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Tutor profile: Maria A.

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Maria A.
Math tutor for 6 years
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Questions

Subject: Trigonometry

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Question:

If $$\sec A = \tan B =-3$$ such that $$A, B \in [\pi,2\pi]$$ find $$\sin(A+B)$$

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Maria A.
Answer:

Let $$\sec A = \tan B =-3$$ where $$A, B \in [\pi,2\pi]$$. Find the value of $$\sin(\A+B) = \sin A\cos B+\cos A\sin B$$. First, determine the values of $$\sin A, \cos A, \sin B$$ and $$\cos B$$. If $$\sec A =-3$$ then by reciprocal identity $$\cos A = \displaystyle {\frac{1}{\sec A} = \frac{1}{-3} = -\frac{1}{3}}$$ Since $$A\in [\pi,2\pi]$$ we assume that $$\sin A < 0$$ Using Pythagorean identity $$\cos^2 A +\sin^2 A =1$$ , $$\displaystyle{ \sin^2 A = 1-\cos^2A \qquad\Rightarrow\qquad \sin A =- \sqrt{1-\cos^2A} = -\sqrt{1-\left(-\frac{1}{3}\right)^2} =-\frac{2\sqrt{2}}{3} }$$ If $$\tan B = -3$$ then by Pythagorean identity $$\tan^2 B +1 =\sec^2 B$$ we have $$\sec^2B = (-3)^2+1 = 10 \qquad \Rightarrow \qquad \sec B = \pm \sqrt{10}$$ Since $$B\in [\pi,2\pi]$$ and $$\tan B<0$$ we assume $$\sec B>0$$ and $$\sin B <0$$. Thus, $$\displaystyle{\sec B =\sqrt{10} \qquad\Rightarrow\qquad \cos B = \frac{1}{\sec B} =\frac{1}{\sqrt{10}}}$$ and by Pythagorean identity, $$\displaystyle{\sin^2A =1-\cos^2B \qquad \Rightarrow \qquad \sin B =-\sqrt{1-\left(\frac{1}{\sqrt{10}}\right)^2}=-\sqrt{\frac{9}{10}} =-\frac{3}{\sqrt{10}}}$$ Therefore, $$\displaystyle{\sin(A+B) = \sin A\cos B + \cos A\sin B}$$ $$\displaystyle{ \qquad\qquad \;\;\;= \left(-\frac{2\sqrt{2}}{3}\right)\left(\frac{1}{\sqrt{10}}\right) +\left(-\frac{1}{3} \right)\left(-\frac{3}{\sqrt{10}} \right) }$$ $$\displaystyle{ \qquad\qquad \;\;\;=\frac{3-2\sqrt{2}}{3\sqrt{10}} }$$

Subject: Calculus

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Question:

Find the limit of $$f(x)$$ if $$3-|x| \leq f(x) \leq 3+|x|$$ by Squeeze Theorem.

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Maria A.
Answer:

Suppose that $$3-|x| \leq f(x) \leq 3+|x|$$ Let $$g(x) = 3-|x|$$ and $$h(x)=3+|x|$$. The Squeeze Theorem states that for a continuous function $$f(x)$$ if $$g(x)\leq f(x)\leq h(x)$$ such that $$\displaystyle{\lim_{x\rightarrow a} g(x) = \lim_{x\rightarrow a} h(x)=L}$$ then $$\lim_{x\rightarrow a} f(x) = L$$. Since $$\displaystyle{\lim_{x\rightarrow 0} g(x) = \lim_{x\rightarrow 0} 3-|x| = 3-0 =3 }$$ and $$\displaystyle{\lim_{x\rightarrow 0} h(x) = \lim_{x\rightarrow 0} 3+|x| = 3+0 =3 }$$ By Squeeze Theorem, $$\displaystyle{ \lim_{x\rightarrow 0} f(x) = 3}$$.

Subject: Algebra

TutorMe
Question:

Solve the given equation by completing the square: $$x^2+6x=-2 $$

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Maria A.
Answer:

Let $$x^2+6x=-2$$. Express the left side of the equation in completed square form. For quadratic equations of the form $$ax^2+bx=c$$, add both sides of the equation by $$\displaystyle{\left(\frac{b}{2}\right)^2}$$ to complete the square. Thus, $$\displaystyle{x^2+6x+\left(\frac{6}{2}\right)^2 =-2+ \left(\frac{6}{2}\right)^2}$$ Simplifying, we get $$\displaystyle{x^2+6x+9 =-2+ 9}$$ Factor out the perfect square at the left side of the equation. $$(x+3)^2 = 7$$ Take the square root on both sides of the equation. $$x+3=\pm\sqrt{7}$$ Solve for $$x$$. $$x=-3\pm\sqrt{7}$$ Therefore, the solution is $$x=\{ -3-\sqrt{7}, -3+\sqrt{7} \}$$

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