Exact value of csc(sin^-1(2/5)+tan^-1(1/3))

[Timcago]

Find the exact value of the following expressions, if they are defined

My first question is about sin(-22.5). I know that i need to use the half angle forumla with by multiplying 22.5 by 2.

Which one is correct:

-sin(45/2)=the square root of ((1-cos(45))/2) ->sin(45/2)= negative square root of (2 minus [the sqaure root of 2]) all divided by 2

Or

sin(-45/2)=the square root of ((1-cos(-45))/2)= positive square root of (2 minus [the square root of 2]) all divided by 2


The main quetions here though is how do i solve this?

csc(sin^-1(2/5)+tan^-1(1/3))

 

[soroban]

Hello, Timcago!

Find the exact value of: \(\displaystyle \,\sin(-22.5^o)\)

We already know that \(\displaystyle -22.5^o\) is in Quadrant 4, where sine is negative.

Identity: \(\displaystyle \,\sin\frac{\theta}{2}\:=\:\frac{1\,=\,\cos\theta}{2}\)

We have: \(\displaystyle \,\sin(-22.5^o) \:=\:\frac{1\,-\,\cos(-45^o)}{2} \;=\;\frac{1\,-\,\frac{\sqrt{2}}{2}}{2} \;=\;\frac{2\,-\,\sqrt{2}}{4}\)

Then: \(\displaystyle \,\sin(-22.5^o)\;=\;-\frac{\sqrt{2\,-\,\sqrt{2}}}{2}\)

Simplify: \(\displaystyle \csc\left[\sin^{-1}\left(\frac{2}{5}\right)\,+\,\tan^{-1}\left(\frac{1}{3}\right]\right]\)

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: \(\displaystyle \,\sin(\alpha\,+\,\beta)\) . . . where \(\displaystyle \alpha\,=\,\sin^{-1}{\left(\frac{2}{5}\right)\) and \(\displaystyle \beta\,=\,\tan^{-1}\left(\frac{1}{3}\right)\)

We find that: \(\displaystyle \,\sin\alpha\,=\,\frac{2}{5},\;\;\cos\alpha\,=\,\frac{\sqrt{21}}{5}\)
\(\displaystyle \;\;\)and that: \(\displaystyle \,\sin\beta\,=\,\frac{1}{\sqrt{10}},\;\;\cos\beta\,=\,\frac{3}{\sqrt{10}}\)

Then: \(\displaystyle \,\sin(\alpha\,+\,\beta)\;=\;\sin\alpha\cos\beta\,+ \,\sin\beta\cos\alpha \;= \;\left(\frac{2}{5}\right)\left(\frac{3}{\sqrt{10}}\right)\,+\,\left(\frac{1}{\sqrt{10}}\right)\left(\frac{\sqrt{21}}{5}\right)\)

\(\displaystyle \;\;\;= \;\frac{6}{5\sqrt{10}}\,+\,\frac{\sqrt{21}}{5\sqrt{10}} \;= \;\frac{6\,+\,\sqrt{21}}{5\sqrt{10}}\)


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:

\(\displaystyle \L\;\;\frac{5\sqrt{10}}{6\,+\,\sqrt{21}}\,\cdot\,\frac{6\,-\,\sqrt{21}}{6\,-\,\sqrt{21}} \;= \;\frac{5\sqrt{10}(6\,-\,\sqrt{21})}{36\,-\,21} \;= \;\frac{5\sqrt{10}(6\,-\,\sqrt{21})}{15} \;= \;\frac{\sqrt{10}(6\,-\,\sqrt{21})}{3}\)
.