Respuesta :
that comes to sin 3 theta + cos 3 theta + 2 sin theta cos theta + 1
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sin theta + cos theta
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sin theta + cos theta
[tex]\bf \textit{difference and sum of cubes}
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a^3+b^3 = (a+b)(a^2-ab+b^2)
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a^3-b^3 = (a-b)(a^2+ab+b^2)
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\textit{also recall that }sin^2(\theta)+cos^2(\theta)=1\\\\
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\cfrac{sin^3(\theta )+cos^3(\theta )}{sin(\theta )+cos(\theta )}+sin(\theta )cos(\theta )[/tex]
[tex]\bf \cfrac{\underline{[sin(\theta )+cos(\theta )]}~~[sin^2(\theta )-sin(\theta )cos(\theta )+cos^2(\theta )]}{\underline{sin(\theta )+cos(\theta )}}+sin(\theta )cos(\theta ) \\\\\\ \cfrac{sin^2(\theta )-sin(\theta )cos(\theta )+cos^2(\theta )}{1}+sin(\theta )cos(\theta ) \\\\\\ sin^2(\theta )\underline{-sin(\theta )cos(\theta )}+cos^2(\theta )\underline{+sin(\theta )cos(\theta )} \\\\\\ sin^2(\theta )+cos^2(\theta )\implies 1[/tex]
[tex]\bf \cfrac{\underline{[sin(\theta )+cos(\theta )]}~~[sin^2(\theta )-sin(\theta )cos(\theta )+cos^2(\theta )]}{\underline{sin(\theta )+cos(\theta )}}+sin(\theta )cos(\theta ) \\\\\\ \cfrac{sin^2(\theta )-sin(\theta )cos(\theta )+cos^2(\theta )}{1}+sin(\theta )cos(\theta ) \\\\\\ sin^2(\theta )\underline{-sin(\theta )cos(\theta )}+cos^2(\theta )\underline{+sin(\theta )cos(\theta )} \\\\\\ sin^2(\theta )+cos^2(\theta )\implies 1[/tex]
