Find θ for given condition?

Question

(sin θ + cos θ - 1)/(2 sin θ cos θ) = √(6) - 2

Find θ ?

Answer 1

multiply both sides by (2 sin θ cos θ)

sin θ + cos θ - 1 = (√(6) - 2)(2 sin θ cos θ)

sin θ + cos θ - 1 = 2√(6)sin θ cos θ - 4 sin θ cos θ

Square both sides

sin^2 θ + cos^2 θ - 1 = 12 sin^2 θ cos^2 θ - 16 sin^2 θ cos θ

sin^2 θ + cos^2 θ - 1 = - 4 sin^2 θ cos^2 θ

Using pythagorean theorem, subbing 1 - sin^2 θ for cos^2 θ

sin^2 θ + 1 - sin^2 θ - 1 = -4 sin^2 θ ( 1 - sin^2 θ)

The right hand side cancels itself out and you are left with

-4 sin^2 θ ( 1 - sin^2 θ) = 0

-4 sin^2 θ + 4 sin^4 θ = 0

4 sin^4 θ = 4 sin^2 θ

sin^4 θ = sin^2 θ

The only numbers which don't change when the exponents change to 2 or 4 are 1, -1 and 0.

Therefore θ can only be equal to pi/2, 3pi/2 or 0 radians.  (90 degrees, 270 degrees or 0 degrees.

Might have gone wrong somewhere, i did it pretty quickly and could of made stupid mistakes, if it is homework or a assignment or something like that i would check over it to see if it adds up.

Answer 2

Just to finish off from gravitate....

sinθ + cosθ = √(6)/2

square both sides

sin²θ + cos²θ + 2sinθcosθ = 6/4

1 + sin2θ = 3/2

sin2θ = 1/2

2θ = π/6 + 2nπ or 5π/6 + 2nπ

θ = π/12 + nπ or 5π/12 + nπ

Answer 3

The first answerer (and third) made mistakes in squaring:

(sinθ + cosθ - 1)²

= sin²θ + 2·sinθ·cosθ - 2·sinθ + cos²θ - 2·cosθ + 1

= 2·sinθ·cosθ - 2·sinθ - 2·cosθ + 2

Using a computer method, I got the solutions:

θ = π/12 + π·n

θ = 5π/12 + π·n

Those could possibly not be the only ones though.

If you try plugging it in, it will give you a correct solution. If you try any of the basic angles (0, π/6, π/4, etc.), you won't find a solution.

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√(6) - 2 = ( sinθ + cosθ - 1 ) / ( 2·sinθ·cosθ )

Multiply and divide by by the conjugate of the numerator:

= ( sinθ + cosθ - 1 ) · ( sinθ + cosθ + 1 )/ [ (2·sinθ·cosθ)·( sinθ + cosθ - 1) ]

= ( sin²θ + cos²θ + sinθ - sinθ + cosθ - cosθ - 1 + 2·sinθ·cosθ)/ [ (2·sinθ·cosθ)·(sinθ+cosθ-1) ]

Sines and cosines cancel:

= ( sin²θ + cos²θ + 0 + 0 - 1 + 2·sinθ·cosθ)/ [ (2·sinθ·cosθ)·(sinθ+cosθ-1) ]

Sine squared plus cosine squared is 1:

= ( 1 - 1 + 2·sinθ·cosθ)/ [ (2·sinθ·cosθ)·(sinθ+cosθ-1) ]

= 2·sinθ·cosθ / [ ( 2·sinθ·cosθ )·( sinθ + cosθ + 1 ) ] = √(6) - 2

Cancel sine times cosine:

= 1 / ( sinθ + cosθ + 1 )

Get the reciprocal of both sides:

1 / ( sinθ + cosθ + 1 ) = √(6) - 2

( sinθ + cosθ + 1 )/1 = 1/[ √(6) - 2 ]

Rationalize the denominator with the conjugate:

sinθ + cosθ + 1 = [ √(6) + 2 ]/{ [ √(6) - 2 ]·[ √(6) + 2 ] }

sinθ + cosθ + 1 = [ √(6) + 2 ]/[ √(6)² - 2² ]

sinθ + cosθ + 1 = [ √(6) + 2 ]/[ 6 - 4 ]

sinθ + cosθ + 1 = √(6)/2 + 1

sinθ + cosθ = √(6)/2

Past here, I'm sure there is some identity that simplifies it...

Answer 4

(sin θ + cos θ - 1)   ={√(6) - 2}*(2 sin θ cos θ)

NOTE (2 sin θ cos θ)  =sin2θ

  sin θ + cos θ - 1  ={√(6) - 2}*sin2θ    square both sides

   sin^2θ+cos^2θ-1^2  =[√(6) - 2}*sin2θ]^2  note sin^2θ+cos^2θ=1

    also  =1-1^2=0

      0 =[{√(6) - 2}*sin2θ]^2  square  root both sides√0=0

  0=[{√(6) - 2}*sin2θ]  I can't simplyfy further now.