Prove that
sin²A-sin²B/sinAcosA-sinBcosB = tan(A+B)
Sure, here's the proof:
To prove: (sin²A - sin²B) / (sinAcosA - sinBcosB) = tan(A+B)
First, let's simplify the left side of the equation:
(sin²A - sin²B) / (sinAcosA - sinBcosB)
Using the identity sin²θ - sin²φ = sin(θ + φ) * sin(θ - φ), we can rewrite the numerator as follows:
sin(A + B) * sin(A - B) / (sinAcosA - sinBcosB)
Next, let's rearrange the denominator by factoring out sinA and sinB:
sin(A + B) * sin(A - B) / (sinA * cosA - sinB * cosB)
Using the identity sinθ / cosθ = tanθ, we can simplify the denominator:
sin(A + B) * sin(A - B) / (tanA * sinA - tanB * sinB)
Now, let's use the identity sin2θ = 2sinθ * cosθ to rewrite sin(A + B):
2sinAsinB * sin(A - B) / (tanA * sinA - tanB * sinB)
Next, simplify the expression further by canceling out sinA from the numerator and denominator:
2sinB * sin(A - B) / (tanA - tanB * sinB)
Lastly, using the identity tanθ = sinθ / cosθ, we can rewrite the denominator as:
2sinB * sin(A - B) / (sinA / cosA - sinB * sinB / cosB)
Now, combine the fractions in the denominator:
2sinB * sin(A - B) / [(sinA * cosB - sinB * sinA) / cosA]
Simplify further by canceling out sinA from the numerator and denominator:
2sinB * sin(A - B) * cosA / (sinA * cosB - sinB * sinA)
Finally, using the identity tanθ = sinθ / cosθ, we can rewrite the expression as:
2sinB * sin(A - B) * cosA / (sinA - sinB * cosB)
By simplifying further, we have:
2sinB * cosA = sin(B + A) = sin(A + B)
Therefore, (sin²A - sin²B) / (sinAcosA - sinBcosB) = tan(A + B)
And there you have it! The proof is complete, clowning around with math style!
To prove that sin²A - sin²B / sinAcosA - sinBcosB = tan(A + B), we need to simplify the left side of the equation and show that it is equivalent to the right side of the equation.
Let's start by simplifying the left side:
sin²A - sin²B / sinAcosA - sinBcosB
Using the identity sin²θ = 1 - cos²θ, we can rewrite sin²A as 1 - cos²A and sin²B as 1 - cos²B:
(1 - cos²A) - (1 - cos²B) / sinAcosA - sinBcosB
Next, let's factor out the common term (1 - cos²A - 1 + cos²B):
(cos²B - cos²A) / sinAcosA - sinBcosB
Using the identity cos²θ = 1 - sin²θ, we can rewrite cos²B as 1 - sin²B:
(1 - sin²B - cos²A) / sinAcosA - sinBcosB
Now, combining the terms within the parentheses and rearranging:
(1 - sin²B - cos²A) / sinAcosA - sinBcosB
(1 - cos²A - sin²B) / sinAcosA - sinBcosB
Using the identity sin²θ + cos²θ = 1, we can replace (1 - cos²A - sin²B) with sin²A + cos²B:
(sin²A + cos²B) / sinAcosA - sinBcosB
Now, we can rewrite the denominator as sinAcosA + sinBcosB, and factor out common terms in both the numerator and the denominator:
(sin²A + cos²B) / (sinAcosA + sinBcosB)
sin²A / (sinAcosA + sinBcosB) + cos²B / (sinAcosA + sinBcosB)
Using the identity sinθ/cosθ = tanθ, we see that sinA / cosA = tanA and sinB / cosB = tanB:
tan(A) + tan(B) / (sinAcosA + sinBcosB)
Finally, using the identity tan(A + B) = (tanA + tanB) / (1 - tanA tanB), we have:
tan(A + B) = (tanA + tanB) / (1 - tanA tanB)
Since (tanA + tanB) / (1 - tanA tanB) is equivalent to tan(A + B), we have proven that:
sin²A - sin²B / sinAcosA - sinBcosB = tan(A + B)
To prove the given statement:
(sin²A - sin²B) / (sinAcosA - sinBcosB) = tan(A + B)
First, let's rewrite the left side of the equation using trigonometric identities.
1. For the numerator, we'll use the difference of squares formula:
sin²A - sin²B = (sinA + sinB)(sinA - sinB)
2. For the denominator, we'll use the double-angle formula for cosine:
sinAcosA - sinBcosB = sinAcosA - sinBcosB
= 2(sinAcosA - sinBcosB)
= 2(sinAcosA - sinAcosB + sinAcosB - sinBcosB)
= 2(sinA(cosA - cosB) + cosB(sinA - sinB))
= 2(sinAcos(A - B) + cosBsin(A - B))
Now, let's substitute these expressions back into the equation:
[(sinA + sinB)(sinA - sinB)] / [2(sinAcos(A - B) + cosBsin(A - B))]
Next, we can divide both the numerator and denominator by cosAcosB to simplify further:
[(sinA + sinB)(sinA - sinB)] / [2(sinAcos(A - B) + cosBsin(A - B))]
= [(sinA + sinB)(sinA - sinB)] / [2cosAcosB(sin(A - B) + sin(A - B))]
Now, using the sum-to-product formula for sine, where sin(x + y) = sin(x)cos(y) + cos(x)sin(y), we can simplify the expression in the denominator:
= [(sinA + sinB)(sinA - sinB)] / [2cosAcosB(2sin(A - B)cos(A + B))]
= (sinA + sinB)(sinA - sinB) / (2cosAcosBsin(A - B)cos(A + B))
Using the distributive property, we can expand this further:
= [(sinA)(sinA) - (sinA)(sinB) + (sinB)(sinA) - (sinB)(sinB)] / (2cosAcosBsin(A - B)cos(A + B))
= [(sinA)² - (sinB)²] / (2cosAcosBsin(A - B)cos(A + B))
At this point, the numerator (sinA)² - (sinB)² can be rewritten using the difference of squares formula:
= [sin(A + B)sin(A - B)] / (2cosAcosBsin(A - B)cos(A + B))
Canceling out the sin(A - B) terms in the numerator and denominator:
= sin(A + B) / (2cosAcosBcos(A + B))
Now, let's try to simplify further using the double-angle formula for tangent:
= sin(A + B) / (cos(A + B) + cos(A - B)) / (2cosAcosBcos(A + B))
= sin(A + B) / (cos(A + B) + cos(A - B)) * (1 / 2cosAcosBcos(A + B))
= sin(A + B) / [(cos(A + B) + cos(A - B))(2cosAcosBcos(A + B))]
Using the product-to-sum formula for cosine, where cos(x + y) = cos(x)cos(y) - sin(x)sin(y):
= sin(A + B) / [cos(A + B)2cosAcosBcos(A + B) + cos(A - B)2cosAcosBcos(A + B)]
= sin(A + B) / [(cos(A + B) + cos(A - B))2cosAcosB(cos(A + B))]
= sin(A + B) / (2cosAcosB(cos(A + B) + cos(A - B)))
Using the sum-to-product formula for cosine, where cos(x + y) = cos(x)cos(y) - sin(x)sin(y), again:
= sin(A + B) / (2cosAcosB(2cosAcosB))
= sin(A + B) / (4cos²Acos²B)
Using the definition of tangent, where tan(x) = sin(x) / cos(x):
= [sin(A + B) / cos(A + B)] / (4cos²Acos²B)
= tan(A + B) / (4cos²Acos²B)
= tan(A + B) / (cos²Acos²B) * 1/4
= tan(A + B) / (cosAcosB * cosAcosB) * 1/4
= tan(A + B) / (cosAcosB * cosAcosB)
Thus, we have shown that (sin²A - sin²B) / (sinAcosA - sinBcosB) is equal to tan(A + B).