Question

Formula of sin square A - sin square B

Answer 1

Answer:

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Step-by-step explanation:

sin^2A+sin^2B=? sin^2A+sin^2B=1-cos(A+B).

Answer 2

Answer:

The formula for given expression is

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

Step-by-step explanation:

Given expression is

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

Using the relation of algebra,

$a {}^{2} - b {}^{2} = (a + b)(a - b)$

We write the given expression as

$sin {}^{2} A - sin {}^{2} B \\ = (sinA + sinB)(sinA - sinB)$

We have some trigonometric relations which are written below

$sin \alpha + sin \beta \\ = 2sin( \frac{ \alpha + \beta }{2} )cos( \frac{ \alpha - \beta }{2} ) \\ \\ sin \alpha - sin \beta \\ = 2cos( \frac{ \alpha + \beta }{2} )sin( \frac{ \alpha - \beta }{2} )$

Based on this relations our expression is transformed as

$sin {A}^{2} - sin {B}^{2} \\ = 2sin( \frac{A + B}{2} )cos( \frac{A - B}{2} ) \times 2cos( \frac{A + B}{2} )sin( \frac{A - B}{2} ) \\ = (2sin( \frac{A + B}{2} )cos( \frac{A + B}{2} ))(2sin( \frac{A - B}{2})(cos( \frac{A + B}{2} ))$

For a fact,we know that

$2sin \alpha cos \alpha = sin2 \alpha$

Hence,the expression becomes

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

Therefore,the formula of the given expression is obtained using algebraic and trigonometric identities.

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