The value of \($\sin \frac{5\pi }{4}$\) and \($\cos \frac{5\pi }{4}$\) is \($-\frac{\sqrt{2}}{2}$\) and \($\tan \frac{5\pi }{4}$\) is \($1$\) .
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The unit circle has a radius of one and is centered at the origin. The unit circle's equation is
\(x^2+y^2=1\)
We have to find the value of \($\sin \frac{5\pi }{4}$\), \($\cos \frac{5\pi }{4}$\) and \($\tan \frac{5\pi }{4}$\).
Write \($\sin \left( \frac{5\pi }{4} \right)$\)as \($\sin \left( \pi +\frac{\pi }{4} \right)$\).
\($=\sin \left( \pi +\frac{\pi }{4} \right)$\)
Use the summation identity,
\($\sin \left( x+y \right)=\sin \left( x \right)\cdot \cos \left( y \right)+\cos \left( x \right)\cdot \sin \left( y \right)$\)
\($=\sin \left( \pi \right)\cdot \cos \left( \frac{\pi }{4} \right)+\cos \left( \pi \right)\cdot \sin \left( \frac{\pi }{4} \right)$\)
Use the following trivial identity,
\($\cos \left( \pi \right)=\left( -1 \right)$\)
\($\sin \left( \pi \right)=0$\)
\($\cos \left( \frac{\pi }{4} \right)=\frac{\sqrt{2}}{2}$\)
And
\($\sin \left( \frac{\pi }{4} \right)=\frac{\sqrt{2}}{2}$\)
So,
\($=0\cdot \frac{\sqrt{2}}{2}+\left( -1 \right)\cdot \frac{\sqrt{2}}{2}$\)
\($\therefore \sin \frac{5\pi }{4}=-\frac{\sqrt{2}}{2}$\)
The value for \($\cos \frac{5\pi }{4}$\) is same as \($\sin \frac{5\pi }{4}$\).
Solve for \($\tan \frac{5\pi }{4}$\).
Rewrite
\($\tan \frac{5\pi }{4}=\tan \frac{\pi }{4}$\)
Use the following trivial identity,
\($\tan \left( \frac{\pi }{4} \right)=1$\)
\($\therefore \tan \left( \frac{5\pi }{4} \right)=1$\)
Hence, the value is,
\($\sin \left( \frac{5\pi }{4} \right)=-\frac{\sqrt{2}}{2}$\),
\($\cos \left( \frac{5\pi }{4} \right)=-\frac{\sqrt{2}}{2}$\),
and
\($\tan \left( \frac{5\pi }{4} \right)=1$\)
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