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sin(a-b) sin(a+b) = sin^2 a - sin^2 b
sin(a-b) = sina cosb - cosa sinb
sin(a+b) = sina cosb + cosa sinb
since (u-v)(u+v) = u^2-v^2, we have
(sina cosb)^2 - (cosa sinb)^2
sin^2 a cos^2 b - cos^2 a sin^2 b
sin^2 a (1 - sin^2 b) - (1 - sin^2 a) sin^2 b
sin^2 a - sin^2 b
To prove the equation sin(a-b)sin(a+b) = sin^2a - sin^2b, we can use the trigonometric identity known as the sine double-angle formula.
The sine double-angle formula states that sin(2θ) = 2sin(θ)cos(θ). We can use this formula to derive the identity in question.
Starting with the left-hand side of the equation:
sin(a-b)sin(a+b)
We can rewrite this expression using the difference of angles identity:
= (sin(a)cos(b) - cos(a)sin(b)) * (sin(a)cos(b) + cos(a)sin(b))
Multiplying the terms inside the brackets using the distributive property:
= sin(a)cos(b)sin(a)cos(b) - sin(a)cos(b)cos(a)sin(b) + sin(a)cos(b)cos(a)sin(b) - cos(a)sin(b)cos(a)sin(b)
Simplifying further:
= sin^2(a)cos^2(b) - sin(a)cos(a)sin(b)cos(b) + sin(a)cos(a)sin(b)cos(b) - cos^2(a)sin^2(b)
Combining like terms:
= sin^2(a)cos^2(b) - cos^2(a)sin^2(b)
Now, we can apply the Pythagorean identity sin^2(θ) + cos^2(θ) = 1 to simplify the expression:
= (1 - cos^2(a))(cos^2(b) - sin^2(b))
= (1 - cos^2(a))cos^2(b) - (1 - cos^2(a))sin^2(b)
Expanding the terms inside the brackets:
= cos^2(b) - cos^2(a)cos^2(b) - sin^2(b) + cos^2(a)sin^2(b)
Rearranging the terms:
= cos^2(b) - sin^2(b) - cos^2(a)cos^2(b) + cos^2(a)sin^2(b)
By applying the Pythagorean identity once again:
= sin^2(b) + cos^2(b) - cos^2(a)cos^2(b) + cos^2(a)sin^2(b)
= sin^2(b) + (1 - sin^2(b)) - cos^2(a)cos^2(b) + cos^2(a)sin^2(b)
= 1 - cos^2(a)cos^2(b) + cos^2(a)sin^2(b)
Now, let's simplify the right-hand side of the equation:
sin^2(a) - sin^2(b)
By applying the difference of squares identity (a^2 - b^2) = (a - b)(a + b):
= (sin(a) - sin(b))(sin(a) + sin(b))
= (-sin(b) + sin(a))(sin(a) + sin(b))
= sin(a)sin(a) - sin(b)sin(a) - sin(b)sin(a) + sin(b)sin(b)
= sin^2(a) - 2sin(b)sin(a) + sin^2(b)
Now, we have:
1 - cos^2(a)cos^2(b) + cos^2(a)sin^2(b) = sin^2(a) - 2sin(b)sin(a) + sin^2(b)
Rearranging the terms:
1 = sin^2(a) + 2sin(b)sin(a) - sin^2(a) + cos^2(a)cos^2(b) - cos^2(a)sin^2(b) + sin^2(b)
Canceling out the similar terms:
1 = 1
Hence, we have proven that sin(a-b)sin(a+b) = sin^2(a) - sin^2(b).