Verified answer

2Θ = arctan(-1) = 3π/4 + nπ : n = 0, 1, 2, 3, ...

Θ = 3π/8 + nπ/2 : n = 0, 1, 2, 3, ...

Θ = 3π/8, 7π/8, 11π/8, 15π/8, 19π/8, ... , 3π/8 + nπ/2

Θ = 67.5°, 157.5°, 247.5°, 337.5°, 427.5°, ... , 67.5° + 90n°

1) Since tan is negative,the angle must lie either in second or fourth quadrants. Taking its absolute value, 2Ө = π/4.

2) For this to be in either second or fourth quadrants, =

= (π - π/4) or (2π - π/4)

= 3π/4 or 7π/4

==> Ө = 3π/8 or 7π/8

3) In general, 2Ө = nπ - π/4, where n is an integer.

==> Ө = (nπ/2 - π/8), where n is an integer.

Example of solutions:

i) When n = 1, Ө = 3π/8

ii) When n = 2, Ө = 7π/8

iii) When n = 3, Ө = 11π/8

iv) When n = 4, Ө = 15π/8

The above four values are the soltion for Ө to be in [0, 2π]

In similar process, you may get infinite values for Ó¨.

tan 2Ó¨ = - 1 ______quadrants 2 and 4

2Ө = 135 ° , 315 ° , 495 ° , 675 °

Ө = 67.5 ° , 157.5 ° , 247.5 ° , 337.5 °

[ ± k 360 ° in each case ]

try to memorize tan,sin,cos using the graph. It easier that way.

Try the link below.

Hope that helps

Ө=3/8*pi+1/2*pi*n, n∈Z

Z is the set of integers

(arctan(-1))/2=Ó¨

Just solve... ill explain if you need help...

get tangent over by using arctan, then divide that by 2.

arctan(-1) = 45 degrees

45/2=22.5

So...

Ó¨= 22.5 degrees