we have the function f:R2→R2, f(x)=cos⁡(x+y).So a function that takes the tuple (x,y) and outputs cos⁡(x+y). Do the formulas like cos2(x)+sin2(x)=1 still make sense but in this version cos⁡(x+y)2+sin⁡(x+y)2=1?The formulas made sense with the vision of a trigonometric circle but what happens if there are 2 variables involved?

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we have the function f:R2→R2, f(x)=cos⁡(x+y).So a function that takes the tuple (x,y) and outputs cos⁡(x+y). Do the formulas like cos2(x)+sin2(x)=1 still make sense but in this version cos⁡(x+y)2+sin⁡(x+y)2=1?The formulas made sense with the vision of a trigonometric circle but what happens if there are 2 variables involved?

amisayq6t · Accepted Answer

For all(!) real numbers a we havecos2(a)+sin2(a)=1If x,y∈R , then put a=x+y , hencecos2(x+y)+sin2(x+y)=1If A,l,e,x,n,d,r,a∈R, thencos2(A+l+e+x+a+n+d+r+a)+sin2(A+l+e+x+a+n+d+r+a)=1

Savanah Wilkerson · Accepted Answer

Yes, it is applicable for any number of real variables. Number of variables in the bracket doesn&#039;t matter as long you take the sine and cosine of the same variables! I have used a very layman language to keep it easy. Hope this helps!

we have the function f :\mathbb{R}^2 \to \mathbb R^2, \ f(x) = \cos(x+y).So a function that ta

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