Answer to Question #223091 in Trigonometry for Zie Rashinda

Question

Solve the following in the interval [0;2 π]

sinx > √3/2

cosx > √3/2

IcosxI ≤ 1/2

Answer 1

(i)

the value of sin is +ve in first and second quadrant so the interval where sinx is positve is $0\space to\space \pi$

now we want sinx to be greater then $\frac {\sqrt3}{2}$ ,now in first quadrant the graph of sin increases from $o\space to\space \frac{\pi}{2}$

and $sin\frac{\pi}{3}$ is $\frac {\sqrt3}{2}$ so in the interval $\frac{\pi}{3}\space to\space \frac{\pi}{2}$ sin should be greater than $\frac {\sqrt3}{2}$

now in second quadrant the graph of sin decreases since the value of $sin\frac{\pi}{2}$ is 1 and the value of $sin\frac{2\pi}{3}$ is $\frac {\sqrt3}{2}$ so the total interval in which sinx is greater than $\frac {\sqrt3}{2}\space is\space (\frac{\pi}{3}\space to\space \frac{2\pi}{3})$

(ii)

the value of cos is +ve in first and fourth quadrant so the interval where cosx is positve is $0\space to\space \frac{\pi}{2} \space and\space\frac{3\pi}{2}\space to\space 2\pi$ since cos0 is 1 and cosx is decreasing graph in first quadrant and $cos\frac{\pi}{6}\space is \space \frac {\sqrt3}{2}$

so in first quadrant cosx greater than $\frac {\sqrt3}{2}$ in interval 0 to $\frac{\pi}{6}$

now let's take fourth quadrant the value of cos increase in fourth quadrant since $cos\frac{11\pi}{6} \space is \space \frac {\sqrt3}{2}$ and $cos2\pi$ is 1, so the interval in fourth quadrant is $\frac{11\pi}{6}$ to $2\pi$

so the total interval in which cosx is greater than $\frac {\sqrt3}{2}\space is\space (0\space to\space \frac{\pi}{6}) \cup (\frac{11\pi}{6}\space to \space 2\pi)$

(iii)

let's take first quadrant graph of cosx is decrease in first quadrant and the value of $cos\frac{\pi}{6}\space is \space \frac{1}{2}$ since we need value less than equal to $\frac{1}{2}$ so the required interval will be $[\frac{\pi}{3}$ , $\frac{\pi}{2}]$

Now let's move forward to second quadrant the graph of cosx is still decrease in second quadrant but we have |cosx| the graph will be increasing the value of $|cos\frac{2\pi}{3}|=\frac{1}{2}$ so the required interval is $[\frac{\pi}{2}\space ,\space \frac{2\pi}{3}]$

Now let's move forward to 3rd quadrant, graph of cosx increase is 3rd quadrant but we have |cosx| so the graph will be decreasing and the value of $|cos\frac{4\pi}{3}|=\frac{1}{2}$ so the required interval is $[\frac{4\pi}{3}\space , \space \frac{3\pi}{2}]$

Now let's move forward to 4th quadrant in this the value of cosx is positive so |cosx| will not effect so the value of $|cos\frac{5\pi}{3}|=\frac{1}{2}$ so the required interval is $[\frac{3\pi}{2}\space ,\space \frac{5\pi}{3}]$

so the total interval in which cosx is less than equal to $\frac{1}{2} \space [\frac{\pi}{3} \space ,\space \frac{2\pi}{3}] \cup [\frac{4\pi}{3}\space , \space \frac{5\pi}{3} ]$