How do I prove : 4sinAcosAcos2Asin15 all over. :sin2A(tan225-2sinsquared A)is equal to. :root 6 - root 2 over 2

Answer 1

First, recognize that tan225 = 1

4sinAcosAcos2Asin15/[sin2A(tan225-2sinsquared A]
= 2 sin(2A)cos(2A)sin15/[sin2A(tan225-2sin^2A]
= 2 cos(2A)sin15/(1-2sin^2A)
Now recognize that 1 - 2sin^2A = cos(2A), which cancels a term in the numerator, leaving
2 sin15 = 0.5176
That equals [6^1/2) - 2^1/2]/2
(Use the formula sin 15 = sin(45 - 30) to prove that.)
2 sin15 = 2sin45cos30 - 2cos45sin30
= sqrt2*sqrt3/2 - sqrt2*(1/2)
= (sqrt6)/2 - (sqrt2)/2

Answer 2

Note that this is an identity that is true no matter what the value of A is. We got all the A terms to cancel out.

Answer 3

I read that as
(4sinA cosA cos (2A) sin15°)/(sin (2A)(tan225° - 2sin^2 A) = (√6-√2)/2

Using the indentities:
sin 2A = 2sinAcosA and
cos 2A = 1 - 2sin^2 A

LS = (4sinAcosA(1- 2sin^2 A) sin15°)/(2sinAcosA(1-2sin^2 A) )
= 2sin15°

now we know that
sin 15° = sin(45-30)°
= sin45cos30 - cos45sin30
= (√2/2)(√3/2) - (√2/2)(1/2)
= (√6 - √2)/4
then
2sin15° = (√6-√2)/2
= RS