Question

sd

Answer 1

Hint: \[\frac{5\pi}{4} = \pi + \frac{\pi}{4}\]

Answer 2

\[\sin(\pi+\theta) = -\sin(\theta)\]\[\cos(\pi+\theta) = -\cos(\theta)\]

Answer 3

@jhalt
The angle is :
\[\frac54\pi\]
This angle is in the 3rd quarter of the plan because :
\[1<\color{red}{\frac54}<\frac32\]
So we can write :
\[\frac54\pi=\pi+\alpha\]
Where alpha is the reference angle .
So :
\[\alpha=\frac54\pi-\pi=\frac\pi4\]
Now for finding the cosine and the sine of the angle we use the reference angle like this :
\[\cos\frac{5\pi}4=\cos\left(\pi+\frac\pi4\right)=-\cos\frac\pi4\\
\sin\frac{5\pi}4=\sin\left(\pi+\frac\pi4\right)=-\sin\frac\pi4\]

Answer 4

and isnt the sin and cosine -sqrt(2)/2

Answer 5

There are 4 quarter :
1st :
\[0<\theta<\frac\pi2\]
2nd :
\[\frac\pi2<\theta<\pi\]
And in this quarter we can write :
\[\theta=\pi-\alpha\]
where alpha is the reference angle. And we have :
\[\cos\theta=\cos(\pi-\alpha)=-\cos\alpha\\
\sin\theta=\sin(\pi-\alpha)=\sin\alpha\]

3rd :
\[\pi<\theta<\frac{3\pi}2\]
And in this quarter we can write :
\[\theta=\pi+\alpha\]
where alpha is the reference angle. And we have :
\[\cos\theta=\cos(\pi+\alpha)=-\cos\alpha\\
\sin\theta=\sin(\pi+\alpha)=-\sin\alpha\]

4th :
\[-\frac\pi2<\theta<0\]
And in this quarter we can write :
\[\theta=-\alpha\]
where alpha is the reference angle. And we have :
\[\cos\theta=\cos(-\alpha)=\cos\alpha\\
\sin\theta=\sin(-\alpha)=-\sin\alpha\]