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

Timcago

Junior Member
Joined
Apr 13, 2006
Messages
77
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.
 
Re: Find sin(x/2), cos2x, and sin3x, when Cotx=(-1/2), cosx&

Hello, Timcago

Given: cotx=12,    cosx<0\displaystyle \cot x \:=\:-\frac{1}{2},\;\;\cos x\,<\,0

Find: sin(x2),    cos(2x),    sin(3x)\displaystyle \,\sin\left(\frac{x}{2}\right),\;\;\cos(2x),\;\;\sin(3x)
You're correct: x\displaystyle x is in quadrant 2.

Since cotx=12=adjopp\displaystyle \,\cot x \:=\:-\frac{1}{2}\:=\:\frac{adj}{opp} in quadrant 2,
    \displaystyle \;\;we have: opp=2,  adj=1,  hyp=5\displaystyle \,opp\,=\,2,\;adj\,=\,-1,\;hyp\,=\,\sqrt{5}

Then:   sinx=25=255,    cosx=15=55\displaystyle \;\sin x\,=\,\frac{2}{\sqrt{5}} \,=\,\frac{2\sqrt{5}}{5},\;\;\cos x\,=\,-\frac{1}{\sqrt{5}} \,= \,-\frac{\sqrt{5}}{5}

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Identity: sin2(x2)  =  1cosx2\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:   sin(x2)  =  5+510\displaystyle \:\,\sin\left(\frac{x}{2}\right) \;= \; \sqrt{\frac{5\,+\,\sqrt{5}}{10}}

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Identity: cos(2x)  =  cos2(x)sin2(x)\displaystyle \,\cos(2x)\;= \;\cos^2(x)\,-\,\sin^2(x)

Hence: cos(2x)  =  (55)2(255)2  =  5252025  =  1525  =  35\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}

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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

So: sin(2x)  =  2(255)(55)  =  2025  =  45\displaystyle \sin(2x)\;=\;2\left(\frac{2\sqrt{5}}{5}\right)\left(-\frac{\sqrt{5}}{5}\right) \;=\;-\frac{20}{25}\;=\;-\frac{4}{5}


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

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

      =  (45)(55)+(255)(35)  =  45256525  =  2525\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}
.
 
Isnt sin positive in the 2nd quadrant though?

wouldnt sin3x be positive?

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