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

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 $\cos(\pi/2+t)\ge 0$.

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The answer in my solution book is : t E [pi,3pi/2] or t E [3pi/2,2pi]. Can anyone tell me how to reach the answer? – Michael Sep 28 '12 at 9:56

## 3 Answers

\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 $[\pi,2\pi]$.

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Thanks for the fast reply, but the answer in my solution book is - t E [pi,3pi/2] or t E [3pi/2,2pi]. I don't know how it get it. – Michael Sep 28 '12 at 9:49
@Michael: your book has the same answer, it's just written complicated. Think about it; $t \in [\pi, 3\pi/2]$ or $t \in [3\pi/2, 2\pi]$ is the same thing as $t \in [\pi, 2\pi]$. – Javier Badia Sep 28 '12 at 12:32

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 $t\in[\pi,2\pi]$

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I double check it. It is. – Michael Sep 28 '12 at 10:20

$\cos(\frac{\pi}2+t)\ge 0$ $\implies 2n\pi-\frac{\pi}2\le \frac{\pi}2+t\le 2n\pi+\frac{\pi}2$ 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, $n=1, \pi \le t \le 2\pi$

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