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Exercise

$\frac{sen^3x+cos^3x}{senx+cosx}+senx\cdot\:\:cosx$

Step-by-step Solution

1

Factor the sum of cubes: $a^3+b^3 = (a+b)(a^2-ab+b^2)$

$\frac{\left(\sin\left(x\right)+\cos\left(x\right)\right)\left(\sin\left(x\right)^2-\sin\left(x\right)\cos\left(x\right)+\cos\left(x\right)^{2}\right)}{\sin\left(x\right)+\cos\left(x\right)}+\sin\left(x\right)\cos\left(x\right)$
2

Simplify the fraction $\frac{\left(\sin\left(x\right)+\cos\left(x\right)\right)\left(\sin\left(x\right)^2-\sin\left(x\right)\cos\left(x\right)+\cos\left(x\right)^{2}\right)}{\sin\left(x\right)+\cos\left(x\right)}$ by $\sin\left(x\right)+\cos\left(x\right)$

$\sin\left(x\right)^2-\sin\left(x\right)\cos\left(x\right)+\cos\left(x\right)^{2}+\sin\left(x\right)\cos\left(x\right)$
3

Cancel like terms $-\sin\left(x\right)\cos\left(x\right)$ and $\sin\left(x\right)\cos\left(x\right)$

$\sin\left(x\right)^2+\cos\left(x\right)^{2}$
4

Applying the pythagorean identity: $\sin^2\left(\theta\right)+\cos^2\left(\theta\right)=1$

$1$
Why is sin(x)^2 + cos(x)^2 = 1 ?

Final answer to the exercise

$1$

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