Trig problem
- Thread starter Cmilner
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Steven G
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You need to try something! cotx = cosx/sinx. Now get a common denominator for subtracting the 1.
confused after this step
If I were to change cscx over into terms of sinx, then the problem would read (cosx/sinx)-1=(1/sinx). If I get the one in terms of sign so that all of the pieces have a common denominator it would be (cosx/sinx)-(sinx/sinx)=(1/sinx). This is as far as I was able to get last time before I majorly confused myself... If you combine the left side you would be left with (cosx-sinx)/sinx=(1/sinx). Where do you go from there?
If I were to change cscx over into terms of sinx, then the problem would read (cosx/sinx)-1=(1/sinx). If I get the one in terms of sign so that all of the pieces have a common denominator it would be (cosx/sinx)-(sinx/sinx)=(1/sinx). This is as far as I was able to get last time before I majorly confused myself... If you combine the left side you would be left with (cosx-sinx)/sinx=(1/sinx). Where do you go from there?
Steven G
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So you have (cosx-sinx)/sinx=(1/sinx). Two well defined equal fractions that have the same denominator must have the same numerator. That is cosx-sinx=1
You should now that when cosx = 1 or -1 then sinx=0 and when sin x=1 or -1 then cos x=0. Does this help you?
D
Deleted member 4993
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The general solution is a bit complex (since the domain is not given)
cos(x) - sin(x) = 1
1/√2 * cos(x) - 1/√2 * sin(x) = 1/√2
sin(π/4 - x) = sin (π/4 ± 2*n*π) or sin (3*π/4 ± 2*n*π)
\(\displaystyle \displaystyle{x \ = \ \mp 2 \cdot n \cdot \pi \ \ or \ \ x \ = \ - \dfrac{\pi}{2} \mp 2 \cdot n \cdot \pi}\)
cos(x) - sin(x) = 1
1/√2 * cos(x) - 1/√2 * sin(x) = 1/√2
sin(π/4 - x) = sin (π/4 ± 2*n*π) or sin (3*π/4 ± 2*n*π)
\(\displaystyle \displaystyle{x \ = \ \mp 2 \cdot n \cdot \pi \ \ or \ \ x \ = \ - \dfrac{\pi}{2} \mp 2 \cdot n \cdot \pi}\)
D
Deleted member 4993
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sin(π/4 - x) = sin (π/4 ± 2*n*π) or sin (3*π/4 ± 2*n*π)
π/4 - x = π/4 ± 2nπ ................ π/4 - x = 3π/4 ± 2nπ
- x = π/4 ± 2nπ - π/4 ............. - x = 3π/4 ± 2nπ - π/4
- x = ± 2nπ ......................... - x = π/2 ± 2nπ
multiply both sides of equation/s by -1 (hence flip sign) to get
\(\displaystyle \displaystyle{x \ = \ \mp 2 \cdot n \cdot \pi \ \ or \ \ x \ = \ - \dfrac{\pi}{2} \mp 2 \cdot n \cdot \pi}\)
Your time to corner
π/4 - x = π/4 ± 2nπ ................ π/4 - x = 3π/4 ± 2nπ
- x = π/4 ± 2nπ - π/4 ............. - x = 3π/4 ± 2nπ - π/4
- x = ± 2nπ ......................... - x = π/2 ± 2nπ
multiply both sides of equation/s by -1 (hence flip sign) to get
\(\displaystyle \displaystyle{x \ = \ \mp 2 \cdot n \cdot \pi \ \ or \ \ x \ = \ - \dfrac{\pi}{2} \mp 2 \cdot n \cdot \pi}\)
Your time to corner
Steven G
Elite Member
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Ok Dennis, you may come out of the corner now. I hope that you learned your lesson.