Write The Subinterval Of (0,Pi) In Which Sin (X+Pi/4) Is Increasing


We know that $cos(x) + sin(x)= sqrt2 sin(pi/4 + x)$. So, $$sin(pi/4 + x) = cos(pi/2 - x)$$ But $cos(pi/2 - x) = sin x$, so we must have $$sin(pi/4 + x)=sin(x).$$Now if I change $sin(x)$ to lớn $sin(pi - x)$ then the correct answer will come, i.e.$$sin(pi/4 + x)=sin(pi - x) implies pi/4 + x = pi -x implies x=3pi/8.$$ But why is $pi/4 + x=x$ not true. My book says, if $sin(x)=sin(y)$ then $ x = (-1)^ny + npi $. If I put $n=0$ then $x=y$ but this is not true in my case. Why?

I also looked on the relevant Wikipedia article but could not understand why these formulas are true for any value of $ heta$ that may be more that $pi/2$ & less than $0$.



My book says, if $sin(x)=sin(y)$ then $ x = (-1)^ny + npi $.

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Yes, that means if $sin(x)=sin(y)$, then there exists some $n$ such that $ x = (-1)^ny + npi $, not that you get to lớn choose which $n$, or that the statement is true for every $n$. For example,$$sin(0)=0=sin(7pi)$$and we vì indeed have that $0=(-1)^npi+npi$ for $n=7$, but that is the only $n$ that works.

More generally, you should be aware that most functions do not have the property that $f(x_1)=f(x_2)implies x_1=x_2$. Take a look at the Wikipedia page on injective functions.


See, If $sin heta =0,$ then $ heta=npi, forall n in Bbb Z$. Similarly if $cos heta=0$, then $ heta=frac (2n+1)pi2, forall n in Bbb Z$.

Now, If $sin x=sin y$, then $sin x - sin y=0$ $Rightarrow$ $2cos(fracx+y2)sin(fracx-y2)=0$ $Rightarrow$ $cos(fracx+y2)=0$ or $sin(fracx-y2)=0$.

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If $cos(fracx+y2)=0$, then $frac x+y2=frac (2n+1)pi2, forall n in Bbb Z$ $Rightarrow$ $x=(2n+1)pi-y, forall n in Bbb Z$.If $sin(fracx-y2)=0$, then $frac x-y2=npi, forall n in Bbb Z$ $Rightarrow$ $x=2npi + y, forall nin Bbb Z$.

Combining these two results, we see that $x=(-1)^ny+npi, forall n in Bbb Z. $


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