Prove the following {sin}^(2) ( (\pi)/(4) - \alpha ) = (1)/(2) (1 - sin2 \alpha ) ​

Question

Prove the following


`{sin}^(2) ( (\pi)/(4) - \alpha ) = (1)/(2) (1 - sin2 \alpha )`
​

Answer 1

trig identity proof

Using the trigonometric identity for the sine of the difference of two angles, we have:

sin(a - b) = sin(a)cos(b) - cos(a)sin(b)

Substituting a = π/4 and b = α, we get:

sin(π/4 - α) = sin(π/4)cos(α) - cos(π/4)sin(α)

sin(π/4 - α) = (1/√2)(cos(α) - sin(α))

Squaring both sides, we get:

sin^2(π/4 - α) = 1/2(cos^2(α) - 2cos(α)sin(α) + sin^2(α))

sin^2(π/4 - α) = 1/2(1 - sin(2α))

This proves the first equation.

For the second equation, we use the double angle formula for the sine:

sin(2x) = 2sin(x)cos(x)

Substituting x = 2π - α, we get:

sin(4π - 2α) = 2sin(2π - α)cos(2π - α)

sin(4π - 2α) = 2(-sin(α))(-cos(α))

sin(4π - 2α) = 2sin(α)cos(α)

Dividing both sides by 2sin^2(α), we get:

sin(4π - 2α)/(2sin^2(α)) = cos(α)/sin(α)

csc(4π - 2α) = cot(α)

Using the identity csc(x) = 1/sin(x) and simplifying, we get:

sin(4π - 2α) = (1 - sin^2(α))/sin(α)

sin(4π - 2α) = cos^2(α)/sin(α)

sin(4π - 2α) = (1 - sin^2(α))(1/sin(α))

sin(4π - 2α) = 1/sin(α) - sin(α)

Substituting the value of sin^2(π/4 - α) we found earlier, we get:

sin(4π - 2α) = 1/sin(α) - (1/2)(1 - sin(2α))

sin(4π - 2α) = (1/2)(1 + sin(2α))/sin(α)

This proves the second equation.