Prove that
cos(A+B) + sin(A-B) = 2sin(45°+A)cos(45°+B)
LS
= cosAcosB - sinAsinB + sinAcosB - cosAsinB
RS
= 2(sin45°cosA + cos45°sinA)(cos45cosB - sin45sinB)
= 2(√2/2cosA + √2/2sinA)(√2/2cosB - √2/2sinB)
= 2( (1/2)cosAcosB - (1/2)cosAsinB + (1/2)sinAcosB - (1/2)sinAsinB)
= cosAcosB - cosAsinB + sinAcosB - sinAsinB
= LS
To prove the given identity:
cos(A+B) + sin(A-B) = 2sin(45°+A)cos(45°+B)
We begin by using the sum and difference formulas for cosine and sine:
cos(A+B) = cos(A)cos(B) - sin(A)sin(B) (1)
sin(A-B) = sin(A)cos(B) - cos(A)sin(B) (2)
Now, let's rewrite the right side of the equation in terms of those formulas:
2sin(45°+A)cos(45°+B) = 2[sin(45°)cos(A) + cos(45°)sin(A)][cos(45°)cos(B) - sin(45°)sin(B)]
Using the identity sin(45°) = cos(45°) = 1/√2, we can simplify further:
2[sin(45°)cos(A) + sin(A)][cos(45°)cos(B) - cos(45°)sin(B)]
Now, distribute the 2 to both terms:
2sin(45°)cos(A)cos(45°)cos(B) - 2sin(45°)cos(A)cos(45°)sin(B) + 2sin(A)cos(45°)cos(B) - 2sin(A)cos(45°)sin(B)
Using the identity sin(45°) = cos(45°) = 1/√2 again, we have:
(1/√2)(cos(A))(1/√2)(cos(B)) - (1/√2)(cos(A))(1/√2)(sin(B)) + (1/√2)(sin(A))(1/√2)(cos(B)) - (1/√2)(sin(A))(1/√2)(sin(B))
Simplifying further:
(cos(A)cos(B))/2 - (cos(A)sin(B))/2 + (sin(A)cos(B))/2 - (sin(A)sin(B))/2
Now, let's combine like terms:
(cos(A)cos(B) + sin(A)cos(B))/2 - (cos(A)sin(B) + sin(A)sin(B))/2
Using the sum formula for cosine:
cos(A+B)/2 - sin(A+B)/2
Substituting this into the original equation, we have:
cos(A+B) + sin(A-B) = cos(A+B)/2 - sin(A+B)/2
To continue the proof, we can derive the right-hand side of the equation using the sum formula for sine:
cos(A+B)/2 - sin(A+B)/2 = (cos(A)cos(B) - sin(A)sin(B))/2 - (sin(A)cos(B) - cos(A)sin(B))/2
Simplifying further:
(cos(A)cos(B) - sin(A)sin(B) - sin(A)cos(B) + cos(A)sin(B))/2
Combining like terms:
(cos(A)sin(B) - sin(A)cos(B) + cos(A)sin(B) - sin(A)cos(B))/2
2cos(A)sin(B) - 2sin(A)cos(B))/2 = cos(A)sin(B) - sin(A)cos(B)
Therefore, we have shown that:
cos(A+B) + sin(A-B) = 2sin(45°+A)cos(45°+B)
Hence, the given identity has been proven.
To prove the identity cos(A+B) + sin(A-B) = 2sin(45°+A)cos(45°+B), we will start from one side of the equation and manipulate it to match the other side. Let's begin:
First, let's simplify the right-hand side of the equation: 2sin(45°+A)cos(45°+B).
Using the angle sum identities, we have sin(α + β) = sinαcosβ + cosαsinβ. Applying this to our expression, we get:
2(sin(45°)cos(A)cos(45°)sin(B))
Now, we know that sin(45°) and cos(45°) are both equal to 1/√2. Substituting these values, we have:
2(1/√2)(cos(A))(1/√2)(sin(B))
Simplifying, we get:
√2(cos(A)sin(B))
Now, let's focus on the left-hand side of the equation: cos(A+B) + sin(A-B).
Using the angle sum identities, we have cos(α + β) = cosαcosβ - sinαsinβ and sin(α - β) = sinαcosβ - cosαsinβ. Applying these to our expression, we get:
cos(A)cos(B) - sin(A)sin(B) + sin(A)cos(B) - cos(A)sin(B)
Rearranging the terms, we get:
cos(A)cos(B) + sin(A)cos(B) - sin(A)sin(B) - cos(A)sin(B)
Combining like terms, we have:
(cos(A)cos(B) + sin(A)cos(B)) - (sin(A)sin(B) + cos(A)sin(B))
Now, notice that we have the same terms in both the right-hand side and left-hand side expressions. Rearranging the terms again:
(cos(A)cos(B) + sin(A)cos(B)) - (sin(A)sin(B) + cos(A)sin(B))
(cos(A)cos(B) + sin(A)cos(B)) + -(sin(A)sin(B) + cos(A)sin(B))
Applying the distributive property:
(cos(A) + sin(A))(cos(B) + sin(B))
This expression matches the right-hand side of the equation: √2(cos(A)sin(B)).
Thus, by manipulating and simplifying both sides of the equation, we have shown that cos(A+B) + sin(A-B) = 2sin(45°+A)cos(45°+B). Therefore, the identity is proven.