Question

What are the sine, cosine, and tangent of 5 pi over 4 radians? sin θ = square root 2 over 2; cos θ = negative square root 2 over 2; tan θ = −1 sin θ = negative square root 2 over 2; cos θ = negative square root 2 over 2; tan θ = 1 sin θ = square root 2 over 2; cos θ = negative square root 2 over 2; tan θ = 1 sin θ = negative square root 2 over 2; cos θ = square root 2 over 2; tan θ = −1

Answer 1

Note that we can draw the angle $\displaystyle{ 5\frac{ \pi }{4}$ by adding 5 $\frac{ \pi}{4}$'s, as shown in the picture showing the unit circle.

Let the coordinates of $\displaystyle{ 5\frac{ \pi }{4}$ be (-a, -b) , then its reflection in the first quadrant is the angle $\frac{ \pi}{4}$ (45°) with coordinates (a, b).


We know that $\displaystyle{ a= \frac{ \sqrt{2} }{2}, \ b= \frac{ \sqrt{2} }{2}$, thus the coordinates (-a, -b) which are the cosine and sine of $\displaystyle{ 5\frac{ \pi }{4}$ respectively, are:


$\displaystyle{ a= \frac{ -\sqrt{2} }{2}, \ b= \frac{ -\sqrt{2} }{2}$.


From the identity tan(x)=sin(x)/cos(x), we easily see that the tangent is 1.


Answer:

$\displaystyle{ cos:\frac{ -\sqrt{2} }{2}, \ sin= \frac{ -\sqrt{2} }{2} \ tan: 1$.