Answer:
The answer is
\(sin(\theta)=-\frac{\sqrt{2}}{2}\)
\(tan(\theta)=-1\)
Step-by-step explanation:
we know that
\(tan(\theta)=\frac{sin(\theta)}{cos(\theta)}\)
\(sin^{2}(\theta)+cos^{2}(\theta)=1\)
In this problem we have
\(cos(\theta)=\frac{\sqrt{2}}{2}\)
\(\frac{3\pi}{2} <\theta < 2\pi\)
so
The angle \(\theta\) belong to the third or fourth quadrant
The value of \(sin(\theta)\) is negative
Step 1
Find the value of \(sin(\theta)\)
Remember
\(sin^{2}(\theta)+cos^{2}(\theta)=1\)
we have
\(cos(\theta)=\frac{\sqrt{2}}{2}\)
substitute
\(sin^{2}(\theta)+(\frac{\sqrt{2}}{2})^{2}=1\)
\(sin^{2}(\theta)=1-\frac{1}{2}\)
\(sin^{2}(\theta)=\frac{1}{2}\)
\(sin(\theta)=-\frac{\sqrt{2}}{2}\) ------> remember that the value is negative
Step 2
Find the value of \(tan(\theta)\)
\(tan(\theta)=\frac{sin(\theta)}{cos(\theta)}\)
we have
\(sin(\theta)=-\frac{\sqrt{2}}{2}\)
\(cos(\theta)=\frac{\sqrt{2}}{2}\)
substitute
\(tan(\theta)=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\)
\(tan(\theta)=-1\)
