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Note: Conversion is based on the latest values and formulas.
How do I find the value of cos 5pi / 6? - Socratic 25 Oct 2015 · Find exact value of cos ((5pi)/6) Ans: sqrt3/2 On the trig unit circle, cos ((5pi)/6) = cos (- pi/6 + pi) = - cos (pi/6) Trig Table of Special Arcs gives --> cos ...
How do you find the exact values of cos 4pi/5? - Socratic 9 Jun 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
What is the value of #cos(pi/7) cos(pi/5)-sin(pi/7) sin(pi/5) 28 May 2016 · Apply the trig identity: cos (a =b) = cos a.cos b - sin a.sin b. #cos (pi/7)cos (pi/5) - sin (pi/7).sin (pi/5) = cos (pi/7 + pi/5) =#
How do you find the exact value of cos 5pi? | Socratic 9 Nov 2016 · 1 Unit circle --> cos (5pi) = cos (pi + 4pi) = cos pi = - 1
How do you find the exact values of cos 2pi/5? - Socratic 16 Mar 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
How do I evaluate cos(pi/5) without using a calculator? 8 Sep 2015 · Now cos 2theta = cos pi/5 = 1 - 2sin^2 theta, gives the result. Trigonometry . Science
How do you find the exact values of Cos(pi/5) * Cos(2pi/5)? 18 Feb 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 ...
How do you evaluate #[cos(pi/5) * Cos(2pi/5)]#? - Socratic 18 Apr 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
What is the period and amplitude for #cos(pi/5)(x)#? - Socratic 25 Jul 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 ...
How do you evaluate #sin(pi/5)#? - Socratic 2 Jul 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 …