How do I find θ from equation n=4(cos(θ))^2+(sin(θ))^2?

In other words, is there a way to explicitly express θ?

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  • TomV
    Lv 7
    4 months ago
    Favorite Answer

    1) Use the Pythagorean Identity to express sinΘ in terms of cosΘ (or cosΘ in terms of sinΘ)

    2) Replace sin²Θ with (1-cos²Θ) 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 = n-1

    cos²Θ = (n-1)/3

    cosΘ = ±√[(n-1)/3] : Real solutions exist only for 1 ≤ n ≤ 4

    For any integer k:

    Θ₁ = arccos(√[(n-1)/3]) ± 2kπ

    Θ₂ = arccos(-√[(n-1)/3]) ± 2kπ

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  • M.
    Lv 7
    4 months ago

    Solve for it!

    How else?

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  • 4 months ago

    n = 4(cos(θ))^2 + (sin(θ))^2

    n = 3 cos^2(θ) + 1

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  • alex
    Lv 7
    4 months ago

    Rule

    cos²Θ + sin²Θ = 1

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  • 4 months ago

    +/-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.

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  • 4 months ago

    n = 4{cos(x)^2 + sin(x)^2}

    cos^2 + sin^2 = 1

    n = 4 for all theta

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  • Anonymous
    4 months ago

    n = 4cos²(θ) + sin²θ

    . .= 3cos²(θ) + (cos²(θ) + sin²θ)

    . .= 3cos²(θ) + 1

    cos²(θ) = (n-1)/3

    cos(θ) = ±√[(n-1)/3]

    Then same as TomV's answer.

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  • 4 months ago

    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))

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  • 4 months ago

    n = 4(cosθ)^2 + (sinθ)^2

    (cosθ)^2 = 1 - (sinθ)^2

    n = 4 - 3(sinθ)^2

    (sinθ)^2 = (4-n)/3

    sinθ = √[(4-n)/3]

    Assuming n between 1 and 4

    θ = 2kπ + arcsin√[(4-n)/3]

    and

    θ = 2(k+1)π - arcsin√[(4-n)/3]

    are solutions to the equation

    (k is any integer)

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  • 4 months ago

    cos²(θ)+sin²(θ)=1

    cos²(θ)=(1-sin²(θ))

    n=4(1-sin²(θ))+sin²(θ)

    n=4-4sin²(θ)+sin²(θ)

    n=4-3sin²(θ)

    3sin²(θ)=(4-n)

    sin²(θ)=(4-n)/3

    sin(θ)=√((4-n)/3)

    Arcsin(sin(θ))=Arcsin(√((4-n)/3))

    (θ)=Arcsin(√((4-n)/3))

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