show that :
sin(A+B).sin(A-B)=Cos^2B - Cos^2A =sin^2A - sin^2B
let's just tackle the LS
LS = sin(A+B)*sin(A-B)
= (sinAcosB + cosAsinB)(sinAcosB - cosAsinB)
= sin^2 Acos^2B - cos^2Asin^2B , the difference of squares pattern
= (1 - cos^2A)(cos^2B) - cos^2A(1-cos^2B)
= cos^2B - cos^2Acos^2b - cos^2A + cos^2Acos^2B
= cos^2B - cos^2A
= RS
In the line in bold, had I replaced cos^2 x with 1 - sin^2 x , I would obtain the second version of the Right Side
thank you . but what about the 3rd part of the question ? sin^2A - sin^2B what do we need to do with it?
Did you not read my post?
I told you how to do it.
oh yeas. sorry for the turmoil
To prove the given trigonometric identity:
sin(A + B) * sin(A - B) = cos^2B - cos^2A = sin^2A - sin^2B,
we can start by using the formulas for the sine and cosine of the sum and difference of two angles.
Formula for the sine of the sum of two angles:
sin(A + B) = sinA * cosB + cosA * sinB.
Formula for the sine of the difference of two angles:
sin(A - B) = sinA * cosB - cosA * sinB.
Now, let's substitute these formulas into the given expression:
(sinA * cosB + cosA * sinB) * (sinA * cosB - cosA * sinB).
To simplify, we'll expand this expression using the distributive property:
sinA * cosB * sinA * cosB - sinA * cosB * cosA * sinB + cosA * sinB * sinA * cosB - cosA * sinB * cosA * sinB.
Next, we can simplify the terms:
(sinA)^2 * (cosB)^2 - (cosA)^2 * (sinB)^2 + (cosA)^2 * (sinB)^2 - (sinB)^2 * (cosA)^2.
Using the commutative property, we can rearrange the terms:
(sinA)^2 * (cosB)^2 + (cosA)^2 * (sinB)^2 - (cosA)^2 * (sinB)^2 - (sinB)^2 * (cosA)^2.
Notice that the terms (cosA)^2 * (sinB)^2 and (sinB)^2 * (cosA)^2 are the same, but have opposite signs. Therefore, they cancel each other out:
(sinA)^2 * (cosB)^2 + (cosA)^2 * (sinB)^2 - (cosA)^2 * (sinB)^2 - (sinB)^2 * (cosA)^2 = (sinA)^2 * (cosB)^2.
Now, let's simplify further:
(sinA)^2 * (cosB)^2 = (cosA)^2 * (cosB)^2 - (1 - (cosA)^2) * (cosB)^2.
Using the identity sin^2A = 1 - cos^2A, we can rewrite (1 - (cosA)^2) as sin^2A:
= (cosA)^2 * (cosB)^2 - sin^2A * (cosB)^2.
Next, we can factor out (cosB)^2 from both terms:
= (cosB)^2 * [(cosA)^2 - sin^2A].
Using the identity cos^2A = 1 - sin^2A, we can rewrite [(cosA)^2 - sin^2A] as cos^2A:
= (cosB)^2 * cos^2A.
Finally, using the commutative property, we get:
= cos^2A * (cosB)^2.
This is equivalent to cos^2B - cos^2A.
Therefore, we have proved that sin(A + B) * sin(A - B) = cos^2B - cos^2A.
Note: To prove the second part of the given identity, sin^2A - sin^2B, you can follow a similar approach and simplify the expression using trigonometric identities.