MATH SOLVE

9 months ago

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 [tex]\displaystyle{ 5\frac{ \pi }{4} [/tex] by adding 5 [tex]\frac{ \pi}{4}[/tex]'s, as shown in the picture showing the unit circle.

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

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

[tex]\displaystyle{ a= \frac{ -\sqrt{2} }{2}, \ b= \frac{ -\sqrt{2} }{2}[/tex].

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

Answer:

[tex]\displaystyle{ cos:\frac{ -\sqrt{2} }{2}, \ sin= \frac{ -\sqrt{2} }{2} \ tan: 1[/tex].

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

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

[tex]\displaystyle{ a= \frac{ -\sqrt{2} }{2}, \ b= \frac{ -\sqrt{2} }{2}[/tex].

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

Answer:

[tex]\displaystyle{ cos:\frac{ -\sqrt{2} }{2}, \ sin= \frac{ -\sqrt{2} }{2} \ tan: 1[/tex].