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# Help! I'm confused with Trigonometry

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sin2θ-csc2θ=tan2θ-cot2θ

Jun 12, 2017

### 4+0 Answers

#1
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CPhill: If you think this is "rediculous", then blame it on "Mathematica 11" !!!

Solve for θ:
sin^2(θ) - csc^2(θ) = tan^2(θ) - cot^2(θ)

Subtract tan^2(θ) - cot^2(θ) from both sides:
cot^2(θ) - csc^2(θ) + sin^2(θ) - tan^2(θ) = 0

The left hand side factors into a product with five terms:
sin^2(θ) tan^2(θ) (cot(θ) - csc(θ)) (cot(θ) + csc(θ)) (cot^2(θ) csc^2(θ) + 1) = 0

Simplify and substitute x = tan^2(θ):
sin^2(θ) tan^2(θ) (cot^2(θ) csc^2(θ) + 1) (cot(θ) - csc(θ)) (cot(θ) + csc(θ)) = (-(1 + tan(θ)^2 + (tan(θ)^2)^2))/(tan(θ)^2 + 1) = -(x^2 + x + 1)/(x + 1) = 0:
-(x^2 + x + 1)/(x + 1) = 0

Multiply both sides by -1:
(x^2 + x + 1)/(x + 1) = 0

Multiply both sides by x + 1:
x^2 + x + 1 = 0

Subtract 1 from both sides:
x^2 + x = -1

Add 1/4 to both sides:
x^2 + x + 1/4 = -3/4

Write the left hand side as a square:
(x + 1/2)^2 = -3/4

Take the square root of both sides:
x + 1/2 = (i sqrt(3))/2 or x + 1/2 = -(i sqrt(3))/2

Subtract 1/2 from both sides:
x = (i sqrt(3))/2 - 1/2 or x + 1/2 = -(i sqrt(3))/2

Substitute back for x = tan^2(θ):
tan^2(θ) = (i sqrt(3))/2 - 1/2 or x + 1/2 = -(i sqrt(3))/2

Take the square root of both sides:
tan(θ) = sqrt((i sqrt(3))/2 - 1/2) or tan(θ) = -sqrt((i sqrt(3))/2 - 1/2) or x + 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:
θ = tan^(-1)(sqrt((i sqrt(3))/2 - 1/2)) + π n_1 for n_1 element Z
or tan(θ) = -sqrt((i sqrt(3))/2 - 1/2) or x + 1/2 = -(i sqrt(3))/2

(i sqrt(3))/2 - 1/2 = 1/4 + i/2 sqrt(3) - 3/4 = (1 + 2 i sqrt(3) - 3)/(4) = (1 + 2 i sqrt(3) + (i sqrt(3))^2)/(4) = ((i sqrt(3) + 1)^2)/(4):
θ = tan^(-1)(1/2 (i sqrt(3) + 1)) + π n_1 for n_1 element Z
or tan(θ) = -sqrt((i sqrt(3))/2 - 1/2) or x + 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:
θ = tan^(-1)(1/2 (i sqrt(3) + 1)) + π n_1 for n_1 element Z
or θ = π n_2 - tan^(-1)(sqrt((i sqrt(3))/2 - 1/2)) for n_2 element Z
or x + 1/2 = -(i sqrt(3))/2

(i sqrt(3))/2 - 1/2 = 1/4 + i/2 sqrt(3) - 3/4 = (1 + 2 i sqrt(3) - 3)/(4) = (1 + 2 i sqrt(3) + (i sqrt(3))^2)/(4) = ((i sqrt(3) + 1)^2)/(4):
θ = tan^(-1)(1/2 (i sqrt(3) + 1)) + π n_1 for n_1 element Z
or θ = π n_2 - tan^(-1)(1/2 (i sqrt(3) + 1)) for n_2 element Z
or x + 1/2 = -(i sqrt(3))/2

Subtract 1/2 from both sides:
θ = tan^(-1)(1/2 (i sqrt(3) + 1)) + π n_1 for n_1 element Z
or θ = π n_2 - tan^(-1)(1/2 (i sqrt(3) + 1)) for n_2 element Z
or x = -(i sqrt(3))/2 - 1/2

Substitute back for x = tan^2(θ):
θ = tan^(-1)(1/2 (i sqrt(3) + 1)) + π n_1 for n_1 element Z
or θ = π n_2 - tan^(-1)(1/2 (i sqrt(3) + 1)) for n_2 element Z
or tan^2(θ) = -(i sqrt(3))/2 - 1/2

Take the square root of both sides:
θ = tan^(-1)(1/2 (i sqrt(3) + 1)) + π n_1 for n_1 element Z
or θ = π n_2 - tan^(-1)(1/2 (i sqrt(3) + 1)) for n_2 element Z
or tan(θ) = sqrt(-(i sqrt(3))/2 - 1/2) or tan(θ) = -sqrt(-(i sqrt(3))/2 - 1/2)

Take the inverse tangent of both sides:
θ = tan^(-1)(1/2 (i sqrt(3) + 1)) + π n_1 for n_1 element Z
or θ = π n_2 - tan^(-1)(1/2 (i sqrt(3) + 1)) for n_2 element Z
or θ = tan^(-1)(sqrt(-(i sqrt(3))/2 - 1/2)) + π n_3 for n_3 element Z
or tan(θ) = -sqrt(-(i sqrt(3))/2 - 1/2)

-(i sqrt(3))/2 - 1/2 = 1/4 - i/2 sqrt(3) - 3/4 = (1 + -2 i sqrt(3) - 3)/(4) = (1 - 2 i sqrt(3) + (i sqrt(3))^2)/(4) = ((-i sqrt(3) + 1)^2)/(4):
θ = tan^(-1)(1/2 (i sqrt(3) + 1)) + π n_1 for n_1 element Z
or θ = π n_2 - tan^(-1)(1/2 (i sqrt(3) + 1)) for n_2 element Z
or θ = tan^(-1)(1/2 (-i sqrt(3) + 1)) + π n_3 for n_3 element Z
or tan(θ) = -sqrt(-(i sqrt(3))/2 - 1/2)

Take the inverse tangent of both sides:
θ = tan^(-1)(1/2 (i sqrt(3) + 1)) + π n_1 for n_1 element Z
or θ = π n_2 - tan^(-1)(1/2 (i sqrt(3) + 1)) for n_2 element Z
or θ = tan^(-1)(1/2 (-i sqrt(3) + 1)) + π n_3 for n_3 element Z
or θ = π n_4 - tan^(-1)(sqrt(-(i sqrt(3))/2 - 1/2)) for n_4 element Z

-(i sqrt(3))/2 - 1/2 = 1/4 - i/2 sqrt(3) - 3/4 = (1 + -2 i sqrt(3) - 3)/(4) = (1 - 2 i sqrt(3) + (i sqrt(3))^2)/(4) = ((-i sqrt(3) + 1)^2)/(4):
Answer: |
| θ = tan^(-1)(1/2 (i sqrt(3) + 1)) + π n_1 for n_1 element Z
or θ = π n_2 - tan^(-1)(1/2 (i sqrt(3) + 1)) for n_2 element Z
or θ = tan^(-1)(1/2 (-i sqrt(3) + 1)) + π n_3 for n_3 element Z
or θ = π n_4 - tan^(-1)(1/2 (-i sqrt(3) + 1)) for n_4 element Z

Jun 12, 2017
edited by Guest  Jun 12, 2017
#2
+98061
0

LOL!!!...I'll take their word for it......!!!!

Jun 12, 2017
#3
0

Thanks for the long time to help me but the way I do it or how I've learned is doing one of the both side to see if it proved it. Like I will do sec20-csc20 to equal the right side. They way you are doing is as if it's a equation of some sort. Maybe i'm wrong or haven't learned to do another way.

Jun 12, 2017
#4
+98061
+2

sin^2θ-csc^2θ=tan^2θ-cot^2θ

This is not an identity

We can see this from the graph, here....

https://www.desmos.com/calculator/lkwcraopqf

The graph of the left side  is not the same as the graph of the right side

Jun 12, 2017