If $\sin(x)=\sin(\pi/4 + x)$, then why isn't $x=x+\pi/4$?

I've been solving a question,

If $\cos(x) + \sin(x)=\sqrt{2} \cos(\pi/2 - x)$ then find the value of $x$.

We know that $\cos(x) + \sin(x)= \sqrt{2} \sin(\pi/4 + x)$. 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 $\sin(\pi - x)$ then the correct answer will come, i.e. $$\sin(\pi/4 + x)=\sin(\pi - x) \implies \pi/4 + x = \pi -x \implies x=3\pi/8.$$ But why is $\pi/4 + x=x$ not true. My book says, if $\sin(x)=\sin(y)$ then $x = (-1)^ny + n\pi$. 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 $\theta$ that may be more that $\pi/2$ and less than $0$.

• Sine is a periodic function. – Error 404 Jul 7 '15 at 16:38
• I meant to say that Sine function is periodic on $\Bbb R$ not on $\Bbb C$.:D – Error 404 Jul 7 '15 at 17:11
• @VikrantDesai I know that $\sin$ is a periodic function with period $2\pi$. – user103816 Jul 7 '15 at 17:27

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

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

• Let $\sin(x)=\sin(0)$, now $x=(−1)^nπ+nπ$ for every integer $n$. Why is this true for every $n$ but in my case only for a specific $n$? Secondly how do we find the value of $n$ for the question given in the book. – user103816 Jul 8 '15 at 10:54
• Your first sentence seems to ignore what I said in my answer (and also uses the formula wrong). Given a fixed number $x$ in advance - "fixed" meaning $x$ doesn't change, it's not being used as a variable, it's some actual number - then $$\sin(x)=\sin(0)\;\implies\; x=(-1)^n\cdot0+n\pi=n\pi\;\textsf{ for SOME integer }n$$ (As I explained in my answer.) Of course, this means is that, if I wanted to find all of the individual numbers $x$ that have the property $\sin(x)=\sin(0)$, then I would know that the set of numbers $$\{(-1)^n\cdot0+n\pi:n\in\mathbb{Z}\}=\{n\pi:n\in\mathbb{Z}\}$$ is it. – Zev Chonoles Jul 8 '15 at 16:27
• No, I didn't ignore your answer. I didn't reckon that you meant $x$ and $y$ to be fixed values. In my question $x$ is not fixed, it can be $2n\pi+3\pi/8$. I still don't get if $x$ is a variable in my question then why $x+\pi/4=x$ is wrong. My guess is that perhaps $x$ is not a complete variable-- it can take only those values which satisfy $\cos(x) + \sin(x)=\sqrt{2} \cos(\pi/2 - x)$ so $x$ is neither completely fixed nor completely variable. And again how do I find $n$ for fixed $x$ and $y$? -- do I try trial and error? – user103816 Jul 9 '15 at 9:24
• @user103816: You're interested in finding every real number $x$ with the property that $$\sin(x+\tfrac{\pi}{4})=\sin(x).$$ So now let $x$ be any particular such number - it's fixed, but we still don't know what it is. What do we know about $x$? As we've discussed, we know that this number $x$ has the property that $$\textsf{there EXISTS an integer }\,n\,\textsf{ such that }\,x+\tfrac{\pi}{4}=(-1)^nx+n\pi$$ We can see that it's impossible for $n$ to be even, but if $n$ is odd, we can rearrange to get $$\textsf{there EXISTS an ODD integer }\,n\,\textsf{ such that }\,2x=n\pi-\tfrac{\pi}{4}$$ – Zev Chonoles Jul 9 '15 at 9:39
• So the set of real numbers $x$ that are solutions to the equation, i.e. the set $$\{x\in\mathbb{R}\,:\,\sin(x+\tfrac{\pi}{4})=\sin(x)\}$$ is equal to the set $$\left\{\frac{n\pi-\tfrac{\pi}{4}}{2}\,:\,n\textsf{ is an odd integer}\right\}$$ which can be written more simply as $$\left\{k\pi+\tfrac{3\pi}{8}\,:\,k\textsf{ is an integer}\right\}$$ – Zev Chonoles Jul 9 '15 at 9:43

See, If $\sin \theta =0,$ then $\theta=n\pi, \forall n \in \Bbb Z$. Similarly if $\cos\theta=0$, then $\theta=\frac {(2n+1)\pi}{2}, \forall n \in \Bbb Z$.

Now, If $\sin x=\sin y$, then $\sin x - \sin y=0$ $\Rightarrow$ $2\cos(\frac{x+y}{2})\sin(\frac{x-y}{2})=0$ $\Rightarrow$ $\cos(\frac{x+y}{2})=0$ or $\sin(\frac{x-y}{2})=0$.

If $\cos(\frac{x+y}{2})=0$, then $\frac {x+y}{2}=\frac {(2n+1)\pi}{2}, \forall n \in \Bbb Z$ $\Rightarrow$ $x=(2n+1)\pi-y, \forall n \in \Bbb Z$. If $\sin(\frac{x-y}{2})=0$, then $\frac {x-y}{2}=n\pi, \forall n \in \Bbb Z$ $\Rightarrow$ $x=2n\pi + y, \forall n\in \Bbb Z$.

Combining these two results, we see that $x=(-1)^{n}y+n\pi, \forall n \in \Bbb Z.$