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An approach. One may write \begin{align} \sum_{k=1}^nk\sin(ka)&=\sum_{k=1}^n\frac{d}{da}(-\cos(ka)) \\\\&=-\frac{d}{da}\sum_{k=0}^n\cos(ka) \\\\&=-\frac{d}{da}\left(\frac{1}{2}+\frac{\sin\left[(n+\frac12)a\right]}{2\sin \frac a2} \right) \\\\&=\frac{(n+1) \sin(na)-n \sin((n+1)a)}{4\sin^2 \frac a2} \end{align} then one may take $a:=2^{\... 24 There is an interesting trick: you may couple the two equations by writing $$e^{ix}+e^{iy} = i \tag{1}$$ hence$e^{ix}$and$e^{iy}$, that are two points on the unit circle, are simmetric with respect to the imaginary axis. By imposing that their sum has unit norm, we clearly get$\{x,y\}=\left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}$: 15 Taking @SiongthyeGoh's comment, in the above picture,$AE=1$,$\angle BAE=x$,$\angle BAC=\beta$, and$\angle BAD=\alpha$. It follows that$DE=\sin(x+\alpha)$and$CE=\sin(x+\beta)$. Applying the cosine theorem to$\triangle DCE$, the given expression is equal to$DC^2$, hence is a constant. 8 Hint. One may write, for$n\ge2$, $$\frac{4n}{n^4-2n^2+2}=\frac{4n}{(n^2-1)^2+1}=\frac{\frac{4n}{(n^2-1)^2}}{1+\frac1{(n^2-1)^2}}=\frac{\frac1{(n-1)^2}-\frac1{(n+1)^2}}{1+\frac1{(n-1)^2(n+1)^2}}$$ giving here $$\arctan\frac{4n}{n^4-2n^2+2}=\arctan\frac1{(n-1)^2}-\arctan\frac1{(n+1)^2}.$$ 8 Let $$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\color {blue}{2\cos(\alpha-\beta)\sin(x+\alpha)}\sin(x+\beta)$$ Now use$2\cos A \sin B = \sin (A+B) - \sin (A-B)$. $$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(\alpha-\beta +x+\alpha\right)-\sin\left(\alpha-\beta -x-\alpha\right)\right]\sin(x+\beta)$$ $$E=\sin^2(x+\alpha)+\sin^2(x+\beta)-\left[\sin\left(... 7 It's sometimes a good idea to turn a sum of sines and cosines into a single trigonometric term when solving equations of the form a \sin x + b\cos x = c, in this case:$$\sqrt{3}\cos v - \sin v \equiv 2\cos \left(v + \frac{\pi}{6}\right)$$So, if you set x = v + \frac{\pi}{6} you need only solve \cos x = \frac{1}{\sqrt{2}} which you can then do, I'm ... 6 HINT: Let A=x+y, B=x-y \tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} Alternatively,$$x+y=m\pi+\arctan3,x-y=n\pi+\arctan2$$where m,n are arbitrary integers. Using Inverse trigonometric function identity doubt: \tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right), when x<0, y<0, and xy>1,$$\arctan3+\arctan2=\... 6 HINT: For real$a>0,$$$(a^2)^{\sin^2\theta}+(a^2)^{\cos^2\theta}=a\left(a^{-\cos2\theta}+a^{\cos2\theta}\right)$$ Now$\dfrac{a^{-\cos2\theta}+a^{\cos2\theta}}2\ge\sqrt{a^{-\cos2\theta}\cdot a^{\cos2\theta}}=1$Can you identify$ahere? 5 Since $$\sin { x } =2\sin { \frac { x }{ 2 } } \cos { \frac { x }{ 2 } } \\ \cos { x } =\cos ^{ 2 }{ \frac { x }{ 2 } } -\sin ^{ 2 }{ \frac { x }{ 2 } }$$ so $$\cos \left( x \right) +\sin \left( x \right) =\frac { 7 }{ 5 } \\ 5\left( \cos ^{ 2 }{ \frac { x }{ 2 } -\sin ^{ 2 }{ \frac { x }{ 2 } } } \right) +10\sin { \frac { x }{ 2 } } \cos { \... 5 Hint:- Let$$I=\displaystyle\int x\sin^2 x\ dxJ=\displaystyle\int x\cos^2 x\ dx$$Now observe that,$$I+J=\displaystyle\int x\ dx$$and$$I-J=\Re\left(\displaystyle\int xe^{2ix}\ dx\right)$$where \Re\left(\displaystyle\int xe^{2ix}\ dx\right) denotes the real part of the integral. 5 Hints: Your sum is related with:$$ S=\sum_{n=1}^{90} n \sin\left(\frac{\pi n}{90}\right) = \sum_{n=0}^{89}(90-n)\sin\left(\frac{\pi n}{90}\right)\tag{1}$$that fulfills:$$ \color{red}{2\,S} = 90\sum_{n=1}^{89}\sin\left(\frac{\pi n}{90}\right) = \color{red}{90\cdot\cot\left(\frac{\pi}{180}\right)}.\tag{2}4 Use \sin(\theta)=\sin(180^\circ-\theta). Let the summation as S. Then \begin{align} 2S&=\sum_{k=1}^{90}(k\sin(2k^\circ)+(90-k)\sin((180-2k)^\circ)) \\ &=90\sum_{k=1}^{90}\sin(2k^{\circ}) \end{align} Now use -2\sin(2k^\circ)\sin(1^\circ)=\cos((2k+1)^\circ)-\cos((2k-1)^\circ), then \begin{align} S=45\sum_{k=1}^{90}\sin(2k^{\circ})=-\frac{45}{2\... 4 Divide the equation by \;2\; :\frac1{\sqrt2}=-\frac12\sin x+\frac{\sqrt3}2\cos x=\sin\frac\pi3\cos x-\cos\frac\pi3\sin x=\sin\left(\frac\pi3-x\right)\implies\frac\pi3-x=\begin{cases}\cfrac\pi4\\{}\\\cfrac{3\pi}4\end{cases}\implies \ldots$$Observe this is similar to the other answer but, perhaps, a little, very little, easier to understand. 4 This simple and elegant solution relies on two results from Closed form for \sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}. and/or from Closed form for \sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}. both of which are derived from the well known Mittag-Leffler expansion of hyperbolic cotangent function (denote \mathbb{W}=\mathbb{Z}/\{0\}) :$$\sum_{n\... 4 In fact, we have \begin{align} I(M,N)&=\int_0^\infty\frac{\sin Nx\sin(N+1)x\sin Mx\cos(M+1)x}{x\sin^2 x\cosh x}\ dx\\[10pt] &=\sum_{m=1}^M\sum_{n=1}^N\left[\arctan\left( e^{(m+n)\pi} \right)-\arctan\left( e^{(m-n)\pi} \right)\right]\\[10pt] &=\frac{1}{2}\sum_{m=1}^M\sum_{n=1}^N\bigg[\operatorname{gd}\!\big((m+n)\pi\big)-\operatorname{gd}\!\big(... 4 Your expression got reduced to 1\mp\sqrt5. But remember, this expression is the value of \dfrac{1}{\sin18^{\circ}}. Since \sin18^{\circ} is not negative, it's reciprocal cannot be negative. Thus, we discard the value 1-\sqrt{5}. Thus, the correct value is - \dfrac{1}{\sin18^{\circ}} = 1+\sqrt{5} \sin18^{\circ} = \dfrac{1}{1+\sqrt{5}} = \dfrac{1}... 4 HINT Use the formulas \sin^2t=\frac {1-\cos2t} 2 and \cos2p+\cos2q=2\cos(p+q)\cos(p-q). 4 Hint: x = \tan \left(\frac{\theta}{2}\right) 4 Using t-formula Let \displaystyle t=\tan \frac{x}{2}, then \displaystyle \cos x=\frac{1-t^2}{1+t^2} and \displaystyle \tan x=\frac{2t}{1-t^2}. Now \begin{align*} \frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2} &=4 \\ \frac{(1+t)^{2}}{1-t^2} &= 4 \\ \frac{1+t}{1-t} &= 4 \quad \quad (t\neq -1) \\ t &= \frac{3}{5} \\ \tan \frac{x}{2} &... 4 One may observe that summing u_{r+1}-u_{r-1} $$may be simplified as telescoping sum:$$ \sum_{r=1}^N\left(u_{r+1}-u_{r-1}\right)=\sum_{r=1}^N\left(u_{r+1}-u_{r}\right)+\sum_{r=1}^N\left(u_{r}-u_{r-1}\right)=u_{N+1}+u_N-u_1-u_0. $$From the identity$$ \arctan \left(\frac{4}{r^2+3} \right)=\arctan \left(\frac{r+1}2 \right)-\arctan \left(\frac{r-1}2 ... 3 $$\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=1}^N\arctan\left(\frac{r-1}2\right)=\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=-1}^{N-2}\arctan\left(\frac{r+1}2\right)$$ $$=\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\arctan\left(0\right)-\arctan\left(\frac{1}2\right)+\arctan\left(\frac{N}2\right)+\... 3$$\text {Let }\quad L=\sin^2 (x+a)+\sin^2(x+b).\text {Let }\quad R=-2\cos (a-b)\sin (x+a) \sin (x+b).\text {We have }\quad L=\frac {1}{2}(1-\cos (2x+2a) +\frac {1}{2}(1-\cos (2x+2b).$$Now with y=2x+a+b and z=a-b we have$$\cos (2x+2a)=\cos (y+z)=\cos y \cos z-\sin y \sin z. \cos (2x+2b)=\cos (y-z)=\cos y \cos z +\sin y \sin z .\... 3 $$F=\sin^2(x+\alpha)+\sin^2(x+\beta)-2\cos(\alpha-\beta)\sin(x+\alpha)\sin(x+\beta)$$ $$=1-\{\underbrace{\cos^2(x+\alpha)-\sin^2(x+\beta)}\}-\cos(\alpha-\beta)\{\underbrace{2\sin(x+\alpha)\sin(x+\beta)}\}$$ Using Prove that\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$and Werner Formula$(2\sin A\sin B=\cdots),$$$D=1-\cos(2x+\alpha+\beta)\cos(\... 3 If a,b\ge 0 then$$a+b\ge 2\sqrt{ab}2^{\sin^2\theta}+2^{\cos^2\theta}\ge 2\sqrt{2^{\sin^2\theta+\cos^2\theta}}=2\sqrt{2}$$3 Observe that$$\tan \frac{7\pi}{8}=-\tan\left( \pi -\frac{7\pi}{8} \right)$$So$$\tan \frac{7\pi}{8}=-\tan\frac{\pi}{8} $$Now use$$\tan 2\theta =\frac{2\tan \theta }{1-\tan^2 \theta}$$Let$$\theta = \frac{\pi}{8}$$So$$\tan \left(2\cdot\frac{\pi}{8} \right)=\frac{2\tan \frac{\pi}{8} }{1-\tan^2 \frac{\pi}{8}}\tan \frac{\pi}{4} =1=\frac{2\tan \... 3 It's easier to use $$\tan\dfrac{7\pi}8=\dfrac{1-\cos\dfrac{7\pi}4}{\sin\dfrac{7\pi}4}$$ Now$\sin\dfrac{7\pi}4=\sin\left(2\pi-\dfrac\pi4\right)=-\sin\dfrac\pi4=-\dfrac1{\sqrt2}$and$\cos\dfrac{7\pi}4=\cos\left(2\pi-\dfrac\pi4\right)=\cos\dfrac\pi4=+\dfrac1{\sqrt2}3 Glad to see that you have made some original discoveries although these are pretty well known results in elementary calculus. It would be interesting to know as to how you arrived at them. Using differential and integral calculus it is easy to prove the following three formulas: \begin{align} \sin x &= x - \frac{x^{3}}{3!} + \cdots + (-1)^{n}\frac{x^{2n +... 3 No, it isn't possible with the identity in your question because\tan \frac\pi2$is not defined. It isn't defined because$\cos \frac\pi 2 = 0$, and$\tan x = \frac{\sin x}{\cos x}$in general. Actually, it's exactly this fact ($\tan x = \frac{\sin x}{\cos x}\$) that you'll want to use. I'll get you started: $$\tan\left(\frac{\pi}{2} - x\right) = \... 3 Since \sin is a monotone increasing function on the interval (0,\frac\pi2), and since 0<\frac35<1, then we know, that there can only be one value of \alpha in this interval for which \sin(\alpha)=\frac35, since \sin will only take that value once. Since \cos is a monotone decreasing function on the interval (0,\pi) and since -1<-\... 3 We have two well-known Pythagorean triples:$$3^2+4^2=5^2 ,\qquad 5^2+12^2=13^2\tag{1}$$hence \sin\alpha=\frac{3}{5} and 0\leq\alpha\leq\frac{\pi}{2} implies \cos\alpha=\frac{4}{5}, as well as \cos\beta=-\frac{12}{13} and \frac{\pi}{2}\leq\beta\leq\pi implies \sin\beta=\frac{5}{13}. So, by the cosine sum formula:$$ \cos(\alpha+\beta)=\cos\alpha\...