Paul Smith
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 Jun 8 awarded Popular Question Jul 2 awarded Curious Apr 2 comment Limit tends to infinity Yeah, thanks. I guess the either the textbook is wrong or my memory failed. I wonder if this exists at first, but I remembered the answer was -3/2, so. Apr 2 comment Trignometry - Cosine Formulae thanks, i did the calculations and realized that Apr 2 asked Limit tends to infinity Mar 21 comment Trignometry - Cosine Formulae thanks, ya, sine rules set limits on that also Mar 21 comment Trignometry - Cosine Formulae yes, sine law can solve it, just curious about how this happens? Mar 21 asked Trignometry - Cosine Formulae Dec 15 comment Evaluate $\lim\limits_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right)$ I realized the necessity to apply square difference identity but just fail to apply that. Thanks! Dec 15 asked Evaluate $\lim\limits_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right)$ Oct 4 asked Find the range of $\cos B \cos C$ if $A+B+C = 180^\circ$ and $A=90^\circ$ Oct 4 comment Trigonometric identities changing sum to product i don't get it, please elaborate, thanks Oct 4 asked Trigonometric identities changing sum to product Sep 29 comment Trigonometric equation $\tan4x=\sqrt 3,\qquad 0\leq x \lt \pi$ ya, thanks, i got that now Sep 29 comment Trigonometric equation $\tan4x=\sqrt 3,\qquad 0\leq x \lt \pi$ Michael. Thank you for your advice, i will pay attention to how I typed latex now. Sep 29 comment Trigonometric equation $\tan4x=\sqrt 3,\qquad 0\leq x \lt \pi$ Abdulhaq. Which two functions are not the same? Sep 28 comment Trigonometric equation $\tan4x=\sqrt 3,\qquad 0\leq x \lt \pi$ verifying textbook now, since it is quite credible, i wonder if i made that wrong Sep 28 comment Trigonometric equation $\tan4x=\sqrt 3,\qquad 0\leq x \lt \pi$ i don't know if i made compound angles right, the original question is $$tanx+tan3x+\sqrt{3}tanxtan3x=tanxtan3x$$, same range for $$\theta$$, does it justifty 3 answers? Sep 28 comment Trigonometric equation $\tan4x=\sqrt 3,\qquad 0\leq x \lt \pi$ i count 4 if I solve for $$\theta$$? $$4\pi$$ is two revolution, the $$1^{st}$$ quadrant, $$3^{rd}$$ quadrant, $$5^{th}$$ quadrant, $$7^{th}$$ quardrant Sep 28 comment Trigonometric equation $\tan4x=\sqrt 3,\qquad 0\leq x \lt \pi$ I am not following, $$\frac {20\pi}{6} = 3.3333333\pi \lt 4\pi$$?