please help/ trigonometry?

Question

the point value of each question is in parenthesis
5.

c. sin ( x+y ) cos ( x-y ) + cos( x+y ) sin ( x-y ) = sin 2x


d. cos 2x = cot^2 x-1 / cot ^2 x+1

Answer 1

sin(x + y) = sin(x)cos(y) + sin(y)cos(x)
sin(x - y) = sin(x)cos(y) - sin(y)cos(x)
cos(x - y) = cos(x)cos(y) + sin(x)sin(y)
cos(x + y) = cos(x)cos(y) - sin(x)sin(y)

sin(x + y)cos(x - y) + cos(x + y)sin(x - y) = sin(2x)

((sin(x)cos(y) + sin(y)cos(x))(cos(x)cos(y) + sin(x)sin(y))) +
((cos(x)cos(y) - sin(x)sin(y))(sin(x)cos(y) - sin(y)cos(x)))

(sin(x)cos(x)cos(y)^2 + sin(y)cos(y)sin(x)^2 + sin(y)cos(y)cos(x)^2 + sin(x)cos(x)sin(y)^2) + (sin(x)cos(x)cos(y)^2 - sin(y)cos(y)cos(x)^2 - sin(y)cos(y)sin(x)^2 + sin(x)cos(x)sin(y)^2)

sin(x)cos(x)cos(y)^2 + sin(y)cos(y)sin(x)^2 + sin(y)cos(y)cos(x)^2 + sin(x)cos(x)sin(y)^2 + sin(x)cos(x)cos(y)^2 - sin(y)cos(y)cos(x)^2 - sin(y)cos(y)sin(x)^2 + sin(x)cos(x)sin(y)^2

(1 + 1)sin(x)cos(x)cos(y)^2 + (1 - 1)sin(y)cos(y)sin(x)^2 + (1 - 1)sin(y)cos(y)cos(x)^2 + (1 + 1)sin(x)cos(x)sin(y)^2

2sin(x)cos(x)cos(y)^2 + 2sin(x)cos(x)sin(y)^2

(2sin(x)cos(x))(cos(y)^2 + sin(y)^2)

cos(y)^2 + sin(y)^2 = 1

2sin(x)cos(x)

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

so

sin(x + y)cos(x - y) + cos(x + y)sin(x - y) = sin(2x)

--------------------------------------------------------------------------

(cot(x)^2 - 1)/(cot(x)^2 + 1)
(((cos(x)/sin(x))^2 - 1)/(((cos(x)/sin(x))^2 + 1)
((cos(x)^2 - sin(x)^2)/sin(x)^2)/((cos(x)^2 + sin(x)^2)/sin(x)^2)
((cos(x)^2 - sin(x)^2)/sin(x)^2) * ((sin(x)^2)/(cos(x)^2 + sin(x)^2))
(cos(x)^2 - sin(x)^2)/(cos(x)^2 + sin(x)^2)

cos(x)^2 - sin(x)^2 = cos(2x)
cos(x)^2 + sin(x)^2 = 1

so

(cot(x)^2 - 1)/(cot(x)^2 + 1) = cos(2x)

Answer 2

you need to use the special angle identitis and substitute them in for both problems. use this site

http://www.sosmath.com/trig/trig.html