Question

sin(75°) = sqrt(1-cos(150°))/2

Answer 1

true or...false?

Answer 2

Type them into your calculator and see if you will get the same thing. You will.

Answer 3

\[\sin(75^\circ)=\sin(45^\circ+30^\circ)
\\=\sin(45^\circ)\cos(30^\circ)+\cos(45^\circ)\sin(30^\circ)
\\=\frac{\sqrt2}2\frac{\sqrt3}{2}+\frac{\sqrt{3}}2\frac12
\\=\frac{\sqrt3(\sqrt2+1)}{4}
\]

Answer 4

Would that be false? I got two different numbers for each one in my calculator.

Answer 5

The 1/2 should be under the radical.

Answer 6

\[\frac{\sqrt{1-\cos(150^\circ})}2=\frac{\sqrt{1-\cos(180^\circ-30^\circ})}2
\\=\frac{\sqrt{1+\cos(30^\circ})}2
\\=\frac{\sqrt{1+\frac{\sqrt3}2}}{2}=\frac{\sqrt{\frac{2+\sqrt3}{2}}}2\]

Answer 7

Take a look at the double angle formula.$$\cos2x=\cos^2x-\sin^2x=1-2\sin^2x\\1-\cos2x=2\sin^2x\\\frac{1-\cos2x}2=\sin^2x\\\sin x=\sqrt{\frac{1-\cos2x}2}$$Let \(x=75^\circ\):$$\sin75^\circ=\sqrt{\frac{1-\cos150^\circ}2}$$If the division by 2 is supposed to be under the radical, they're equal; if not, they're not.

Answer 8

Thank you!