How do you simplify the expression $\frac{{sin}^{3} t + {cos}^{3} t}{1 - sin t cos t}$ $(sin^3t+cos^3t)/(1-sintcost)$ ?

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Answer 1 · 2016-10-26T10:30:09

$\frac{{sin}^{3} t + {cos}^{3} t}{1 - sin t cos t} = sin t + cos t$ $(sin^3t+cos^3t)/(1-sintcost)=sint+cost$

To simplify $\frac{{sin}^{3} t + {cos}^{3} t}{1 - sin t cos t}$ $(sin^3t+cos^3t)/(1-sintcost)$ , we use the identity

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

$\frac{{sin}^{3} t + {cos}^{3} t}{1 - sin t cos t}$ $(sin^3t+cos^3t)/(1-sintcost)$

= $\frac{(sin t + cos t) ({sin}^{2} t + {cos}^{2} t - sin t cos t)}{1 - sin t cos t}$ $((sint+cost)(sin^2t+cos^2t-sintcost))/(1-sintcost)$

= $\frac{(sin t + cos t) (1 - sin t cos t)}{1 - sin t cos t}$ $((sint+cost)(1-sintcost))/(1-sintcost)$

= $(sin t + cos t)$ $(sint+cost)$

How do you simplify the expression sin3t+cos3t1−sintcost(sin^3t+cos^3t)/(1-sintcost)?