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You're making the mistake of not converting everything into degrees and just the $\pi$ into degrees. $$(3\pi-10) rad = (540-10\times 57.296)^{\circ}$$ $$= -32.96^{\circ} \in [-90^{\circ},90^{\circ}]$$

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$\sin\theta+\cos\theta=\dfrac{\sqrt3}{\sqrt2}$ and $\sin\theta\cos\theta=\dfrac k{\sqrt2}$. $(\sin\theta+\cos\theta)^2-2\sin\theta\cos\theta=1$ $\dfrac32-\sqrt2 k=1$ $k=\dfrac 1{2\sqrt2}$

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This requires the use of the Laplace transform. We set $$J(t;q)=q\int_0^\infty \frac{\cos(tx)dx}{x^2+q^2}.$$ Then we lake the Laplace transform of it: \begin{align} \mathcal{L}\{J(t;q)\}(s)&=\int_0^\infty e^{-st}J(t;q)dt\\ &=q\int_0^\infty \int_0^\infty \frac{e^{-st}\cos(tx)}{q^2+x^2}dxdt\\ &=q\int_0^\infty \frac{1}{x^2+q^2}\int_0^\infty e^{-... 3 I'll leave it to you to prove by induction that the partial sum is \arctan\left(1-\frac{1}{n^2+n+1}\right), so the limit is \pi/4. One approach to obtaining this partial sum is that of @achillehui's telescope, viz.\left[\arctan(2r^2+2r+1)\right]_{-1}^n=\arctan\frac{n^2+n}{n^2+n+1}.$$Edit: just to spell it out, the definition f(r):=\arctan(2r^2+2r+1)... 2 For the minimum, note that since x,y\ge0\implies y\le\dfrac{2\pi}3, we have$$\dfrac{x-y}2=\dfrac{x+y-2y}2=\dfrac{\dfrac{2\pi}3-2y}2=\frac\pi3-y$$so$$\sin x+\sin y = \sqrt{3}\cos\frac{x-y}{2}=\sqrt3\cos\left(\frac\pi3-y\right)\ge\begin{cases}\sqrt3\cos\left(\frac\pi3-0\right)\\\sqrt3\cos\left(\frac\pi3-\frac{2\pi}3\right)\end{cases}=\frac{\sqrt3}2.$$2 Your mistake is at this step: \sin(2\theta)=-\frac{1}{\sqrt{2}}, hence 2\theta = \frac{5\pi}{4} or \frac{7\pi}{4}. The correct step is: \sin(2\theta)=-\frac{1}{\sqrt{2}}, hence 2\theta = \frac{5\pi}{4}+2k\pi or \frac{7\pi}{4}+2k\pi, where k is any integer. So dividing by 2 we get \theta = \frac{5\pi}{8}+k\pi or \frac{7\pi}{8}+k\pi, where ... 2 Hence, 2\theta = \frac{5\pi}{4} or \frac{7\pi}{4} Don't forget +2\pi n. Since the signs of \cos\theta and \sin\theta place \theta in the fourth quadrant, only two candidates hold up. You can choose between them by testing whether \cos\theta > \cos\frac{7\pi}{4} 2 If 0\le \theta<2\pi, then 0\le 2\theta<4\pi. You should have 4 possible values of \theta. \cos (\theta) = \sqrt{\frac{1}{2}+\frac{1}{2\sqrt{2}}} and \sin (\theta) = -\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{2}}} imply that \sin(2\theta)=-\dfrac1{\sqrt2}, but not the other way round. Note that \displaystyle \tan\theta=\frac{-\sqrt{\frac{1}{2}-\... 2 Remember that if the range of \theta is 0 \leq \theta \lt 2 \pi then the range od 2\theta will be 0\leq\theta \lt 4\pi. So 2\theta = \ldots 2 Suppose \theta is an angle in radians. Let's create a new unit for angles called the \mathrm{Moytaba}, or \mathrm{Moy} for short. Let us define it by 1 \,\,\mathrm{Moy} = c\,\,\mathrm{rad} where c is some constant scaling factor (of course, any transformation between units of the same dimension has to be a constant scaling factor, for obvious ... 2 The fact that \cos \frac{\pi}{3} = \frac{1}{2} is not something that requires memorization, but like many identities in mathematics, it is convenient and efficient to memorize because the proof is more sophisticated than the result. In an equilateral triangle \triangle ABC, draw the altitude \overline{AD} from A to \overline{BC}. Since \angle A =... 2 Simply use that \sin\frac{\pi}3=\frac{\sqrt3}2 and \cos\frac{\pi}3=\frac12. Then \tan \frac{\pi}3=\dfrac{\frac{\sqrt3}2}{\frac12}=\sqrt3. Thus \arctan \sqrt3=\frac{\pi}3. Adjust for the sign: \arctan -\sqrt3=-\frac{\pi}3. 2 I agree with one of the comments to your question that it’s worth memorizing some of the basic triangles and the associated sines and cosines. However, a 30-60-90 triangle can be quickly reverse-engineered from \tan\theta = -\sqrt3: we have$$\frac yx = -\sqrt3 \\ x^2+y^2=1$$from which 4x^2=1, so x=\pm\frac12 and y=\mp\frac12\sqrt3. The hypotenuse ... 2$$\log\frac{1+\cos x}{1-\cos x}=\log\frac{(1+\cos x)^2}{\sin^2 x}=2\log(1+\cos x)-2\log|\sin x|.$$The first term is constant, while the sine is asymptotic to |x| (for x\to0). 2 I think, it's better$$\left|\sin\frac{\alpha}{2}\right|=\sqrt{\frac{1-\cos\alpha}{2}}$$and$$\left|\cos\frac{\alpha}{2}\right|=\sqrt{\frac{1+\cos\alpha}{2}}$$Your mistake is that \sqrt{x^2}=\pm x is wrong. The right identity it's:$$\sqrt{x^2}=|x|.$$For example, after$$\sin^4\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}+\frac{\sin^2\alpha}{4}=0$$we ... 2 Use Prosthaphaeresis Formulas,$$0=\sin x-\sin y=2\sin\dfrac{x-y}2\cos\dfrac{x+y}2 If $\sin\dfrac{x-y}2=0,\dfrac{x-y}2=m\pi$ $\implies x-y=2m\pi$ As $0\le x+y\le\pi, -\pi\le x-y\le\pi\implies m=0$ If $\cos\dfrac{x+y}2=0,\dfrac{x+y}2=(2r+1)\dfrac\pi2\implies x+y=(2r+1)\pi$ But $0\le x+y\le\pi,r=0$

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