Question

Identity transformations of Trigonometric Expressions.
prove the following identities.
$\frac{si{n}^4\alpha -co{s}^4\alpha +co{s}^2\alpha }{2(1-cos\alpha )}=co{s}^2\frac{\alpha }{2}.$

Answer 1

$\frac{si{n}^4\alpha -co{s}^4\alpha +co{s}^2\alpha }{2(1-cos\alpha )}=co{s}^2\alpha /2$

take $LHS=\frac{si{n}^4\alpha -co{s}^4\alpha +co{s}^2\alpha }{2(1-cos\alpha )}$

$LHS=\frac{(si{n}^2\alpha {)}^2-(co{s}^2\alpha {)}^2+co{s}^2\alpha }{2(1-cos\alpha )}$ $a^2=b^2=(a-b)(a+b)$

$LHS=\frac{(si{n}^2\alpha -co{s}^2\alpha )(si{n}^2\alpha +co{s}^2\alpha )+co{s}^2\alpha }{2(1-cos\alpha )}$

$LHS=\frac{(si{n}^2\alpha -co{s}^2\alpha )+co{s}^2\alpha }{2(1-cos\alpha )}$

$LHS=\frac{si{n}^2\alpha }{2(1-cos\alpha )}=\frac{4si{n}^2\alpha /2co{s}^2\alpha /2}{2(1-1+2si{n}^2\alpha /2)}$

$sinA=2sinA/2cos\frac{cosA}{2}$

$cosA=1-2sinA/2$

$LHS=\frac{4si{n}^2\alpha /2co{s}^2\alpha /2}{4si{n}^2\alpha /2}=co{s}^2\alpha /2=RHS$