Hello, Timcago!
Find the exact value of:
sin ( − 22. 5 o ) \displaystyle \,\sin(-22.5^o) sin ( − 2 2 . 5 o )
We already know that
− 22. 5 o \displaystyle -22.5^o − 2 2 . 5 o is in Quadrant 4, where sine is negative.
Identity:
sin θ 2 = 1 = cos θ 2 \displaystyle \,\sin\frac{\theta}{2}\:=\:\frac{1\,=\,\cos\theta}{2} sin 2 θ = 2 1 = cos θ
We have:
sin ( − 22. 5 o ) = 1 − cos ( − 4 5 o ) 2 = 1 − 2 2 2 = 2 − 2 4 \displaystyle \,\sin(-22.5^o) \:=\:\frac{1\,-\,\cos(-45^o)}{2} \;=\;\frac{1\,-\,\frac{\sqrt{2}}{2}}{2} \;=\;\frac{2\,-\,\sqrt{2}}{4} sin ( − 2 2 . 5 o ) = 2 1 − cos ( − 4 5 o ) = 2 1 − 2 2 = 4 2 − 2
Then:
sin ( − 22. 5 o ) = − 2 − 2 2 \displaystyle \,\sin(-22.5^o)\;=\;-\frac{\sqrt{2\,-\,\sqrt{2}}}{2} sin ( − 2 2 . 5 o ) = − 2 2 − 2
Simplify:
csc [ sin − 1 ( 2 5 ) + tan − 1 ( 1 3 ] ] \displaystyle \csc\left[\sin^{-1}\left(\frac{2}{5}\right)\,+\,\tan^{-1}\left(\frac{1}{3}\right]\right] csc [ sin − 1 ( 5 2 ) + tan − 1 ( 3 1 ] ]
We have: \(\displaystyle \L\,\frac{1}{\sin\left[\sin{-1}\left(\frac{2}{5}\right)\,+\,\tan^{-1}\left(\frac{1}{3}\right)\right]}\)
We will evaluate the denominator . . .
We have:
sin ( α + β ) \displaystyle \,\sin(\alpha\,+\,\beta) sin ( α + β ) . . . where \(\displaystyle \alpha\,=\,\sin^{-1}{\left(\frac{2}{5}\right)\) and \(\displaystyle \beta\,=\,\tan^{-1}\left(\frac{1}{3}\right)\)
We find that:
sin α = 2 5 , cos α = 21 5 \displaystyle \,\sin\alpha\,=\,\frac{2}{5},\;\;\cos\alpha\,=\,\frac{\sqrt{21}}{5} sin α = 5 2 , cos α = 5 2 1 \displaystyle \;\; and that:
sin β = 1 10 , cos β = 3 10 \displaystyle \,\sin\beta\,=\,\frac{1}{\sqrt{10}},\;\;\cos\beta\,=\,\frac{3}{\sqrt{10}} sin β = 1 0 1 , cos β = 1 0 3
Then:
sin ( α + β ) = sin α cos β + sin β cos α = ( 2 5 ) ( 3 10 ) + ( 1 10 ) ( 21 5 ) \displaystyle \,\sin(\alpha\,+\,\beta)\;=\;\sin\alpha\cos\beta\,+ \,\sin\beta\cos\alpha \;= \;\left(\frac{2}{5}\right)\left(\frac{3}{\sqrt{10}}\right)\,+\,\left(\frac{1}{\sqrt{10}}\right)\left(\frac{\sqrt{21}}{5}\right) sin ( α + β ) = sin α cos β + sin β cos α = ( 5 2 ) ( 1 0 3 ) + ( 1 0 1 ) ( 5 2 1 ) = 6 5 10 + 21 5 10 = 6 + 21 5 10 \displaystyle \;\;\;= \;\frac{6}{5\sqrt{10}}\,+\,\frac{\sqrt{21}}{5\sqrt{10}} \;= \;\frac{6\,+\,\sqrt{21}}{5\sqrt{10}} = 5 1 0 6 + 5 1 0 2 1 = 5 1 0 6 + 2 1
Therefore: \(\displaystyle \L\,\csc(\alpha\,+\,\beta)\;=\;\frac{5\sqrt{10}}{6\,+\,\sqrt{21}}\)
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If we are expected to rationalize the denominator:
\(\displaystyle \L\;\;\frac{5\sqrt{10}}{6\,+\,\sqrt{21}}\,\cdot\,\frac{6\,-\,\sqrt{21}}{6\,-\,\sqrt{21}} \;= \;\frac{5\sqrt{10}(6\,-\,\sqrt{21})}{36\,-\,21} \;= \;\frac{5\sqrt{10}(6\,-\,\sqrt{21})}{15} \;= \;\frac{\sqrt{10}(6\,-\,\sqrt{21})}{3}\)
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