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7

The following contour plots hint us why the third equation defies many attempts. Another possible explanation is that, if we substitute $x = \tan(\alpha/2)$ and $y = \tan(\beta/2)$ then the formulas $$ \sin \alpha = \frac{2x}{1+x^2}, \quad \cos \alpha = \frac{1-x^2}{1+x^2}, \quad \tan \alpha = \frac{2x}{1-x^2} $$ show that \begin{align*} ...


6

Hint: $$\cos^2\theta=\frac{1}{2}(1+\cos(2\theta))$$ Edit: Also, after some calculations, I am pretty sure you made an error (can anyone double check this?) You have got $\displaystyle\frac{4}{3}\int\cos^2\theta d\theta$, but the substitution $9x^2=16\sin^2\theta$ gives us the following: ...


5

Hint So far so good! The standard technique for handling this integrand is to invoke the double angle identity $$\cos 2 \theta = 2 \cos^2 \theta - 1.$$


5

Hint: recall that $$ \cos (2x) = 1-2\sin^2 x $$


5

so $$cos (2) +cos(178) =cos(2)+cos(180-2)=0\\ cos (4) +cos(176) =cos(4)+cos(180-4)=0\\...\\$$


4

Three relevant identities: $\cos(90^{\circ} - x) = \sin(x)$, $\sin(90^{\circ} - x) = \cos(x)$, $\sin(-x) = - \sin(x)$. Therefore, \begin{align*} \cos(180^{\circ} - x) &= \cos(90^{\circ} - (x - 90^{\circ}))\\ &= \sin(x - 90^\circ)\\ &= -\sin(90^\circ - x)\\ &= -\cos(x). \end{align*}


4

hint: $\cos x + \cos (180^{\circ}-x) = 2\cos 90^{\circ}\times \cos(x-90^{\circ}) = 2\times 0 \times \sin x=....?$


4

Let $AD=y$ so that $\cos C=\frac{2y}{12}$ Then using the cosine rule, $$x^2=y^2+3^2-6y\cos C$$ So $x=3$


4

HINT: For every argument $n^\circ$ we have $\sin^2n^\circ=\cos^2(90^\circ-n^\circ)$, where $n\in\{1,2,\dots, 89\}$.


4

$RHS = \dfrac{\cos^3 x}{\sin^2 x}+ \cos x=\dfrac{\cos^3 x+\cos x \sin^2x}{\sin^2 x}=\dfrac{\cos x (\sin^2x+\cos^2 x)}{\sin^2 x}=\dfrac{\cos x}{\sin^2x}$ BECAUSE $\sin^2 x + \cos^2 x =1$... Now, $\dfrac{\cos x}{\sin^2x}=\dfrac{\cos x}{\sin x} \dfrac{1}{\sin x}=\cot x \csc x.$


3

The Gelfond-Schneider theorem says if $a\notin \{0,1\}$ is algebraic and $b$ is algebraic and irrational, then $a^b$ is transcendental. Apply this to $a = e^{i\pi/180}$, and you find that if $b$ is an irrational algebraic, $\omega = \cos(b^\circ) + i \sin(b^\circ)$ is transcendental. Then $\cos(b^\circ) = (\omega + \omega^{-1})/2$ and $\sin(b^\circ) = ...


3

I'm not sure what your difficulty is, so I'll use the geometrical definition to get the requested values. This definition means that when going a distance of $\theta$ counter-clockwise along the unit circle starting at $(1,0)$, the coordinates of the point on the circle are $(\cos \theta, \sin \theta)$. Note that $0$ corresponds to the point $(1,0)$ on the ...


3

HINT: $$(\cos A+\cos B+\cos C)^2+(\sin A+\sin B+\sin C)^2$$ $$=3+2\sum\cos(A-B)=0$$ Now use the fact that sum of squares of two real numbers is zero


3

\begin{align} \tan(2x) & = \frac{2\tan x}{1-\tan^2 x} = \frac{2(-4/3)}{1-(-4/3)^2} = \frac{-8/3}{1-16/9} = \frac{-24}{-7} = \frac{24} 7. \\[30pt] \tan(2x) & = \frac{\sin(2x)}{\cos(2x)} = \frac{-24/25}{-7/25} = \frac{24} 7. \end{align}


3

Recall the sine angle identity $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta.$$ For the choice $\beta = \pi/4$, we observe $$\sin \beta = \cos \beta = \frac{1}{\sqrt{2}},$$ thus we have as a special case $$\sin\left( \alpha + \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \left(\sin \alpha + \cos \alpha\right).$$ Consequently, for ...


3

Hint: Sketch the graph of $\sin^2$ and $\cos^2$ between $0$ and $90$ degrees. There is a symmetry between the numerator and the denominator. To formalize that, use $\sin n = \cos (90^o - n)$. The final answer is simple or would be if $\cos^2 0^o$ were also in the denominator.


3

By setting $u=\tan\frac{x}{2}$, we have to solve: $$ A\left(\frac{2u}{1+u^2}\right)^3+B\left(\frac{1-u^2}{1+u^2}\right)^3 = -c $$ that is equivalent to finding the roots of the sixth-degree polynomial: $$ p(u)=(c+B)+3(c-B)u^2+8A u^3+3(c+B)u^4+(c-B)u^6. $$ I do not agree that this polynomial has in general six real roots: by looking at the coefficients of ...


3

Since, as it was pointed out, $\cos(2x) = 1-2\sin^2 x$, you can see that $$\sin^2(x) + c = -\frac12 \cos(2x) + (\frac12 + c)$$ This means that both integrals describe the same set of solutions. Remember that anytime you are integrating a function $f$, you get a set of functions for which $F'=f$. The set always contains functions that differ by only a ...


3

Let $x^2+1=t\implies 2xdx=dt$ $$\int \frac{x^3}{(x^2+1)^3}dx=\frac{1}{2}\int \frac{(t-1)dt}{t^3}$$ $$=\frac{1}{2}\int \frac{(t-1)dt}{t^3}dt$$ $$=\frac{1}{2}\int (t^{-2}-t^{-3}) dt$$ $$=\frac{1}{2}\left(-1\frac{1}{t}+\frac{1}{2t^2}\right) $$ $$=\frac{1}{4t^2}-\frac{1}{2t}+C_1$$ $$=\frac{1}{4(x^2+1)^2}-\frac{1}{2(x^2+1)}+C_1$$ Now, let $x=\tan \theta\implies ...


3

Notice that: $x = \tan^{-1}\left(\dfrac{1}{3}\right)\Rightarrow \tan x = \dfrac{1}{3}, \text{ and } x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\Rightarrow \cos x > 0 \Rightarrow \sin x > 0\Rightarrow \csc x > 0\Rightarrow \tan^2 x = \dfrac{1}{9}\Rightarrow \sec^2 x = 1+ \tan^2 x = 1+\dfrac{1}{9} = \dfrac{10}{9}\Rightarrow \cos ^2 x = ...


3

hint: $x^2+y^2 = 9(\cos^2 \theta+\sin^2\theta)+4(\sin^2\theta + \cos^2\theta) = ....$


2

We have $$2+ \cos{\frac{3x}{2}} + \sqrt{3} \sin{\frac{3x}{2}} = 4\sin^2{\frac{x}{4}}$$ $$1+\frac{1}{2}\cos{\frac{3x}{2}} + \frac{\sqrt{3}}{2} \sin{\frac{3x}{2}} = \frac{4}{2}\sin^2{\frac{x}{4}}$$ $$\cos{\frac{3x}{2}}\cos \frac{\pi}{3} +\sin{\frac{3x}{2}}\sin\frac{\pi}{3} =-\left(1-2\sin^2{\frac{x}{4}}\right)$$ ...


2

I'm not sure I see what's so important about the suggested triangle, either. Here's a diagram that illustrates the relation in question: For the sake of completeness, here's the fairly obvious proof: With $O$, $P$, $Q$, $R$ as shown, we see that $\overline{PQ}$ is both parallel and congruent to $\overline{OR}$, so that $\square ORPQ$ is a parallelogram. ...


2

Let $\cos(\theta)=x_0$, so that $\sqrt{1-x_0^2}=\sin(\theta)$. From known trigonometry, we have $$x_{r+1}=\cos\left(\frac{\arccos(x_r)}2\right),$$ and by recurrence $$x_r=\cos\left(\frac{\arccos(x_0)}{2^r}\right)=\cos\left(\frac{\theta}{2^r}\right).$$ Hence the denominator is the infinite product ...


2

Notice, the following formula $$\color{blue}{\cos A-\cos B=2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{B-A}{2}\right)}$$ & $$\color{blue}{\sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}$$ Now, we have $$\frac{\cos 7x-\cos x+{\sin 3x}}{\sin 7x+\sin x-\cos 3x}=-\tan 3x$$ $$\implies LHS=\frac{(\cos 7x-\cos x)+{\sin ...


2

You are almost there (I think you have a typo in the expression you got) : Using $$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$ $$\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$$ we have $$\begin{align}\frac{\cos(7x)-\cos x+\sin(3x)}{\sin(7x)+\sin ...


2

We have: $6\cos (2\cos^{-1}x) = 6(2\cos^2(\cos^{-1}x)-1)=6(2x^2-1)$


2

The trigonometric functions relate an angle to a length. More precisely, in a rectangle triangle, the sine, cosine and tangent of an angle are the ratio of two of the sides. $$\cos(\theta)=\frac xr,\ \sin(\theta)=\frac yr,\ \tan(\theta)=\frac yx.$$ For convenience, we will now assume that $r=1$, as this doesn't change the proportions. ...


2

hint: $\dfrac{a}{b} = \dfrac{\sin \beta}{\sin \alpha}$, and divide both sides by $b$ and use the above substitution to deduce the equation.


2

Re-writing the equation as $$0 = 2\prod (\cos A + 1) - \sum \cos(B-C) - \sum \cos A - 2$$ we begin by multiplying-out the product, and carrying-on from there: $$\begin{align} 0 &= 2\cos A \cos B \cos C + 2\sum \cos B \cos C \color{blue}{+ 2\sum \cos A} \color{red}{+ 2} \\[4pt] &\quad-\sum\cos(B-C) \color{blue}{- \sum\cos A} \color{red}{- 2} \\[8pt] ...



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