Question

prove that -
Cos³x-sin³x/cosx - sinx = 1/2( 1+ sin2x)​

Answer 1

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prove that -

Cos³x-sin³x/cosx - sinx = 1/2( 1+ sin2x)

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$we \: know \: that \: =$

$(a ^{3} - {b}^{3} ) =(a - b) ( {a}^{2} + {b}^{2} + ab)$

$so - - - - >$

(cos³x - sin³x) = (cosx-sinx)(cos²x+sin²x+cosxsinx)

(cos³x - sin³x) = (cosx-sinx)(1+cosxsinx)

putting this in Question we get,

(cosx-sinx)(1+cosxsinx)/(cosx-sinx)

= (1+cosxsinx)

Multiply and divide by 2 now we get,

$\frac{1}{2} (2 + 2sinx \times cosx)$

$\frac{1}{2} (2 + sin2x)$

$\huge\blue{</strong><strong>PROVED</strong><strong>}$