Find sin(x/2), cos2x, and sin3x, when Cotx=(-1/2), cosx<0

[Timcago]

For the following, find sin(x/2), cos2x, and sin3x

Cotx=(-1/2), cosx<0

Alright so its in the second quadrant and in order to find sin(x/2) i must find the cos to fulfill the equation.

1+cot^2x=csc^2x

cscx=(1+(-1/2)^2)^(1/2) ->cscx=[(5)^(1/2)/2] square root of 5 over 2

sinx={1/[(5)^(1/2)/2]} -> sinx= {[3(5)^(1/2)]/5} 3 square root of 5 over 2

cosx=(1-({[3(5)^(1/2)]/5})^2)^(1/2) the square root of 1 minus 3 square root of 5 over 2 to the second = a non real result

A non real result! What does that make the answer for sin(x/2), cos2x, and sin3x?

Please correct me if i made a mistake.

 

[soroban]

Re: Find sin(x/2), cos2x, and sin3x, when Cotx=(-1/2), cosx&

Hello, Timcago

Given: \(\displaystyle \cot x \:=\:-\frac{1}{2},\;\;\cos x\,<\,0\)

Find: \(\displaystyle \,\sin\left(\frac{x}{2}\right),\;\;\cos(2x),\;\;\sin(3x)\)

You're correct: \(\displaystyle x\) is in quadrant 2.

Since \(\displaystyle \,\cot x \:=\:-\frac{1}{2}\:=\:\frac{adj}{opp}\) in quadrant 2,
\(\displaystyle \;\;\)we have: \(\displaystyle \,opp\,=\,2,\;adj\,=\,-1,\;hyp\,=\,\sqrt{5}\)

Then: \(\displaystyle \;\sin x\,=\,\frac{2}{\sqrt{5}} \,=\,\frac{2\sqrt{5}}{5},\;\;\cos x\,=\,-\frac{1}{\sqrt{5}} \,= \,-\frac{\sqrt{5}}{5}\)

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Identity: \(\displaystyle \,\sin^2\left(\frac{x}{2}\right)\;=\;\frac{1\,-\,\cos x}{2}\)

Hence: \(\displaystyle \L\,\sin^2\left(\frac{x}{2}\right) \;=\;\frac{1\,-\,\left(-\frac{\sqrt{5}}{5}\right)}{2} \;= \;\frac{5\,+\,\sqrt{5}}{10}\)

Therefore: \(\displaystyle \:\,\sin\left(\frac{x}{2}\right) \;= \; \sqrt{\frac{5\,+\,\sqrt{5}}{10}}\)

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Identity: \(\displaystyle \,\cos(2x)\;= \;\cos^2(x)\,-\,\sin^2(x)\)

Hence: \(\displaystyle \,\cos(2x)\;=\;\left(-\frac{\sqrt{5}}{5}\right)^2\,-\,\left(\frac{2\sqrt{5}}{5}\right)^2 \;= \;\frac{5}{25}\,-\,\frac{20}{25}\;=\;-\frac{15}{25}\;=\;-\frac{3}{5}\)

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Identity: \(\displaystyle \,\sin(2x)\;=\;2\sin(x)\cos(x)\)

So: \(\displaystyle \sin(2x)\;=\;2\left(\frac{2\sqrt{5}}{5}\right)\left(-\frac{\sqrt{5}}{5}\right) \;=\;-\frac{20}{25}\;=\;-\frac{4}{5}\)


Identity: \(\displaystyle \,\sin(A\,+\,B)\;=\;\sin(A)\cos(B)\,+\,\sin(B)\cos(A)\)

We have: \(\displaystyle \,\sin(3x)\;=\;\sin(2x\,+\,x)\;=\;\sin(2x)\cos(x)\,+\,\sin(x)\cos(2x)\)

\(\displaystyle \;\;\;= \;\left(-\frac{4}{5}\right)\left(-\frac{\sqrt{5}}{5}\right) \,+ \,\left(\frac{2\sqrt{5}}{5}\right)\left(-\frac{3}{5}\right) \;= \;\frac{4\sqrt{5}}{25}\,-\,\frac{6\sqrt{5}}{25}\;=\;-\frac{2\sqrt{5}}{25}\)
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[Timcago]

Isnt sin positive in the 2nd quadrant though?

wouldnt sin3x be positive?

ANd if not then why is sin(x/2) positive?