Simplifying a trigonometric function

Aug 2016
6
0
Salt Lake City
How does -sin²(x+y) simplify to -x²/x²+1?
 

skeeter

MHF Helper
Jun 2008
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North Texas
... is there more context to this simplification that you have not posted, say the relationship between x and y?
 
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Aug 2016
6
0
Salt Lake City
This was part of an implicit differentiation question. However, the book showed the step -sin²(x+y) and then the next step -x²/x²+1 with both equal to y' just one being a simplification of the other, which is why I posted it on this Trigonometry thread.
 

skeeter

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Jun 2008
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North Texas
Please post the entire problem from the beginning, including the original equation.
 
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Dec 2013
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757
Colombia
This was part of an implicit differentiation question. However, the book showed the step -sin²(x+y) and then the next step -x²/x²+1 with both equal to y' just one being a simplification of the other, which is why I posted it on this Trigonometry thread.
This is almost certainly something to do with substituting something in from the original equation.
 

HallsofIvy

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Apr 2005
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To answer your original question, \(\displaystyle -sin^2(x+ y)\) does not simplify to \(\displaystyle -\frac{x^2}{x^2+ 1}\). For one thing, the first is a function of x and y, the second a function only of x.
 
Aug 2016
6
0
Salt Lake City
The original equation being implicitly differentiated is tan(x+y)=x. The steps are as follows:
(1+y')sec²(x+y)=1
y'=1-sec²(x+y)/sec²(x+y)
=-tan²(x+y)/tan²(x+y)
=sin²(x+y)
=-x²/x²+1
 

skeeter

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Jun 2008
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6,765
North Texas
$\tan(x+y)=x \implies \tan^2(x+y)=x^2$

line 3 of the steps shown should be ...

$\dfrac{-\tan^2(x+y)}{\tan^2(x+y)+1} = \dfrac{-x^2}{x^2+1}$

... they didn't have to change it to $-\sin^2(x+y)$ to get to the final result.
 
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