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2h
revised Find $\tan x $ if $\sin x+\cos x=\frac12$
added 147 characters in body
2h
answered Antiderivative of $\frac{1}{1+\sin {x} +\cos {x}}$
2h
comment Trigonometry question: $\sin^2(A) + \sin^2(B) - \sin^2(C) = 2\sin(A)\sin(B)\cos(C).$
@Kanishk, Please find the edited answer
2h
revised Trigonometry question: $\sin^2(A) + \sin^2(B) - \sin^2(C) = 2\sin(A)\sin(B)\cos(C).$
added 48 characters in body
4h
answered Trigonometry question: $\sin^2(A) + \sin^2(B) - \sin^2(C) = 2\sin(A)\sin(B)\cos(C).$
5h
answered Find $\tan x $ if $\sin x+\cos x=\frac12$
5h
comment Trigonometry Question: find Value of…
@Sudhanshu, How about this?
5h
answered Trigonometry Question: find Value of…
6h
comment Trigonometry Question - Tough one
@Sudhanshu, math.stackexchange.com/questions/469599/…
6h
comment Solve for x: sin2 x − cos2 x = sin x, −π ≤ x ≤π
We have $$\sin x=\sin^2x-\cos^2x=-\cos2x=\sin\left(2x-\frac\pi2\right)$$ $$2x-\frac\pi2=n\pi+(-1)^nx$$
6h
answered Solve for $x$, $\tan x +\sec x = 2\cos x$ ; $−∞ < x < ∞$
8h
awarded  Good Answer
17h
answered Given a satisfactory real number = [any integer]/(2b) where a and b are integers, how would one find the minimum value of b?
17h
answered Valid way of evaluating limits?
18h
comment How prove this $\sum_{cyc}\frac{x+y-2z}{(x+y)^2+z^2}=0$
Write $$x+y-2z=y-z-(z-x), y+z-2x=z-x-(x-y)$$ then factor out $z-x$ etc.
18h
comment Calculate $\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$
Related: math.stackexchange.com/questions/533039/… and math.stackexchange.com/questions/633784/…
1d
answered Prove that $17$ divides $9a + 5b$
1d
comment Which one is correct for $\sqrt{-16} \times \sqrt{-1}$? $4$ or $-4$?
@cirpis, en.wikipedia.org/wiki/…
1d
comment Which one is correct for $\sqrt{-16} \times \sqrt{-1}$? $4$ or $-4$?
@cirpis, Also, when we write $a\ge0,$ we implicitly assume $a$ to be real, right as we can not say $P+iQ\ge$ or $\le0$ for real $P,Q$
1d
comment Which one is correct for $\sqrt{-16} \times \sqrt{-1}$? $4$ or $-4$?
@cirpis, Even in complex domain, for $a,b>0 \sqrt{-a}\sqrt{- b}=i\sqrt a\cdot i\sqrt b=-ab$ right? and not $\sqrt{-a}\sqrt{- b}=\sqrt{(-a)(-b)}=\sqrt{ab}$