Question

Will fan and medal need help asap please!
tantheta=-1.
Part 1. Write an equation that expressed the value of theta in terms of an appropriate inverse trigonometric expression
Part 2. On the interval [0,2pi], what values of theta satisfy your equation in Part 1?
Part 3. Write an expression for all solutions to the equation.

Answer 1

@jim_thompson5910 Can you give me some help? thanks!

Answer 2

we have tan(theta) = -1

what trig function is the opposite of tangent?

Answer 3

arctan? or are they referring to cotangent?

Answer 4

arctangent or inverse tangent, yes

cotangent is NOT the inverse. It is the reciprocal. We're not going to use cotangent here

Answer 5

Okay, great, that is what i thought.

Answer 6

Apply arctan to both sides to get...


tan(theta) = -1
arctan(tan(theta)) = arctan(-1)
theta = arctan(-1)

Answer 7

Okay! So then it would be theta=3pi/4 +/- npi?

Answer 8

correct

Answer 9

that will answer part 3

Answer 10

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

what did you get for part 2?

Answer 11

3pi/4 and 7pi/4?

Answer 12

correct on both

Answer 13

sweet! so part 3 would be x=3pi/4 +/- 2npi ?

Answer 14

Also, if you have time, could you quickly help me make sure I am doing the next problems right?

Answer 15

Summary:

part 1:
theta = arctan(-1)

part 2:
theta = 3pi/4 and theta = 7pi/4

part 3:
x=3pi/4 +/- n*pi

notice how it's n*pi and NOT 2n*pi

Answer 16

the n*pi has you doing half rotations
contrast that with 2n*pi is where you're doing full rotations

Answer 17

Right! thanks for that.

Answer 18

and you could drop the plus minus to just say 3pi/4 + n*pi. You'll have to state that n can be any integer (positive or negative)

Answer 19

So in another question, it has sin(theta/2)=1/2... So it would be theta/2=sin^-1(1/2), on [0,2pi] it would be theta/2= pi/6 and 5pi/6... and for theta it would = pi/12 and 5pi/12?

Answer 20

Don't apply the restriction just yet

Answer 21

sin(theta/2) = 1/2

theta/2 = arcsin(1/2)

theta/2 = pi/6+2n*pi .... or .... theta/2 = 5pi/6+2n*pi

theta = 4pi/6+4n*pi .... or .... theta = 10pi/6+4n*pi

theta = 2pi/3+4n*pi .... or .... theta = 5pi/3+4n*pi

Answer 22

Now plug in integer values of n to see which outputs are between 0 and 6.28 (2pi = 6.28 approx)

Answer 23

From here, what I did was set up a table using geogebra. The cells in blue represent the solutions. The solutions only occur when n = 0. The other values of n make solutions that are outside the interval [0,2pi)

See attached

Answer 24

Plug in n = 0 back into the formulas to get...

theta = 2pi/3+4n*pi .... or .... theta = 5pi/3+4n*pi
theta = 2pi/3+4*0*pi .... or .... theta = 5pi/3+4*0*pi
theta = 2pi/3 .... or .... theta = 5pi/3

which are the only two exact solutions in the interval [0,2pi)

Answer 25

Oh wow! That's awesome. I didn't realize that I was supposed to multiply the 2 (operation mistake). Okay, that makes sense. So if I am instructed to write an expression for all solutions to the equation, what would that be ?

Answer 26

` So if I am instructed to write an expression for all solutions to the equation, what would that be ? `

I would write `theta = 2pi/3+4n*pi .... or .... theta = 5pi/3+4n*pi`

Answer 27

and state how n is any integer (positive or negative)

Answer 28

once you state how n can be positive or negative, you don't have to worry about the plus/minus. It can simply be a plus.

Answer 29

Fantastic. Thank you ever so much!

Answer 30

oh wait, I messed up on the first piece

for some reason, I went from `theta/2 = pi/6+2n*pi` to `theta = 4pi/6+4n*pi`

when I should have gone from `theta/2 = pi/6+2n*pi` to `theta = 2pi/6+2n*pi`

my mistake

Answer 31

sin(theta/2) = 1/2

theta/2 = arcsin(1/2)

theta/2 = pi/6+2n*pi .... or .... theta/2 = 5pi/6+2n*pi

theta = 2pi/6+4n*pi .... or .... theta = 10pi/6+4n*pi

theta = pi/3+4n*pi .... or .... theta = 5pi/3+4n*pi

Answer 32

ok fixed the error

Answer 33

Great! I didn't catch the error either, so thanks so much!

Answer 34

no problem