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Albert T.
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Trigonometry
TutorMe
Question:

The positive integer value of n>3 satisfying the equation:$$\frac{1}{sin(\frac{\pi}{n})}=\frac{1}{sin(\frac{2\pi}{n})}+\frac{1}{sin(\frac{3\pi}{n})} $$ is

Albert T.
Answer:

$$\frac{1}{sin(\frac{\pi}{n})}=\frac{1}{sin(\frac{2\pi}{n})}+\frac{1}{sin(\frac{3\pi}{n})} $$ $$\Rightarrow \frac{1}{sin(\frac{\pi}{n})}-\frac{1}{sin(\frac{3\pi}{n})}=\frac{1}{sin(\frac{2\pi}{n})}$$ $$\Rightarrow \frac{sin(\frac{3\pi}{n})-sin(\frac{\pi}{n})}{sin(\frac{3\pi}{n})sin(\frac{\pi}{n})}=\frac{1}{sin(\frac{2\pi}{n})}$$ $$\Rightarrow \frac{2cos(\frac{\frac{3\pi}{n}+\frac{\pi}{n}}{2})sin(\frac{\frac{3\pi}{n}-\frac{\pi}{n}}{2})}{sin(\frac{3\pi}{n})sin(\frac{\pi}{n})}=\frac{1}{sin(\frac{2\pi}{n})}$$ $$\Rightarrow \frac{2cos(\frac{2\pi}{n})sin(\frac{\pi}{n})}{sin(\frac{3\pi}{n})sin(\frac{\pi}{n})}=\frac{1}{sin(\frac{2\pi}{n})}$$ $$\Rightarrow \frac{2cos(\frac{2\pi}{n})}{sin(\frac{3\pi}{n})}=\frac{1}{sin(\frac{2\pi}{n})}$$ $$\Rightarrow 2cos(\frac{2\pi}{n})sin(\frac{2\pi}{n})=sin(\frac{3\pi}{n})$$ $$\Rightarrow sin(\frac{4\pi}{n})=sin(\frac{3\pi}{n})$$ $$\Rightarrow \frac{4\pi}{n}=\pi-\frac{3\pi}{n}$$ $$\Rightarrow \frac{7\pi}{n}=\pi$$ $$\Rightarrow n=7 $$

Calculus
TutorMe
Question:

Solve the integral $$\displaystyle \int\limits_0^1 4x \cdot e^{x^2+3} dx$$

Albert T.
Answer:

$$ u=x^2+3, $$so that $$\frac{du}{dx}=2x $$ $$\displaystyle \int 4x \cdot e^{u} \frac{du}{2x} = 2\int e^u \, du = 2e^{x^2+3}+C $$ &\\ We now evaluate the limits $$\displaystyle \int\limits_0^1 4x \cdot e^{x^2+3} dx = 2e^4-2e^3 $$

Algebra
TutorMe
Question:

Find the minimum value of y, if for any real x $$y=(15-x)(17-x)(15+x)(17+x)$$

Albert T.
Answer:

$$y=(15-x)(17-x)(15+x)(17+x)$$ $$ =(x^{2}-225)(x^{2}-289)$$ $$=x^{4}-514x^{2}+65025$$ $$=x^{4}-2(257)x^{2}+257^{2}-257^{2}+65025$$ $$=(x^{2}-257)^{2}-66049+65025$$ $$=(x^{2}-257)^{2}-1024$$ We note that $$=(x^{2}-257)^{2}\geq 0$$, and $${min}(y)=-1024$$ when $$(x^{2}-257)^{2}=0$$.

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