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Use the identity!!!!!!!!!!!!!!!!!!!!!!!!!!

+1 vote
Solve sin*2theta-costheta=1/2 for values between 0 degrees to 360.
asked Jan 18, 2013 in TRIGONOMETRY by linda Scholar

2 Answers

+1 vote

sin2θ - cosθ = 1/2

Note : sin2θ = 1 - cos2θ

(1 - cos2θ) -cosθ = 1/2

Multiply each side by 2.

[(1 - cos2θ) -cosθ](2) = (1/2)(2)

Simplify

2 - 2cos2θ - 2cosθ = 1

Subtract 1 from each side.

2 - 2cos2θ - 2cosθ - 1 = 1 - 1

1 - 2cos2θ - 2cosθ  = 0

Multiply each side by negative one.

-1 + 2cos2θ + 2cosθ  = 0

2cos2θ + 2cosθ - 1 = 0

Let cosθ = t

2t2 + 2t - 1 = 0

Now slove the equation factor method

x = [-b + √(b2-4ac)] / 2a

Compire the eqations ax2 + bx + c = 0

substitute a = 2, b = 2 and c = -1 in the formula

t = [-2 + √(22-4(2)(-1))] / 2(2)

t = [-2 + √(4 + 8) ] / 4

t = [-2 + √(12) ] / 4

t = [-2 + √(4)(3) ] / 4

t = [-2 + 2√(3) ] / 4

t =2 [-1 + √(3) ] / 4

t = [-1 + √(3) ] / 2

t = [-1 + √(3) ] / 2  or   t = [-1 - √(3) ] / 2

But t = cosθ

So,  cosθ= [-1 + √(3) ] / 2  or   cosθ = [-1 - √(3) ] / 2

answered Jan 19, 2013 by richardson Scholar

In the interval [0, 2π], the values of θ are π/3, π and 5π/3.

0 votes

The equation of the sin2θ - cosθ = 1/2.

1 - cos2θ - cosθ = 1/2

cos2θ + cosθ - 1 + 1/2 = 0

(2cos2θ + 2cosθ - 2 + 1)/2 = 0

2cos2θ + 2cosθ - 2 + 1 = 0

2cos2θ + 2cosθ - 1 = 0

2cos2θ + 2cosθ - cosθ - 1 = 0

2cosθ(cosθ + 1) - 1(cosθ + 1) = 0

(2cosθ - 1) (cosθ + 1) = 0

2cosθ - 1 = 0 or cosθ + 1 = 0

Take 2cosθ - 1 = 0

2cosθ = 1

cosθ = 1/2

The function cos(θ) has a period of , first find all the solutions in the interval [0, 2π]

The function cos(θ) is positive in first and fourth quadrants.

In first quadrant 0 ≤ θ ≤ π/2.

cosθ = cos(π/3)

Apply inverse of cosine to each side.

θ = π/3

In fourth quadrant 3π/2 ≤ θ ≤ 2π.

cosθ = cos(2π -  π/3) = cos((6π - π)/3) = cos(5π/3)

Apply inverse of cosine to each side.

θ = 5π/3

Take cosθ + 1 = 0.

cosθ = -1

cosθ = cos(π)

Apply inverse of cosine to each side.

θ = π.

Therefore in the interval [0, 2π], the values of θ are π/3, π and 5π/3.

answered Jul 7, 2014 by joly Scholar

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