sin\ 75^\circ = \frac{\sqrt{6}+\sqrt{2}}{4} I don't know where to start.
M manasan New member Joined Jul 21, 2019 Messages 8 Jul 28, 2019 #1 \(\displaystyle sin\ 75^\circ = \frac{\sqrt{6}+\sqrt{2}}{4} \) I don't know where to start. Last edited: Jul 28, 2019
MarkFL Super Moderator Staff member Joined Nov 24, 2012 Messages 3,021 Jul 28, 2019 #2 I would begin with: [MATH]75^{\circ}=45^{\circ}+30^{\circ}[/MATH] And then apply an angle sum identity for the sine function.
I would begin with: [MATH]75^{\circ}=45^{\circ}+30^{\circ}[/MATH] And then apply an angle sum identity for the sine function.
M manasan New member Joined Jul 21, 2019 Messages 8 Jul 29, 2019 #3 MarkFL said: I would begin with: [MATH]75^{\circ}=45^{\circ}+30^{\circ}[/MATH] And then apply an angle sum identity for the sine function. Click to expand... Thank You Sir for the hint I literally feel like an idiot for missing that
MarkFL said: I would begin with: [MATH]75^{\circ}=45^{\circ}+30^{\circ}[/MATH] And then apply an angle sum identity for the sine function. Click to expand... Thank You Sir for the hint I literally feel like an idiot for missing that