I need exact value by using a half angle identity of sin pi/12
Mind looking up and then posting here the "half angle identity for the sine?" That'd give you plenty with which to work.
i think you can rewriting it bc. pi/12 = 15 so you can writing sin15 what you can calculi rweriting it in this form sin15 = sin(45-30) = ... and there use formula sin(a-b) = ... hope you know it
after you distributed these so you need using these knowladge again sin45 = sqrt2 /2 sin30 = 1/2 cos45 = sqrt2 /2 cos30 = sqrt3 /2 hope these will help you
Half angle for sin is sin a/2=plus or minus sqrt1-cosa/2
\[\cos 2 \theta=1-2\sin ^2 \theta\] \[\sin \theta=\pm \frac{ 1 }{ 2 }\sqrt{1-\cos 2 \theta}\] \[when ~\theta=\frac{ \pi }{ 12 }<\frac{ \pi }{ 2 },\sin \theta ~is~positive.\] \[\sin \theta=\sqrt{1-\cos 2\theta }\] substitute \[\theta=\frac{ \pi }{ 12 }\] \[\cos 2\theta=\cos \frac{ \pi }{ 6 }=?\]
Half angle for sin is sin a/2=plus or minus sqrt1-cosa/2 \[\sin \frac{ a }{ 2 }= \pm \frac{ \sqrt{1-\cos a} }{ 2 }\]
Note that \[\frac{ \pi }{ 12 }=\frac{ \frac{ \pi }{ 6 } }{ 2 }\]
so that "a" in the formula immediately above is pi/6. It happens that cos pi/6 is \[\frac{ \sqrt{3} }{ 2 }\]
Substitute this value for "cos a" in the formula immediately above. Simplify your result as much as possible.
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