Answer:
a) For this case we can use the fact that [tex] sin (\pi/3) = \frac{\sqrt{3}}{2}[/tex]
And for this case since we ar einterested on [tex]-\frac{\pi}{3}[/tex] and we know that the if we are below the y axis the sine would be negative then:
[tex] sin (-\pi/3) = -\frac{\sqrt{3}}{2}[/tex]
b) From definition we can use the fact that [tex] tan x= \frac{sin x}{cos x}[/tex] and we got this:
[tex] tan (5\pi/4) = \frac{sin(5\pi/4)}{cos(5\pi/4)}[/tex]
We can use the notabl angle [tex] \pi/4[/tex] and we know that :
[tex] sin (\pi/4) = cos(\pi/4) = \frac{\sqrt{2}}{2}[/tex]
Then we know that [tex]5\pi/4[/tex] correspond to 225 degrees and that correspond to the III quadrant, and we know that the sine and cosine are negative on this quadrant. So then we have this:
[tex] tan (5\pi/4) = \frac{sin(5\pi/4)}{cos(5\pi/4)}= \frac{\frac{sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1[/tex]
Step-by-step explanation:
For this case we can use the notable angls given on the picture attached.
Part a
For this case we can use the fact that [tex] sin (\pi/3) = \frac{\sqrt{3}}{2}[/tex]
And for this case since we ar einterested on [tex]-\frac{\pi}{3}[/tex] and we know that the if we are below the y axis the sine would be negative then:
[tex] sin (-\pi/3) = -\frac{\sqrt{3}}{2}[/tex]
Part b
From definition we can use the fact that [tex] tan x= \frac{sin x}{cos x}[/tex] and we got this:
[tex] tan (5\pi/4) = \frac{sin(5\pi/4)}{cos(5\pi/4)}[/tex]
We can use the notabl angle [tex] \pi/4[/tex] and we know that :
[tex] sin (\pi/4) = cos(\pi/4) = \frac{\sqrt{2}}{2}[/tex]
Then we know that [tex]5\pi/4[/tex] correspond to 225 degrees and that correspond to the III quadrant, and we know that the sine and cosine are negative on this quadrant. So then we have this:
[tex] tan (5\pi/4) = \frac{sin(5\pi/4)}{cos(5\pi/4)}= \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1[/tex]
