Solve zero of a trigonometric equation

Question

This is the equation I have:

$$\sin \left(\frac{\pi t+\frac{\pi }{2}}{2}\right)-\sin \left(\frac{\pi t}{2}\right)=0$$

Also the variable t has a requirement $0\le t\le \frac{3}{2}$. I have tried $\frac{\pi t+\frac{\pi }{2}}{2}=\frac{\pi t}{2}$ but when trying to solve for t, it t gets removed. I'm not sure what to do.

Answer 1

Just use the identity $$ \sin (a+b)=\sin(a) \cos (b) +\cos(a) \sin (b). $$

Then, in your case, you reduce your equation to: $$ \sin(\tfrac{\pi}{2}x) (\tfrac{\sqrt{2}-2}{2})+\tfrac{\sqrt{2}}{2}\cos (\tfrac{\pi}{2}x)=0 $$ or, equivalently $$ \tan y = \tfrac{\sqrt{2}}{2-\sqrt{2}} $$ where $y=\tfrac{\pi}{2}x$.

This yields that the unique solution in $(0,\tfrac{3}{2})$ is $x=\tfrac{3}{4}$.

Answer 2

Take $(\pi t/2)=x$. Use the derived formula $$\sin A-\sin B=2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac {A+B}{2}\right)$$ to get $$\sin \left(x+\frac{\pi}{4}\right)-\sin x=0 \Rightarrow2\sin\frac{\pi}{8}\cos\left(x+\frac{\pi}{8}\right)=0\\\Rightarrow\cos\left(\frac{\pi}{2}\left[ t+\frac{1}{4}\right]\right)=0$$ We know that $\cos (n\pi)=0$ (only) for all odd multiples of $\pi/2$ i.e., $$t+\frac{1}{4}=2n+1\Rightarrow t=2n+\frac{3}{4}\quad (n\in\mathbb{Z}).$$ This is the general solution. If the minimum positive value is desired, put $n=0$.

NOTE. If you can't recall this formula, derive it. Hint: $$\sin(X+Y)-\sin(X-Y)=2\sin Y\cos X$$ Put $X=(A+B)/2$ and $Y=(A-B)/2$.