How do you prove $sin (A + B) \cdot sin (A - B) = {sin}^{2} A - {sin}^{2} B$ $sin(A+B) * sin(A-B) = sin^2 A - sin^2 B$ ?

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Answer 1 · 2015-05-25T08:47:03

The standard formula for $sin (A + B)$ $sin(A+B)$ is:

$sin (A + B) = sin (A) cos (B) + cos (A) sin (B)$ $sin(A+B) = sin(A)cos(B)+cos(A)sin(B)$

Now $sin (- B) = - sin (B)$ $sin(-B) = -sin(B)$ and $cos (- B) = cos (B)$ $cos(-B) = cos(B)$ , so

$sin (A - B) = sin (A) cos (B) - cos (A) sin (B)$ $sin(A-B) = sin(A)cos(B)-cos(A)sin(B)$

So:

$sin (A + B) \cdot sin (A - B)$ $sin(A+B)*sin(A-B)$

$= (sin A cos B + cos A sin B) (sin A cos B - cos A sin B)$ $= (sin A cos B + cos A sin B)(sin A cos B - cos A sin B)$

$= {(sin A cos B)}^{2} - {(cos A sin B)}^{2}$ $= (sin A cos B)^2 - (cos A sin B)^2$

...using the identity $(p + q) (p - q) = p^{2} - q^{2}$ $(p+q)(p-q) = p^2-q^2$ .

$= {sin}^{2} A {cos}^{2} B - {sin}^{2} B {cos}^{2} A$ $= sin^2Acos^2B-sin^2Bcos^2A$

$= {sin}^{2} A (1 - {sin}^{2} B) - {sin}^{2} B (1 - {sin}^{2} A)$ $= sin^2A(1-sin^2B)-sin^2B(1-sin^2A)$

...using ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2 theta + cos^2 theta = 1$ (Pythagoras)

$= {sin}^{2} A - {sin}^{2} B - {sin}^{2} A {sin}^{2} B + {sin}^{2} B {sin}^{2} A$ $= sin^2A-sin^2B-sin^2Asin^2B+sin^2Bsin^2A$

$= {sin}^{2} A - {sin}^{2} B$ $= sin^2A - sin^2B$

How do you prove sin(A+B)⋅sin(A−B)=sin2A−sin2Bsin(A+B) * sin(A-B) = sin^2 A - sin^2 B?