# 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?

### 10 Answers

- TomVLv 74 months agoFavorite 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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- KrishnamurthyLv 74 months ago
n = 4(cos(θ))^2 + (sin(θ))^2

n = 3 cos^2(θ) + 1

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- az_lenderLv 74 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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- Steve ALv 74 months ago
n = 4{cos(x)^2 + sin(x)^2}

cos^2 + sin^2 = 1

n = 4 for all theta

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- Anonymous4 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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- lenpol7Lv 74 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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- RealProLv 74 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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- no sea naboLv 64 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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