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Tan(theta)= -1

Part1. Write an equation that expresses the value of theta in terms of an appropriate inverse trigonometric expression

Part2. On the interval [0,2pi], what values of theta satisfy your equation in part 1?

Part3. Write an expression for all solutions to the equation

Respuesta :

we are given

[tex] tan(\theta)=-1 [/tex]

(I)

Since, we have to solve for theta

so, we will take tan^-1 both sides

[tex] tan^{-1}(tan(\theta))=tan^{-1}(-1) [/tex]

we can cancel tan with tan^-1

we get

[tex] \theta=tan^{-1}(-1) [/tex]

(II)

we can use unit circle

we know that

tan is negative in second and fourth quadrant

so, we get

[tex] \theta=\frac{3\pi}{4} [/tex]

[tex] \theta=\frac{7\pi}{4} [/tex]

(III)

We know that time period of tan is pi

It means after pi , it repeats itself

so, we can add pi*k to get general solution

where k can be any integers

so, our solution is

[tex] \theta=\frac{3\pi}{4}+\pi k [/tex]

[tex] \theta=\frac{7\pi}{4}+\pi k [/tex].................Answer

we are given

tan(\theta)=-1

(I)

Since, we have to solve for theta

so, we will take tan^-1 both sides

tan^{-1}(tan(\theta))=tan^{-1}(-1)

we can cancel tan with tan^-1

we get

\theta=tan^{-1}(-1)

(II)

we can use unit circle

we know that

tan is negative in second and fourth quadrant

so, we get

\theta=\frac{3\pi}{4}  

\theta=\frac{7\pi}{4}  

(III)

We know that time period of tan is pi

It means after pi , it repeats itself

so, we can add pi*k to get general solution

where k can be any integers

so, our solution is

\theta=\frac{3\pi}{4}+\pi k  

\theta=\frac{7\pi}{4}+\pi k  .................Answer

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