Trigonometry

Question

If 270º <x< 360º , show that √(2+√(2+2cosx)) = 2 sin x/4.

 

Thank You! :)

Answer 1

square both side to remove one square root sign

2+ sqr(2+2cosx) = 4sin^2(x/4)

move the 2 to the right side to isolate the square root term

sqr(2+2cosx) = 4sin^2(x/4) -2

square both sides again

2+2cosx = 16sin^4(x/4) - 16sin^2(x/4) + 4

subtract 2 from both sides

2cosx = 16sin^4(x/4) - 16sin^2(x/4) + 2

divide by 2

cosx = 8sin^4(x/4) - 8sin^2(x/4) + 1 use the double angle formula on the left side

cos^2(x/2) - sin^2(x/2) = 8sin^4(x/4) - 8sin^2(x/4) + 1

1-sin^2(x/2) - sin^2(x/2) = 8sin^4(x/4) - 8sin^2(x/4) + 1

-2sin^2(x/2) = 8sin^4(x/4) - 8sin^2(x/4)

divide by 2

-sin^2(x/2) = 4sin^4(x/4) - 4sin^2(x/4)

-4sin^2(x/4)(cos^2(x/4) = 4sin^4(x/4) - 4sin^2(x/4)

divide by -4sin^2(x/4) to get

cos^2(x/4) = 1-sin^2(x/4)

sin^2(x/4) + cos^2(x/4) = 1

this is an identity, the equation of a circle where any value of x satisfies the equation.

but in deriving this equation, we multiplied by variables introducing potential extraneous solutions.

Try plugging 2pi into the original equation which is the same as zero

sqr(2+sqr(2+2cosx) = 2sin(x/4)

sqr(2+sqr(4) = 0

2 = 0

one counterexample is enough to disprove an "identity"

this would have worked, if we took the negative sqr(4) and that's how we introduced extraneous solutions.