Answer:
[tex]\sin\theta=-\frac{\sqrt2}{2}}[/tex] and [tex]\tan\theta=-1[/tex]
Step-by-step explanation:
Given : [tex]\cos\theta=\frac{\sqrt2}{2}[/tex] and [tex]\frac{3\pi}{2}<\theta<2[/tex]
To find : Evaluate [tex]\sin\theta[/tex] and [tex]\tan\theta[/tex] ?
Solution :
As [tex]\frac{3\pi}{2}<\theta<2[/tex] means [tex]\theta[/tex] belong to the third or fourth quadrant .
So, [tex]\sin\theta[/tex] is negative and [tex]\tan\theta[/tex] is positive in third and negative in fourth.
Using the formula of trigonometric,
[tex]\sin^2\theta+\cos^2\theta=1[/tex]
[tex]\sin\theta=\sqrt{1-\cos^2\theta}[/tex]
Substitute [tex]\cos\theta=\frac{\sqrt2}{2}[/tex] ,
[tex]\sin\theta=\pm\sqrt{1-(\frac{\sqrt2}{2})^2}[/tex]
[tex]\sin\theta=\pm\sqrt{1-\frac{2}{4}}[/tex]
[tex]\sin\theta=\pm\sqrt{1-\frac{1}{2}}[/tex]
[tex]\sin\theta=\pm\sqrt{\frac{1}{2}}[/tex]
[tex]\sin\theta=-\frac{\sqrt2}{2}}[/tex]
Now using another formula,
[tex]\tan\theta=\frac{\sin\theta}{\cos\theta}[/tex]
[tex]\tan\theta=\frac{-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}[/tex]
[tex]\tan\theta=-1[/tex]
Therefore, [tex]\sin\theta=-\frac{\sqrt2}{2}}[/tex] and [tex]\tan\theta=-1[/tex].
Answer:
The first one is B.... -sqrt(2)/ 2
The Second one is -1
Step-by-step explanation:
