Jul 2, 2016 · sin(pi/5)=sqrt(10-2sqrt5)/4 Let theta=pi/5, then 5theta=pi and 3theta=pi-2theta. Note theta) is an acute angle. Hence sin3theta=sin(pi-2theta) but as sin(pi-A)=sinA This can be written as sin3theta=sin2theta expanding them or 3sintheta-4sin^3theta=2sinthetacostheta as theta=pi/5 we have sintheta!=0 and dividing by it we get 3-4sin^2theta=2costheta or 3-4(1 …

Sep 8, 2015 · Cos (pi /5) = cos 36° = (sqrt5 + 1)/4. If theta = pi/10, then 5theta = pi/2 => cos3theta = sin2theta.[ cos (pi /2 - alpha) = sinalpha}. => 4 cos ^3 theta - 3costheta ...

Oct 24, 2017 · Q1 I would convert to degrees. That helps me a lot more. We can use the conversion factor 180/pi. pi/5 = pi/5 * 180/pi = 36˚ This is clearly in the first quadrant, because 0 < 36 < 90. Hopefully this helps!

Jun 2, 2015 · pi -> 180 deg 5pi -> 5(180) = 900 deg. 11204 views around the world

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 18, 2016 · 1/4 Let A = Cos( pi/5 )*Cos( 2*pi/5 ) But Sin( 2*X ) = 2Sin( X )*Cos( X ) => Cos( X ) = Sin( 2*X ) / [ 2*Sin( X ) ] ----- > (1 ) When X = Pi / 5 Then Cos( Pi/5 ...

Jun 9, 2016 · How do you use the ordered pairs on a unit circle to evaluate a trigonometric function of any angle

Feb 10, 2009 · I am trying to prove that cos(pi/5)= (1+sqrt(5))/4 I tried finding the 5th roots of unity and then just using the real part cis(pi/5) but then i dont know... Math Help Forum Search

Jul 25, 2018 · As below. Standard form of cosine function is y = A cos (Bx - C) + D Given y = cos ((pi/5) x) A = 1, B = pi/5, C = D = 0 Amplitude = |A| = 1 Period = (2 pi) / |B ...

Apr 18, 2016 · 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