Trigonometry: Angles with sine, cosine and tangent
Special values of trigonometric functions
We have seen that the unit circle is symmetrical and that we therefore only need to know the first eighth. We will now look at special values for the first quarter. We then see the symmetry in the sine and cosine as we saw earlier. It is important to memorize these values.
#{\bf \alpha}# (in radians) |
#{\bf 0}# |
#{\bf \dfrac{\pi}{6}}# |
#{\bf \dfrac{\pi}{4}}# |
#{\bf \dfrac{\pi}{3}}# |
#{\bf \dfrac{\pi}{2}}# |
#{\bf \sin(\alpha)}# |
#0# |
#\dfrac{1}{2}# |
#\dfrac{1}{\sqrt{2}}# |
#\dfrac{\sqrt{3}}{2}# |
#1# |
#{\bf\cos(\alpha)}# |
#1# |
#\dfrac{\sqrt{3}}{2}# |
#\dfrac{1}{\sqrt{2}}# |
#\dfrac{1}{2}# |
#0# |
#{\bf \tan(\alpha)}# |
#0# |
#\dfrac{\sqrt{3}}{3}# |
#1# |
#\sqrt{3}# |
- |
Determine #\cos(\frac{7 \pi}{6})# without using your calculator.
#\cos(\frac{7 \pi}{6})=# #-\frac{1}{2}\sqrt{3}#
By use of reflections across the #x#-axis and the #y#-axis, we find #\cos(\frac{7 \pi}{6})=-\cos(\frac{\pi}{6})=-\dfrac{\sqrt{3}}{2}#.
By use of reflections across the #x#-axis and the #y#-axis, we find #\cos(\frac{7 \pi}{6})=-\cos(\frac{\pi}{6})=-\dfrac{\sqrt{3}}{2}#.
Unlock full access
Teacher access
Request a demo account. We will help you get started with our digital learning environment.
Student access
Is your university not a partner?
Get access to our courses via Pass Your Math independent of your university. See pricing and more.
Or visit omptest.org if jou are taking an OMPT exam.
Or visit omptest.org if jou are taking an OMPT exam.