For which values of a is y = sin(x + pi/4 + a) an odd function?

Step-by-step solution

1. 1

   Write the condition for oddness. A function is odd when $f(-x)=-f(x)$ for every $x$. Collecting the constants into a single phase $c=\tfrac{\pi}{4}+a$, the function is $f(x)=\sin(x+c)$ and the question becomes: for which $c$ is $\sin(x+c)$ odd?

2. 2

   Find which phases keep the sine odd. The bare $\sin x$ is odd, and adding a multiple of $\pi$ preserves that:

   $\sin(x+k\pi)=(-1)^{k}\sin x,$

   which is $\pm\sin x$ — still an odd function. Any other phase mixes in a cosine, since $\sin(x+c)=\sin x\cos c+\cos x\sin c$, and the $\cos x\sin c$ part is even, destroying the symmetry unless $\sin c=0$.

3. 3

   Confirm the condition is also necessary. Testing at $x=0$: an odd function must satisfy $f(0)=-f(0)$, hence $f(0)=0$. Here $f(0)=\sin c$, so $\sin c=0$ and therefore

   $c=k\pi,\qquad k\in\mathbb{Z}.$

   The single value $x=0$ is enough to pin the condition down completely, and step 2 shows it is sufficient as well.

4. 4

   Solve for $a$. From $c=\tfrac{\pi}{4}+a=k\pi$:

   $a=k\pi-\frac{\pi}{4},\qquad k\in\mathbb{Z}.$

   The smallest positive value is $k=1$, giving $a=\tfrac{3\pi}{4}$; taking $k=0$ gives $a=-\tfrac{\pi}{4}$, which returns the function to plain $\sin x$.

5. 5

   Verify numerically. With $a=\tfrac{3\pi}{4}$ the function is $\sin(x+\pi)=-\sin x$. Checking oddness at $x=0.7$: $f(0.7)+f(-0.7)=2\times10^{-16}\approx0$ ✓. For contrast, $a=0$ gives $\sin\!\left(x+\tfrac{\pi}{4}\right)$, and there $f(0)=\tfrac{\sqrt2}{2}\ne0$, so it is neither odd nor even.