Please help me prove all these identities...thanks!!!?

Question

1. sin(pi/2+y) + sin (pi/2+[-y])= 2 cos y

2. sin (x+pi/2) / cos(x+pi/2)= -cot x

3. sin ( 90 degrees+y)+sin(-3 pi/2 -y) = 0

4. cos(y-3pi/2)= -sin y

5. sin (x-y) / sin x sin y =tan x - tan y

Answer 1

1) Choose the LHS. It's more complex

LHS = sin (pi/2 + y) + sin(pi/2 - y)
LHS = sin(pi/2)cos(y) + sin(y)cos(pi/2) +
sin(pi/2)cos(y) - sin(y)cos(pi/2)

Remember that cos(pi/2) is equal to 0, effectively eliminating them.

LHS = sin(pi/2)cos(y) + sin(pi/2)cos(y)
LHS = 2sin(pi/2)cos(y)

sin(pi/2) = 1.

LHS = 2(1)cos(y) = 2cos(y) = RHS.

2. Choose the left hand side.

LHS = [sin (x + pi/2)] / cos (x + pi/2)]
= [sin(x)cos(pi/2) + sin(pi/2)cos(x)] / [cos(x)cos(pi/2) - sin(x)sin(pi/2)]

Again, note that cos(pi/2) = 0, which simplifies things a great deal.

= [sin(pi/2)cos(x)] / [-sin(x)sin(pi/2)]

And diddo for sin(pi/2) = 1.

= cos(x) / [-sin(x)] = -cot(x) = RHS

3. Choose the LHS.

LHS = sin (90 + y) + sin (-3pi/2 - y)

Note that 90 degrees is equal to pi/2 in radians.

LHS = sin (pi/2 + y) + sin (-3pi/2 - y)
= sin(pi/2)cos(y) + sin(y)cos(pi/2) + sin(-3pi/2)cos(y) - sin(y)cos(-3pi/2)

sin(pi/2) = 1, cos(pi/2) = 0

= cos(y) + sin(-3pi/2)cos(y) - sin(y)cos(-3pi/2)

Note that -3pi/2 on the unit circle is the same as pi/2

= cos(y) + sin(pi/2)cos(y) - sin(y)cos(pi/2)
= cos(y) + (1)(cos(y)) = 2cos(y).

I didn't quite get the same answer as you. Maybe I made an error.

4. Choose the LHS.

LHS = cos (y - 3pi/2)
= cos(y)cos(3pi/2) + sin(y)sin(3pi/2)
= 0 + sin(y) (-1) = -sin(y) = RHS

Do #5 on your own because you need to practice these to do well on tests.

Answer 2

Use the identity sin(x+-y) = sin(x)cos(y)+-cos(x)sin(y)

Now, sin(y+pi/2) = sin(y)cos(pi/2) + cos(y)sin(pi/2)
= 0*sin(y)+1*cos(y) = cos(y)
Note that sin(pi/2) = 1 and cos(pi/2) = 0.

sin(pi/2-y) = sin(pi/2)cos(y) - cos(pi/2)sin(y)
= 1*cos(y)-0*sin(y) = cos(y)

so sin(pi/2+y) + sin (pi/2+[-y] = 2cos(y) from last derivations.

2) Using the derived identities above, you get
sin (x+pi/2) / cos(x+pi/2)= cos(x)/-sin(x) = -cot(x)
because cos(x+pi/2) = cos(x)cos(pi/2)-sin(x)sin(pi/2) =
cos(x)*0-sin(x)*1=-sin(x)

This is using cos(x+-y)= cos(x)cos(y)-+sin(x)sin(y)

3.sin(-3 pi/2 -y) = sin(-3pi/2) cos(y)-cos(-3pi/2)sin(y)
= -1*cos(y)-0*sin(y)= - cos(y).
so,
sin ( 90 degrees+y)+sin(-3 pi/2 -y) = cos(y)-cos(y) =0.