How to solve $\cos(\pi/2+t)\ge 0$?

Question

I have a trig question. How do you I solve this. I appreciate much if you could show it step by step. Find all the value of in the interval $[0,2\pi]$ for which $\cos(\pi/2+t)\ge 0$.

Answer 1

\begin{align} \cos(\frac{\pi}{2}+t)&=-\sin(t)\ge0\\ \sin(t)&\le0 \end{align}

From the sine graph, the solution is $[\pi,2\pi]$.

Or if you plot $\cos(\frac{\pi}{2}+t)$ as shown in the following graph,

the solution is also $[\pi,2\pi]$.

Answer 2

Either something from your book got copied wrong or the answer given was incorrect. Let's find the answer through a simple substition. Let $u=t+\frac\pi2$. If $t\in[0,2\pi]$, then $u\in[\frac\pi2,\frac{5\pi}2]$. Now where on this interval is $\cos u$ positive? From $\frac{3\pi}2$ to $\frac{5\pi}2$. $u\in[\frac{3\pi}2,\frac{5\pi}2]$ corresponds to $t\in[\pi,2\pi]$

Answer 3

$\cos(\frac{\pi}2+t)\ge 0 $ $\implies 2n\pi-\frac{\pi}2\le \frac{\pi}2+t\le 2n\pi+\frac{\pi}2$ where $n$ is any integer as the angle must lie in the 1st and 4th quadrant.

$\implies (2n-1)\pi\le t\le 2n\pi$

The special values are

$-\pi\le t\le 0$ for $n=0$,

$\pi \le t \le 2\pi$ for $n=1$,

$3\pi \le t \le 4\pi$ for $n=2$,

As $t$ lies in $[0,2\pi]$, the solution should be $\pi \le t \le 2\pi$.

---

Alternatively,

as $t\ge 0, (2n-1)\pi \ge 0 \implies n\ge 1$

as $t \le 2\pi, 2n\pi\le 2\pi \implies n\le 1$

So, $n=1, \pi \le t \le 2\pi $