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Mutaja
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I have another two problems I find difficult. They both involve trigonometry, so I thought I could fit both under the same post. Also, if possible, I'd like some help in regards to confirming that one problem I've solved is done correctly.
First, the derivative. Find y' when y = ##arctan^2(x)##
2nd problem. Solve the equation ##sin^2x##+[itex]\frac{\sqrt{3}}{2}[/itex]sin2x = 1- ##2cos^2x##, xe[0, 2[itex]\pi[/itex])
Lastly, I've solved this problem, and hopefully someone can take a quick look at it and verify if I've done it correctly: find the limit when x->∞ (y - [itex]\sqrt{y^2-y}[/itex])
I'm not sure what to write here. Relevant equations are how trigonometric functions relate to each other. For example ##sin^2x + cos^2x## = 1. There are a lot of them, and I think it would be confusing to write them all.
First problem:
Find y' when y = ##arctan^2(x)##
y = arctanx * arctanx
y' = ([itex]\frac{1}{x^2+1}[/itex])##^2##
I feel that this is way too easy, although it makes sense in a way.
Edit: in the above problem, does arctanx*arctanx translate to ##arctanx^2## and not ##arctan^2x## as I've written? That might explain a lot, and if that's true, please let me know how I can begin to attempt to solve this problem.
Problem 2:
Solve the equation ##sin^2x##+[itex]\frac{\sqrt{3}}{2}[/itex]sin2x = 1- ##2cos^2x##, xe[0, 2[itex]\pi[/itex])
##sin^2x##+##2cos^2x## + [itex]\frac{\sqrt{3}}{2}[/itex]sin2x=1
##sin^2x##+##2cos^2x## + [itex]\frac{\sqrt{3}}{2}[/itex](2sinxcosx)=1
##sin^2x##+##cos^2x##+##cos^2x##+[itex]\frac{\sqrt{3}}{2}[/itex]2(sinxcosx)=1
##cos^2x##+[itex]\frac{\sqrt{3}}{2}[/itex]2(sinxcosx)=0
cosx+[itex]\frac{\sqrt{3}}{2}[/itex]2sinx=0
cosx+[itex]\sqrt{3}[/itex]sinx=0
And here I am stuck. I'm not even sure what the answer is going to be. I'm imagining I should be getting something like x= [itex]\pi[/itex] + [itex]\pi[/itex]n -> ne[+,2[itex]\pi[/itex]) or something. I'm really confused about this, unfortunately.
3rd 'problem':
x->∞ (y - [itex]\sqrt{y^2-y}[/itex])
[itex]\frac{(y-\sqrt{y^2-y})(y+\sqrt{y^2-y})}{y+\sqrt{y^2-y}}[/itex]
[itex]\frac{y}{y+\sqrt{y^2-y}}[/itex] -> divide by y.
[itex]\frac{1}{1+\frac{\sqrt{y^2-y}}{y}}[/itex]
[itex]\frac{1}{1+\sqrt{\frac{y^2-y}{y^2}}}[/itex]
using l'hopitals
[itex]\frac{1}{1+\sqrt{\frac{2y-1}{2y}}}[/itex]
= [itex]\frac{1}{2}[/itex]
Any help is appreciated, and just let me know if I can do something different with how I post or something. I'm new here so I still have a lot to learn in regards to the forum as well as math.
Homework Statement
First, the derivative. Find y' when y = ##arctan^2(x)##
2nd problem. Solve the equation ##sin^2x##+[itex]\frac{\sqrt{3}}{2}[/itex]sin2x = 1- ##2cos^2x##, xe[0, 2[itex]\pi[/itex])
Lastly, I've solved this problem, and hopefully someone can take a quick look at it and verify if I've done it correctly: find the limit when x->∞ (y - [itex]\sqrt{y^2-y}[/itex])
Homework Equations
I'm not sure what to write here. Relevant equations are how trigonometric functions relate to each other. For example ##sin^2x + cos^2x## = 1. There are a lot of them, and I think it would be confusing to write them all.
The Attempt at a Solution
First problem:
Find y' when y = ##arctan^2(x)##
y = arctanx * arctanx
y' = ([itex]\frac{1}{x^2+1}[/itex])##^2##
I feel that this is way too easy, although it makes sense in a way.
Edit: in the above problem, does arctanx*arctanx translate to ##arctanx^2## and not ##arctan^2x## as I've written? That might explain a lot, and if that's true, please let me know how I can begin to attempt to solve this problem.
Problem 2:
Solve the equation ##sin^2x##+[itex]\frac{\sqrt{3}}{2}[/itex]sin2x = 1- ##2cos^2x##, xe[0, 2[itex]\pi[/itex])
##sin^2x##+##2cos^2x## + [itex]\frac{\sqrt{3}}{2}[/itex]sin2x=1
##sin^2x##+##2cos^2x## + [itex]\frac{\sqrt{3}}{2}[/itex](2sinxcosx)=1
##sin^2x##+##cos^2x##+##cos^2x##+[itex]\frac{\sqrt{3}}{2}[/itex]2(sinxcosx)=1
##cos^2x##+[itex]\frac{\sqrt{3}}{2}[/itex]2(sinxcosx)=0
cosx+[itex]\frac{\sqrt{3}}{2}[/itex]2sinx=0
cosx+[itex]\sqrt{3}[/itex]sinx=0
And here I am stuck. I'm not even sure what the answer is going to be. I'm imagining I should be getting something like x= [itex]\pi[/itex] + [itex]\pi[/itex]n -> ne[+,2[itex]\pi[/itex]) or something. I'm really confused about this, unfortunately.
3rd 'problem':
x->∞ (y - [itex]\sqrt{y^2-y}[/itex])
[itex]\frac{(y-\sqrt{y^2-y})(y+\sqrt{y^2-y})}{y+\sqrt{y^2-y}}[/itex]
[itex]\frac{y}{y+\sqrt{y^2-y}}[/itex] -> divide by y.
[itex]\frac{1}{1+\frac{\sqrt{y^2-y}}{y}}[/itex]
[itex]\frac{1}{1+\sqrt{\frac{y^2-y}{y^2}}}[/itex]
using l'hopitals
[itex]\frac{1}{1+\sqrt{\frac{2y-1}{2y}}}[/itex]
= [itex]\frac{1}{2}[/itex]
Any help is appreciated, and just let me know if I can do something different with how I post or something. I'm new here so I still have a lot to learn in regards to the forum as well as math.