2018年3月8日 · See Below Use the Property: sin(x-y)=sinxcosy-cosxsiny LHS: sin(2pi-theta) =sin2picostheta-cos2pisintheta =0*costheta-1*sintheta =0-sin theta =-sintheta =RHS

2018年3月9日 · color(green)(sqrt3 / 2) sin ((2pi) / 3) = sin (pi - ((2pi)/3) )= sin ((3pi - 2pi) / 3) = sin (pi)/3 sin ((pi)/3) = sin 60 = sqrt3 / 2

You would need an expression to work with. For example: Given #sinalpha=3/5# and #cosalpha=-4/5#, you could find #sin2 alpha# by using the double angle identity

2016年11月16日 · How do you know if #sin 30 = sin 150#? How do you show that #(costheta)(sectheta) = 1# if #theta=pi/4#? See all questions in Trigonometric Functions of Any Angle

2016年5月5日 · 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#?

What is the period function of #y = sin (4x)# and the x value being 0 to 2pi? Whats the equation for a sine function with a period of 3/7, in radians? How do you find the amplitude and period for #s = 1/2 cos (pit - 8)#?

2018年4月17日 · sin at (2pi)/3 = sqrt3/2 The exact value for sin(2pi)/3 = sqrt3/2 For the unit circle sin(2pi)/3 is in the 2nd quadrant making sine positive. For the unit circle values on the 60^o angles all have a value of (1/2, sqrt3/2) where x= cosine and y=sine (2pi)/3 = …

2016年5月25日 · #sin ((2pi)/3)=sin(pi-(2pi)/3)=sin(pi/3)=sqrt3/2# #cos((2pi)/3)=-cos(pi-(2pi)/3)=-cos(pi/3)=-1/2# #tan((2pi)/3)=-tan(pi-(2pi)/3)=-tan(pi/3)=-sqrt3#

2017年6月11日 · Let: #alpha = cos ((2pi)/7) + i sin ((2pi)/7)# Then: #alpha^7 = cos (2pi) + i sin (2pi) = 1# So: #0 = alpha^7-1 = (alpha-1)(alpha^6+alpha^5+alpha^4+alpha^3+alpha^2 ...

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 ...