Q:

# 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

Accepted Solution

A:
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.

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