Question

Compute \[\sin(15^{\Large\textrm{o}})\cdot\sin(22.5^{\Large\textrm{o}})\cdot\sin(67.5^{\Large\textrm{o}})\cdot\sin(75^{\Large\textrm{o}})\]

Answer 1

no thx

Answer 2

Calculator time

Answer 3

Looks daunting :)
\[\Large \sin(45^o-30^o)\cdot \sin\left(\frac{45^o}{2}\right)\cdot\sin\left(\frac{135^o}2\right)\cdot \sin (45^o+30^o)\]

Answer 4

\[\large \sin(45^o-30^o)=\sin(45^o)\cos(30^o)-\cos(45^o)\sin(30^o)\]\[\large = \frac{\sqrt 2}{ 2}\cdot\frac{\sqrt3}2-\frac{\sqrt2}{2}\cdot\frac{1}{2}\]
\[\Large =\frac{\sqrt6-\sqrt2}{4}\]

Answer 5

On second thought, product-to-sum formulas may be more bearable :)

Answer 6

What is it for \(\sin\) ?

Answer 7

Well, \[\Large \sin(\alpha)\sin(\beta) = \frac12\left[\cos(\alpha-\beta)-\cos(\alpha +\beta)\right]\]

Answer 8

Pair up 15 and 75
Pair up 22.5 and 67.5

Answer 9

\[\begin{align}
\sin15^{\Large\textrm{o}}\cdot\sin75^{\Large\textrm{o}}
&=\frac{1}{2}\left[-\cos60^{\Large\textrm{o}}-\cos90^{\Large\textrm{o}}\right] \\
&=\frac{1}{2}\left[-\frac{1}{2}\right] \\
&=-\frac{1}{4}
\end{align}\]\[\begin{align}
\sin22.5^{\Large\textrm{o}}\cdot\sin67.5^{\Large\textrm{o}}
&=\frac{1}{2}\left[-\cos45^{\Large\textrm{o}}-\cos90^{\Large\textrm{o}}\right] \\
&=\frac{1}{2}\left[-\frac{\sqrt{2} }{2}\right] \\
&=-\frac{\sqrt{2} }{4}
\end{align}\]\[
\sin(15^{\Large\textrm{o}})\cdot\sin(22.5^{\Large\textrm{o}})\cdot\sin(67.5^{\Large\textrm{o}})\cdot\sin(75^{\Large\textrm{o}}) = -\frac{1}{4} \left(-\frac{\sqrt{2} }{4}\right) = \frac{\sqrt{2} }{16}\]
Does that sound right?

Answer 10

Not only does it sound right, it IS right :D