin(a-pi/2)=sin(-(pi/2-a)) = -sin(pi/2-a)=-cosa=-m