pls. help:

prove the identity
sin2x+sin2y=2sin(x+y)cos(x-y)

LS = 2sinxcosx + 2 sinycosy

= 2[sinxcosx + sinycosy}

RS = 2[sinxcosy + cosxsiny][cosxcosy + sinxsiny]
= 2[ sinxcosxcos^2 y + sin^2 xcosysiny + cos^2 xsinycosy + sinxcosxsin^2 y]
= 2[sinxcosx(cos^2y + sin^2y) + cosysiny(sin^2x + cos^2x) ]
= 2[ sinxcosx + sinycosy]
= LS

please check my typing

To prove the identity sin2x + sin2y = 2sin(x+y)cos(x-y), we will start with the left-hand side (LHS) and simplify it to match the right-hand side (RHS).

LHS: sin2x + sin2y

Now, let's rewrite sin2x and sin2y using the double-angle formula:

sin2x = 2sinx*cosx
sin2y = 2siny*cosy

Substituting these values back into the LHS, we have:

LHS = 2sinx*cosx + 2siny*cosy

Now, we will simplify the RHS using the sum and difference identities:

2sin(x+y)cos(x-y) = 2[sinx*cosy + cosx*siny]

Now, let's compare the LHS and RHS:

LHS = 2sinx*cosx + 2siny*cosy
RHS = 2[sinx*cosy + cosx*siny]

We can see that the LHS and RHS are identical. Therefore, we have proven the identity sin2x + sin2y = 2sin(x+y)cos(x-y).

To prove the identity sin(2x) + sin(2y) = 2sin(x+y)cos(x-y), we will use the trigonometric identities for sine and cosine.

Starting with the left-hand side (LHS):

LHS = sin(2x) + sin(2y)

Using the double-angle formula for sine, sin(2θ) = 2sin(θ)cos(θ), we can rewrite the LHS as:

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

Now, we'll focus on the right-hand side (RHS):

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

Using the sum-to-product formula for sine, sin(A+B) = sin(A)cos(B) + cos(A)sin(B), we can rewrite the RHS as:

RHS = 2[sin(x)cos(y) + cos(x)sin(y)][cos(x)cos(y) + sin(x)sin(y)]

Expanding the brackets, we get:

RHS = 2[sin(x)cos(y)cos(x)cos(y) + sin(x)cos(y)sin(x)sin(y) + cos(x)sin(y)cos(x)cos(y) + cos(x)sin(y)sin(x)sin(y)]

Simplifying further:

RHS = 2[sin(x)cos(x)cos(y)cos(y) + sin(x)sin(y)cos(x)cos(y) + cos(x)sin(y)cos(x)cos(y) + sin(x)sin(y)sin(x)sin(y)]

Using the identity sin^2(θ) + cos^2(θ) = 1, we can rewrite the RHS:

RHS = 2[1 - cos^2(x)sin^2(y) - sin^2(x)sin^2(y) + cos^2(x)cos^2(y) + sin^2(x)cos^2(y)]

Simplifying further:

RHS = 2[1 - sin^2(x)sin^2(y) - sin^2(x)sin^2(y) + cos^2(x)cos^2(y) + sin^2(x)cos^2(y)]

RHS = 2[1 - 2sin^2(x)sin^2(y) + cos^2(x)cos^2(y) + sin^2(x)cos^2(y)]

Combining like terms:

RHS = 2[1 + cos^2(x)cos^2(y) + sin^2(x)cos^2(y) - 2sin^2(x)sin^2(y)]

RHS = 2[1 + (cos^2(x) + sin^2(x))cos^2(y) - 2sin^2(x)sin^2(y)]

Using the identity cos^2(θ) + sin^2(θ) = 1, we can simplify the RHS further:

RHS = 2[1 + cos^2(y)cos^2(x)cos^2(y) - 2sin^2(x)sin^2(y)]

RHS = 2[1 + cos^2(y) - 2sin^2(x)sin^2(y)]

RHS = 2[1 + cos^2(y) - 2sin^2(x)(1 - cos^2(y))]

RHS = 2[1 + cos^2(y) - 2sin^2(x) + 2sin^2(x)cos^2(y)]

RHS = 2[1 - 2sin^2(x) + 2sin^2(x)cos^2(y) + cos^2(y)]

RHS = 2[1 + cos^2(y) - 2sin^2(x) + 2sin^2(x)cos^2(y)]

RHS = 2[sin^2(y) + cos^2(y) - 2sin^2(x) + 2sin^2(x)cos^2(y)]

RHS = 2[sin^2(x) + cos^2(x)][sin^2(y) + cos^2(y)] - 2sin^2(x)

Using the identity sin^2(θ) + cos^2(θ) = 1, we can further simplify the RHS:

RHS = 2(1)(1) - 2sin^2(x)

RHS = 2 - 2sin^2(x)

Now, comparing the LHS and RHS, we see that they are equal:

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

RHS = 2 - 2sin^2(x)

Since the LHS is equal to the RHS, we have proven that sin(2x) + sin(2y) = 2sin(x+y)cos(x-y) holds true.