OpenStudy (anonymous):

sin(π/4)cos(3π/4)-cos(π/4)sin(3π/4)]

5 years ago
OpenStudy (anonymous):

Use your special triangles. Draw them out first.

5 years ago
OpenStudy (anonymous):

You should know off the top of your head that: pi/4 = 45 degrees

5 years ago
OpenStudy (anonymous):

To find 3π/4, simply find the beta angle of that, which is also pi/4. Depending on which quadrant it's in will determine whether it's negative or positive.

5 years ago
OpenStudy (anonymous):

Use \[\sin( a-b) = \sin(a)\cos(b) - \sin(b) \cos(b)\\ \sin(\pi/4- 3 \pi/4)=\sin(π/4)\cos(3π/4)-\cos(π/4)\sin(3π/4)]\\ \sin(- \pi/2 )=\sin(π/4)\cos(3π/4)-\cos(π/4)\sin(3π/4)]\\ -1=\sin(π/4)\cos(3π/4)-\cos(π/4)\sin(3π/4)]\\ \]

5 years ago
OpenStudy (anonymous):

\[\sin \left(\frac{\pi }{4}\right) \cos \left(\frac{3 \pi }{4}\right)-\cos \left(\frac{\pi }{4}\right) \sin \left(\frac{3 \pi }{4}\right)= \]\[\frac{1}{\sqrt{2}} \left(-\frac{1}{\sqrt{2}}\right)-\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}= \]\[-\frac{1}{2}-\frac{1}{2} \]\[-1 \]

5 years ago