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I know that the cos can be changed into 1-sin^2 theta but I dont know what to do after how do I get everything on the right hand side and simplify it?

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I know that the cos can be changed into 1-sin^2 theta but I dont know what to do after how do I get everything on the right hand side and simplify it?

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Hurkyl

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Hurkyl

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A sin x + B cos x = C

So that I may find a y such that

A = sin y

B = cos y

which allows me to use the addition of angles formula for sine.

(Note that you need to have A^2 + B^2 = 1 to be able to do this... so you may need to multiply through by a constant so that this will be true)

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xanthym

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√2*sin(θ) = √3 - cos(θ)aisha said:

I know that the cos can be changed into 1-sin^2 theta but I dont know what to do after how do I get everything on the right hand side and simplify it?

⇒ √2*{1 - cos^2(θ)}^(1/2) = √3 - cos(θ) (<--- sin^2() + cos^2() = 1)

⇒ 2*{1 - cos^2(θ)} = 3 - 2*√3*cos(θ) + cos^2(θ) (<--- Squaring Both Sides)

⇒ 3*cos^2(θ) - 2*√3*cos(θ) + 1 = 0

⇒ {cos(θ) - 1/√3}^2 = 0

⇒ θ = arccos(1/√3)

⇒ Possible solutions: θ = {(0.9553 rad)=(54.736 deg)} or {(5.328 rad)=(305.264 deg)}

Checking original equation, only first possibility is solution:

~~

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the solutions are 55 degrees and 305 degrees I found the last post to be a little confusing

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If you do understand the last post then all you need to do is take the 55° and then minus it from 360° and then you will end up with both answers.aisha said:the solutions are 55 degrees and 305 degrees I found the last post to be a little confusing

The Bob (2004 ©)

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Ouabache

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Where did you become confused? Did you understand the substitutionxanthym said:√2*sin(θ) = √3 - cos(θ)

⇒ √2*{1 - cos^2(θ)}^(1/2) = √3 - cos(θ) (<--- sin^2() + cos^2() = 1)

⇒ 2*{1 - cos^2(θ)} = 3 - 2*√3*cos(θ) + cos^2(θ) (<--- Squaring Both Sides)

⇒ 3*cos^2(θ) - 2*√3*cos(θ) + 1 = 0

⇒ {cos(θ) - 1/√3}^2 = 0

⇒ θ = arccos(1/√3)

⇒ Possible solutions: θ = {(0.9553 rad)=(54.736 deg)} or {(5.328 rad)=(305.264 deg)}

Checking original equation, only first possibility is solution:

θ = {(0.9553 rad)=(54.736 deg)}

[tex]\sin\theta = [1 - cos^2\theta]^\frac{1}{2} [/tex]

With that substitution, the equation became in terms of [tex]\cos\theta [/tex].

After squaring both sides and rearranging terms, xanthym rewrote the equation so it looks like:

[tex] ax^2+bx+c = 0 [/tex] ....... [tex] 3cos^2\theta - 2\sqrt{3}\cos\theta + 1 = 0 [/tex]

Can you factor that equation directly?

If that looks too tricky, :surprised

do you know another formula you can use to find roots of an equation in that form?

Do you understand taking the arccos (inverse cosine) of a value to find the angle?

(hint: to find both angles, remember there will be more than one quadrant between 0 and 2pi, where cosθ has the same sign).

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Daniel.

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Thanks it took me a little time to absorb but i do understand

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Daniel.

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