Yahoo Answers: Answers and Comments for How do I find θ from equation n=4(cos(θ))^2+(sin(θ))^2? [Mathematics]
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From Doctor
enUS
Sun, 18 Aug 2019 14:38:35 +0000
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Yahoo Answers: Answers and Comments for How do I find θ from equation n=4(cos(θ))^2+(sin(θ))^2? [Mathematics]
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From lenpol7: Remember the Trig. Identity
Cos^ = 1  Sin^2 ...
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Sun, 18 Aug 2019 15:10:57 +0000
Remember the Trig. Identity
Cos^ = 1  Sin^2
Substitute
n/4 = (1  Sin^2(The) ) + Sin^2(Th)
n = 4(1  Sin^2(Th) ) + Sin^2(Th)
n = 4  4Sin^2)Th) + Sin^2(Th)
n = 4  3Sin^2(Th)
n  4 =  3Sin^2(Th)
(4  n)/3 = Sin^2(Th)
sqrt((4  n) / 3) = Sin(Th)
Theta = Sin^1(sqrt((4  n) / 3))

From Krishnamurthy: n = 4(cos(θ))^2 + (sin(θ))^2
n = 3 cos^2(θ) + 1
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Mon, 19 Aug 2019 04:27:02 +0000
n = 4(cos(θ))^2 + (sin(θ))^2
n = 3 cos^2(θ) + 1

From az_lender: +/sqrt(n)/2 = cos(t)^2 + sin(t)
= 1  sin^2(t...
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Sun, 18 Aug 2019 17:30:39 +0000
+/sqrt(n)/2 = cos(t)^2 + sin(t)
= 1  sin^2(t) + sin(t) =>
sin^2(t)  sin(t)  1 +/sqrt(n)/2 = 0.
So now you have a quadratic equation whose only unknown is sin(t).
sin(t) = (1/2) +/ (1/2)*sqrt[1 + 4 /+ 2*sqrt(n)].
This will be a real number if (a) you take the + in /+ or (b) you take the minus, but n is quite small.
Finally,
theta = arcsin[(1/2) +/ (1/2)*sqrt(5 /+ 2*sqrt(n))].
Do check my arith/algebra, but my general idea is the right one.

From RealPro: n = 4(cosθ)^2 + (sinθ)^2
(cosθ)^2 = 1  (sinθ...
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Sun, 18 Aug 2019 14:58:04 +0000
n = 4(cosθ)^2 + (sinθ)^2
(cosθ)^2 = 1  (sinθ)^2
n = 4  3(sinθ)^2
(sinθ)^2 = (4n)/3
sinθ = √[(4n)/3]
Assuming n between 1 and 4
θ = 2kπ + arcsin√[(4n)/3]
and
θ = 2(k+1)π  arcsin√[(4n)/3]
are solutions to the equation
(k is any integer)

From no sea nabo: cos²(θ)+sin²(θ)=1
cos²(θ)=(1sin²(θ))
n=4(1si...
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Sun, 18 Aug 2019 14:55:29 +0000
cos²(θ)+sin²(θ)=1
cos²(θ)=(1sin²(θ))
n=4(1sin²(θ))+sin²(θ)
n=44sin²(θ)+sin²(θ)
n=43sin²(θ)
3sin²(θ)=(4n)
sin²(θ)=(4n)/3
sin(θ)=√((4n)/3)
Arcsin(sin(θ))=Arcsin(√((4n)/3))
(θ)=Arcsin(√((4n)/3))

From TomV: 1) Use the Pythagorean Identity to express sin...
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Sun, 18 Aug 2019 14:49:34 +0000
1) Use the Pythagorean Identity to express sinΘ in terms of cosΘ (or cosΘ in terms of sinΘ)
2) Replace sin²Θ with (1cos²Θ) to obtain a quadratic equation in cosΘ
3) Use your favorite technique to solve the quadratic equation for cosΘ in terms of n
4) Θ = arccos(cosΘ)
4cos²Θ + sin²Θ = n
4cos²Θ + 1  cos²Θ = n
3cos²Θ = n1
cos²Θ = (n1)/3
cosΘ = ±√[(n1)/3] : Real solutions exist only for 1 ≤ n ≤ 4
For any integer k:
Θ₁ = arccos(√[(n1)/3]) ± 2kπ
Θ₂ = arccos(√[(n1)/3]) ± 2kπ

From alex: Rule
cos²Θ + sin²Θ = 1
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Sun, 18 Aug 2019 23:45:58 +0000
Rule
cos²Θ + sin²Θ = 1

From Anonymous: n = 4cos²(θ) + sin²θ
. .= 3cos²(θ) + (cos²(θ) ...
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Sun, 18 Aug 2019 15:35:55 +0000
n = 4cos²(θ) + sin²θ
. .= 3cos²(θ) + (cos²(θ) + sin²θ)
. .= 3cos²(θ) + 1
cos²(θ) = (n1)/3
cos(θ) = ±√[(n1)/3]
Then same as TomV's answer.

From M.: Solve for it!
How else?
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Mon, 19 Aug 2019 23:37:05 +0000
Solve for it!
How else?

From Steve A: n = 4{cos(x)^2 + sin(x)^2}
cos^2 + sin^2 = 1
n...
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Sun, 18 Aug 2019 15:36:59 +0000
n = 4{cos(x)^2 + sin(x)^2}
cos^2 + sin^2 = 1
n = 4 for all theta