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Tutor profile: David S.

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David S.
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Questions

Subject: Pre-Calculus

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Question:

Prove the following relationship using trigonometric identities: $(\sin(107^{\circ})\cos(39^{\circ})-\cos(107^{\circ})\sin(39^{\circ})=\cos(22^{\circ})$)

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David S.
Answer:

We can see that the left side of the equation is in the form of the sine angle difference identity: $$\sin(A-B)=\sin A\cos B-\cos A\sin B$$. $(\sin(107^{\circ}-39^{\circ})=\cos(22^{\circ})$) Simplifying, we get $(\sin(68^{\circ})=\cos(22^{\circ})$) Using the cofunction identity between sine and cosine $(\sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta\\\sin(90^{\circ}-x^{\circ})=\cos(x^{\circ})$) $(\sin(68^{\circ})=\sin(90^{\circ}-22^{\circ})\quad\implies\quad \cos(22^{\circ})$)

Subject: Calculus

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Question:

Solve the following indefinite integral: $({\displaystyle\int}-\mathrm{e}^{\cos\left(x\right)}\sin\left(x\right)\cos\left(\cos\left(x\right)\right)\,\mathrm{d}x$)

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David S.
Answer:

First, we do a simple u-sub where $$u=\cos(x)$$ and $$\mathrm{d}u=-\sin(x)\, \mathrm{d}x$$ $({\displaystyle\int}\mathrm{e}^u\cos\left(u\right)\,\mathrm{d}u$) Then, we integrate by parts with $$f=\cos(u)$$ and $$g'=\mathrm{e}^u\, \mathrm{d}u$$. This means $$f'=-\sin(u)\,\mathrm{d}u$$ and $$g=\mathrm{e}^u$$. $({\displaystyle\int}fg'\implies fg-{\displaystyle\int}f'g$) Substituting, we get $(\mathrm{e}^u\cos\left(u\right)-{\displaystyle\int}-\mathrm{e}^u\sin\left(u\right)\,\mathrm{d}u\\\mathrm{e}^u\cos\left(u\right)+{\displaystyle\int}\mathrm{e}^u\sin\left(u\right)\,\mathrm{d}u$) Then, we integrate by parts again, with $$j=\sin(u)$$ and $$k'=\mathrm{e}^u\,\mathrm{d}u$$. This means $$j'=\cos(u)\,\mathrm{d}u$$ and $$k=\mathrm{e}^u$$. $(\mathrm{e}^u\cos\left(u\right)+{\displaystyle\int}jk'\implies \mathrm{e}^u\cos\left(u\right)+jk-{\displaystyle\int}j'k$) Substituting, we get $(\mathrm{e}^u\cos\left(u\right)+\mathrm{e}^u\sin(u)-{\displaystyle\int}\mathrm{e}^u\cos(u)\,\mathrm{d}u$) Now, we appear stuck, since we've ended up with the same integral we started with (after the initial u-sub). However, if we realize that we can set what we started with equal to our result: $({\displaystyle\int}\mathrm{e}^u\cos\left(u\right)\,\mathrm{d}u=\mathrm{e}^u\cos\left(u\right)+\mathrm{e}^u\sin(u)-{\displaystyle\int}\mathrm{e}^u\cos(u)\,\mathrm{d}u$) We can add the integral to both sides, yielding $(2{\displaystyle\int}\mathrm{e}^u\cos\left(u\right)\,\mathrm{d}u=\mathrm{e}^u\cos\left(u\right)+\mathrm{e}^u\sin(u)$) Which means our u-sub integral equals: $({\displaystyle\int}\mathrm{e}^u\cos\left(u\right)\,\mathrm{d}u=\frac{\mathrm{e}^u\cos\left(u\right)+\mathrm{e}^u\sin(u)}{2}$) Finally, substituting in our original u-sub values gives $({\displaystyle\int}-\mathrm{e}^{\cos\left(x\right)}\sin\left(x\right)\cos\left(\cos\left(x\right)\right)\,\mathrm{d}x=\dfrac{\mathrm{e}^{\cos\left(x\right)}\sin\left(\cos\left(x\right)\right)+\mathrm{e}^{\cos\left(x\right)}\cos\left(\cos\left(x\right)\right)}{2}$) Lastly, to complete our answer, since the integral is indefinite (no bounds) we must add a constant of integration, which we'll call $$C$$. $(\mathrm{Answer:}\qquad \dfrac{\mathrm{e}^{\cos\left(x\right)}\sin\left(\cos\left(x\right)\right)+\mathrm{e}^{\cos\left(x\right)}\cos\left(\cos\left(x\right)\right)}{2}+C$)

Subject: Algebra

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Question:

Find all values of $$x$$ that make the following statement true. Your answer can be in terms of $$y$$ (assume $$y$$ is some real number). $(-5x^2+2x^3=18xy^2-45y^2$)

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David S.
Answer:

First, we move all the terms to one side so that we have a polynomial equation equal to zero. $(-5x^2+2x^3-18xy^2+45y^2=0$) Next, we can check whether any of our factoring techniques can be applied here. From the order it's in currently, we can clearly see that there are common factors that can be taken out by grouping (namely an $$x^2$$ and a $$9y^2$$). $(x^2 (-5+2x)+9y^2(-2x+5)=0$) We can now see that our second group looks $$\textit{similar}$$ but has the wrong signs. This means we simply take out a negative from one of the groups. It doesn't matter which we choose, so let's go with the first one. $(-x^2 (5-2x)+9y^2(-2x+5)=0\\-x^2 (-2x+5)+9y^2(-2x+5)=0$) Now, we continue our factoring to get $((-x^2+9y^2)\: (-2x+5)=0\\(9y^2-x^2)\: (-2x+5)=0$) The first set of parenthesis can be factored, because it is a difference of perfect squares. $((3y+x)\: (3y-x)\: (-2x+5)=0$) Setting each parenthesis equal to zero yields $(3y+x=0\\3y-x=0\\-2x+5=0$) So, our three answers are $(x=\frac{5}{2},\quad x=-3y,\quad x=3y$)

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