C2 trig help please Watch

gildartz
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#1
Report Thread starter 8 years ago
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I'm using the alpha C2 workbook and this one question confused me a bit.
solve for X:

2cos^2 X = 1

therefore

cos^2X = 1/2

(I can't show it very well on the computer but its meant to be cos squared then a 2, it's not cos to the power of 2x)

in the answers section, they went from this step to:
cosX = +1/2 or cosX = -1/2

but I don't see how they did this.

any help will be appreciated
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Farhan.Hanif93
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#2
Report 8 years ago
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(Original post by gildartz)
I'm using the alpha C2 workbook and this one question confused me a bit.
solve for X:

2cos^2 X = 1

therefore

cos^2X = 1/2

(I can't show it very well on the computer but its meant to be cos squared then a 2, it's not cos to the power of 2x)

in the answers section, they went from this step to:
cosX = +1/2 or cosX = -1/2

but I don't see how they did this, shouldn't it be:
cosX = +1/4 or -1/4

any help will be appreciated
\sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt 2}
So you're both wrong.
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Clarity Incognito
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#3
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(Original post by gildartz)
I'm using the alpha C2 workbook and this one question confused me a bit.
solve for X:

2cos^2 X = 1

therefore

cos^2X = 1/2

(I can't show it very well on the computer but its meant to be cos squared then a 2, it's not cos to the power of 2x)

in the answers section, they went from this step to:
cosX = +1/2 or cosX = -1/2

but I don't see how they did this, shouldn't it be:
cosX = +1/4 or -1/4

any help will be appreciated
 cos^2(x) = \dfrac{1}{2}

 \sqrt{(cos(x))^2} = \sqrt{\dfrac{1}{2}}

 cos(x) = \pm \dfrac{\sqrt{1}}{\sqrt{2}}

 cos(x) = \pm \dfrac{1}{\sqrt{2}}
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cmhcgs815
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it should be cosx = 1/(2^0.5) or -1/(2^0.5)
i.e, root everything to get 1 over plus or minus root 2.
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gildartz
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#5
Report Thread starter 8 years ago
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(Original post by Farhan.Hanif93)
\sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt 2}
So you're both wrong.
yeah i just noticed that so i edited my post XD
So the workbook is wrong? I thought it was a typo at first but then they solved X for 1/2 and -1/2 not 1/root2 and -1/root2

well thanks everyone, now I know to not trust the book XD
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Farhan.Hanif93
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(Original post by Clarity Incognito)
 cos^2(x) = \dfrac{1}{2}

 \sqrt{(cos(x))^2} = \sqrt{\dfrac{1}{2}}

 cos(x) = \pm \dfrac{\sqrt{1}}{\sqrt{2}}

 cos(x) = \pm \dfrac{1}{\sqrt{2}}
I love using unnecessary amount of latex, makes me happy :cool:
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