# Simplifying a trigonometric function

#### theusher

How does -sin²(x+y) simplify to -x²/x²+1?

#### skeeter

MHF Helper
... is there more context to this simplification that you have not posted, say the relationship between x and y?

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#### theusher

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

MHF Helper
Please post the entire problem from the beginning, including the original equation.

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#### Archie

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

MHF Helper
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.

#### theusher

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

MHF Helper
$\tan(x+y)=x \implies \tan^2(x+y)=x^2$

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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