How do I find Sin75 using half-angle identity? x --=150/2 ? 2
J Jaskaran Junior Member Joined May 5, 2006 Messages 67 Jun 15, 2007 #1 How do I find Sin75 using half-angle identity? x --=150/2 ? 2
M morson Full Member Joined Apr 12, 2007 Messages 263 Jun 15, 2007 #2 sin(150)=sin(30)=1/2\displaystyle sin(150) = sin(30) = 1/2sin(150)=sin(30)=1/2 \(\displaystyle cos(150) = -cos(30) = -\frac{sqrt{3}}{2}\\) Use the half-angle formula: \(\displaystyle \L\ sin^2(a) = \frac{1}{2}\(1 - cos(2a))\) Put a = 75, so: \(\displaystyle \L\ sin^2(75) = \frac{1}{2}\(1 + \frac{sqrt{3}}{2}\)\) \(\displaystyle \L\ 4sin^2(75) = (2 + sqrt{3})\) Solve for sin(75)\displaystyle sin(75)sin(75), but remember that it is in the 1st quadrant.
sin(150)=sin(30)=1/2\displaystyle sin(150) = sin(30) = 1/2sin(150)=sin(30)=1/2 \(\displaystyle cos(150) = -cos(30) = -\frac{sqrt{3}}{2}\\) Use the half-angle formula: \(\displaystyle \L\ sin^2(a) = \frac{1}{2}\(1 - cos(2a))\) Put a = 75, so: \(\displaystyle \L\ sin^2(75) = \frac{1}{2}\(1 + \frac{sqrt{3}}{2}\)\) \(\displaystyle \L\ 4sin^2(75) = (2 + sqrt{3})\) Solve for sin(75)\displaystyle sin(75)sin(75), but remember that it is in the 1st quadrant.
