# 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. \$ Thanks for contributing an answer lớn goodsmart.com.vnematics Stack Exchange!

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