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Sep
18
answered Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer.
Sep
18
comment If xsinθ = ysin(θ + 2π/3) = zsin(θ + 4π/3) then prove that Σxy = 0?
@Shubham, We have $$xy+yz+zx=xyz\left(\frac1x+\frac1y+\frac1z\right)=\frac{xyz}k\left(\sin A+\sin\left(\dfrac{2\pi}3+A\right)+\sin\left(\dfrac{4\pi}3+A\right)\right)=?$$
Sep
18
comment If xsinθ = ysin(θ + 2π/3) = zsin(θ + 4π/3) then prove that Σxy = 0?
@Shubham, If $x\sin\theta=\cdots=k, \frac1x=\frac{\sin\theta}k$ So, $$\sum xy=xyz(\sum\frac1z)=0$$
Sep
18
revised Modular calculus and square
added 59 characters in body
Sep
18
answered Modular calculus and square
Sep
18
comment Sum of the infinite series $\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \dots$
How have you determined the exponent to be $$-\frac23?$$
Sep
18
comment Sum of the series $\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\dots$
Related : math.stackexchange.com/questions/746388/…
Sep
18
answered Proving $\left(\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}\right)^n=\cos n\left(\frac{\pi }{2}-x\right)+i\sin n\left(\frac{\pi }{2-x}\right)$
Sep
18
comment Sum of the infinite series $\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \dots$
Related : math.stackexchange.com/questions/746388/…
Sep
18
answered Trig Equation - 2 years out of math & lost
Sep
18
answered Math question about solutions on intervals in trig.
Sep
18
answered If xsinθ = ysin(θ + 2π/3) = zsin(θ + 4π/3) then prove that Σxy = 0?
Sep
18
answered Showing solution of normal line
Sep
17
awarded  Nice Question
Sep
17
comment Find the remainder when $2^{561}$ is divided by $561$ using simple congruence properties.
@rsadhvika, How about this?
Sep
17
answered Find the remainder when $2^{561}$ is divided by $561$ using simple congruence properties.
Sep
17
answered Solving a question by mathematical induction
Sep
17
answered Congruences in number theory
Sep
17
comment Solve the following integration
math.stackexchange.com/questions/425603/…
Sep
17
answered If $n \in \mathbb{N} - \{1\}$, $ a \in \mathbb{Z}$, and $gcd(a,n)=1$, show there is $1 \leq i<n$ with $n|(a^i -1)$.