Question

What is the sign of trigonometric value of cosec(13π/12)?

Answer 1

Answer

It has a negative sign.

Value is $\csc(\dfrac{13}{12} \pi )=\dfrac{1}{-\sin(\dfrac{\pi }{12} )}=-2$.

Explanation

$\dfrac{13\pi }{12}$ is an angle of the 3rd Quadrant.

$\dfrac{\pi }{12}$ on the 1st Quadrant forms a symmetry against the origin with $\dfrac{13\pi }{12}$.

Hence, $\sin(\dfrac{13\pi }{12} )=-\sin(\dfrac{\pi }{12} )=-\dfrac{1}{2}$, and hence, $\csc(\dfrac{13\pi }{12} )=\dfrac{1}{-\frac{1}{2} } =-2$.

More information

Interpreting angle-converting formulas graphically

For a given point $(x,y)$,

• $(-x,-y)$ is symmetric against the origin with it.
• $(x,-y)$ is symmetric against the x-axis with it.
• $(-x,y)$ is symmetric against the y-axis with it.

Assume $(x,y)$ lies on the 1st Quadrant and the unit circle, which forms an angle $\theta$ with the positive x-axis

since

• $\theta+\pi$ is symmetric against the origin
• $-\theta$ is symmetric against the x-axis
• $\pi -\theta$ is symmetric against the y-axis

we can substitute the vertices and calculate using the definition,

since the definitions are

• $\sin(\theta)=\dfrac{y}{1}$
• $\cos(\theta)=\dfrac{x}{1}$
• $\tan(\theta)=\dfrac{y}{x}$

General Cases

• $\theta+\pi$, half-rotation.
• $-\theta$, negative angle.
• $\pi-\theta$, supplementary angle.
• $\dfrac{\pi }{2} -\theta$, complementary angle.

For $\dfrac{\pi }{2} -\theta$, we can come up with a right triangle.

It is in the opposite direction with $\theta$ in a right triangle. Sine becomes cosine, cosine becomes sine, tangent becomes inverse.

Answer 2

Answer:

It has a negative sign.

Value is \csc(\dfrac{13}{12} \pi )=\dfrac{1}{-\sin(\dfrac{\pi }{12} )}=-2csc(

12

13

π)=

−sin(

12

π

)

1

=−2 .

Explanation

\dfrac{13\pi }{12}

12

13π

is an angle of the 3rd Quadrant.

\dfrac{\pi }{12}

12

π

on the 1st Quadrant forms a symmetry against the origin with \dfrac{13\pi }{12}

12

13π

.

Hence, \sin(\dfrac{13\pi }{12} )=-\sin(\dfrac{\pi }{12} )=-\dfrac{1}{2}sin(

12

13π

)=−sin(

12

π

)=−

2

1

, and hence, \csc(\dfrac{13\pi }{12} )=\dfrac{1}{-\frac{1}{2} } =-2csc(

12

13π

)=

−

2

1

1

=−2 .

More information

Interpreting angle-converting formulas graphically

For a given point (x,y)(x,y) ,

(-x,-y)(−x,−y) is symmetric against the origin with it.

(x,-y)(x,−y) is symmetric against the x-axis with it.

(-x,y)(−x,y) is symmetric against the y-axis with it.

Assume (x,y)(x,y) lies on the 1st Quadrant and the unit circle, which forms an angle \thetaθ with the positive x-axis

since

\theta+\piθ+π is symmetric against the origin

-\theta−θ is symmetric against the x-axis

\pi -\thetaπ−θ is symmetric against the y-axis

we can substitute the vertices and calculate using the definition,

since the definitions are

\sin(\theta)=\dfrac{y}{1}sin(θ)=

1

y

\cos(\theta)=\dfrac{x}{1}cos(θ)=

1

x

\tan(\theta)=\dfrac{y}{x}tan(θ)=

x

y

General Cases

\theta+\piθ+π , half-rotation.

-\theta−θ , negative angle.

\pi-\thetaπ−θ , supplementary angle.

\dfrac{\pi }{2} -\theta

2

π

−θ , complementary angle.

For \dfrac{\pi }{2} -\theta

2

π

−θ , we can come up with a right triangle.

It is in the opposite direction with \thetaθ in a right triangle. Sine becomes cosine, cosine becomes sine, tangent becomes inverse.