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.