2016年6月27日 · How do you find the value of #sin ((2pi)/3)#? Trigonometry Trigonometric Identities and Equations Double Angle Identities. 1 Answer

2017年12月17日 · "see explanation" >"using the "color(blue)"difference expansion for sin" •color(white)(x)sin(A-B)=sinAcosB-cosAsinB rArrsin(2pi-x) =sin(2pi)cosx-cos(2pi)sinx ...

2016年3月22日 · How do you find the exact value of #sin((2pi)/3)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle. 1 Answer

2017年8月27日 · #sin^2((2pi)/3+x)=(sin((2pi)/3)cosx+cos((2pi)/3)sinx)^2# = #(sqrt3/2cosx+1/2sinx)^2# = #3/4cos^2x+1/4sinx+sqrt3/2sinxcosx#

2016年4月7日 · Zero Apply the trig identity: cos (a - b) = cos a.cos b + sin a.sin b cos ((2pi)/3 - pi/6) = cos ((2pi)/3).cos (pi/6) + sin ((2pi)/3).sin (pi/6) Trig table gives ...

2015年11月9日 · The coefficient in front of sin is 1, so amplitude =1. Amplitude =1 Find the period by dividing the coefficient in front of x by 2pi Period =((2pi)/(2pi))=1

2016年4月7日 · 0, pi, (2pi)/3, (4pi)/3, 2pi sin 2x + sin x = 0 Apply the identity: sin 2x = 2sin x.cos x 2sin x.cos x + sin x = 0. sin x(2cos x + 1) = 0 a. sin x = 0 --> x = 0, x = pi and x = 2pi b/ 2cos x …

2016年2月14日 · What is arcsin(sin 2pi/3)? Trigonometry Inverse Trigonometric Functions Basic Inverse Trigonometric Functions.

#sin((2pi)/3)=sqrt(3)/2# so the problem we're solving is really: #arcsin(sqrt(3)/2)#. We're looking for the angle between #-pi/2# and #pi/2# that has a sine value of #sqrt(3)/2#. We know it has …

2017年2月3日 · int sin^2(2pix)dx = x/2 - 1/(8pi)sin(4pix) + C Use the trigonometric identity: sin^2theta = (1-cos2theta ...