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trigonometrical
trigonometrical, a. (ˌtrɪgənəʊˈmɛtrɪkəl) [f. trigonometry or mod.L. trigonometria + -ic + -al1; after geometrical, etc.] Of, pertaining to, or performed by trigonometry. trigonometrical functions, those functions of an angle, or of an abstract quantity, used in trigonometry, viz. the sine, tangent, ...
Oxford English Dictionary
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Great Trigonometrical Survey
The Great Trigonometrical Survey was a project that aimed to survey the entire Indian subcontinent with scientific precision. Walker to amalgamate the Great Trigonometrical, Topographical and Revenue Surveys into the Survey of India.
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Trigonometrical equations Find the general solution of the equation $\sin x + \sin 2x + \sin 3x = 0$. I have started doing this problem by applying the formula of $\sin A + \sin B$ but couldn't generalise it. Please s...
HINT: Using Prosthaphaeresis Formulas, $$\sin x+\sin3x=2\sin\frac{3x+x}2\cos\frac{3x-x}2$$ We can also use $\sin x=\sin(2x-x),\sin3x=\sin(2x+x)$ Now $\sin y=0\implies y=n\pi$ and $\cos A=\cos B\implies A=2m\pi\pm B$ where $m,n$ are arbitrary integers
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Trigonometrical solution of complex equation Need present in trigonometrical form the solution of the complex equation: $x^6 = 1 + \sqrt3 + (1-\sqrt3)i$ To take out the coefficients of the real & imaginary parts re...
So, you know that $\cos\theta=\frac{1+\sqrt3}{2\sqrt2}$ and that $\sin\theta=\frac{1-\sqrt3}{2\sqrt2}$. Therefore, $2\sin(\theta)\cos(\theta)=-\frac12$, which means that $\sin(2\theta)=\sin\left(-\frac\pi6\right)$. This suggests (and it is easy to prove) that $\theta=-\frac\pi{12}$. So$$x^6=2\sqrt2\...
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Trigonometrical Solve There are 2 different values of $ \ \theta \ $. They are $ \ a \ $ and $ \ b \ $, such that $ \ 0 \ < \ a,b \ < \ 360^\circ \ $. If $ \ \sin(\theta+\phi) = \frac{1}{2} \sin2\phi \ $ , prove tha...
We have $\displaystyle\sin\theta\cos\phi=\sin\phi(\cos\phi-\cos\theta)$ Squaring we get $\displaystyle\sin^2\theta\cos^2\phi=\sin^2\phi(\cos\phi-\cos\theta)^2$ $\displaystyle\implies \sin^2\phi(\cos^2\phi+\cos^2\theta-2\cos\phi\cos\theta)=(1-\cos^2\theta)\cos^2\phi$ $\displaystyle\iff \cos^2\theta-2...
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Trigonometrical relation (searching a easy way to see it). In the figure I want to know $cos(\phi)$. I only know the cosines $cos(\theta)$ and $cos(\eta)$. A is in the xy plane. $, which says $OC\perp BC$. From here we get $\cos\varphi=\cos\theta\cos\eta$ immediately: $$\cos\varphi=\frac{OC}{OB}=\frac{AO}{OB}\cdot\frac{OC}{OA}=\cos\theta\cos\eta.$$
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Express a trigonometrical expression as an integral polynomial in $\alpha$ & $\beta$ How could we express this =?$$
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The product of two Trigonometrical functions I have the question: Express the following as the product of two trigonometrical functions: $$\sin(4 \theta) - 6 \sin(6 \theta)$$ Here is my attempt, is this correct ? ...
Using that $$\sin (2a)=2\sin (a)\cos (a) $$ and $$\sin (3b)=3\sin(b)-4\sin^3 (b) $$ we find $$\sin (4t)=2\sin (2t)\cos (2t) $$ and $$\sin (6t)=3\sin(2t)-4\sin^3(2t) $$ the difference gives $$\sin (2t)(2\cos (2t)-3+4(1-\cos^2 (2t)) $$ $$=\sin (2t)\left(4\cos^2 (2t)+2\cos (2t)+1\right) $$
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Trigonometrical integral of $1/(b+\cos\theta)$ Show that: if $b>1$ $$2\int_0^\pi \frac{1}{b+\cos\theta} \, d\theta= \int_0^{2\pi}\frac{1}{b+\cos\theta} \, d\theta$$ Thanks for your help!!
Well, ${\cos\theta}={\cos(-\theta)}={\cos(2\pi-\theta)}$, where ${\theta}\in[0,2\pi]$ which means ${\cos\theta}$ is symmetry besides ${\theta}=\pi$ therefore , ${b+cosθ}={b+cos(2\pi-\theta)}$ The requirement b>1 exists in order to avoid poles for $\frac{1}{b+cosθ}$, since the min of $cos{\theta}$ is...
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Converst trigonometrical sum in product I'm starting to work with some trigonometrical properties and I got the following problem: Convert the following sum into a product: $\cos{x} + \cos{3x} + \cos{5x} + \cos{7x}$ ...
Using Prosthaphaeresis Formulas, $$\cos x+\cos7x=2\cos3x\cos4x$$ and $$\cos3x+\cos5x=2\cos x\cos4x$$ Taking out $2\cos4x,$ $$\cos x+\cos3x=?$$
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Find minimum value that the trigonometric expression may take For $x\in\left(0, \frac{\pi}{2}\right)$ find a minimal value, which the expression $$\sec x+\csc x+\sec^{2}x+\csc^{2}x$$ can take. My attempt: I follow...
Let $\sin{x}=a$ and $\cos{x}=b$. Hence, $a^2+b^2=1$ and by AM-GM we obtain: $$\sec x+\csc x+\sec^{2}x+\csc^{2}x=$$ $$=\frac{a+b}{ab}+\frac{1}{a^2b^2}\geq\frac{2\sqrt2}{\sqrt{a^2+b^2}}+\frac{4}{(a^2+b^2)^2}=4+2\sqrt2.$$ The equality occurs for $a=b=\frac{1}{\sqrt2}$, which says that we got a minimal ...
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Isolating a variable trapped in trigonometrical function $\frac{p}{r}=\pm1 + \varepsilon \cdot\cos\sqrt{K\frac{M}{r^3}t^2}$. Is there any technique or simple way to isolate $r$ ?
There's no way to do that. One thought some people have is to use complex arithmetic and try the Lambert W function, but as has been shown in other posts here, allowing complex numbers throws the equation off. You will have to solve numerically.
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How were trigonometrical functions of $\dfrac{2\pi}{17}$ calculated? I know they were calculated by Gauss, but how? Is there a method for calculating them?
If I remember correctly(from what I read), Gauss proved that We can factor a rational multiple of $x^{17} - 1$ as $(x-1)(P(x)^2 + aP(x) + b)$ Where $P(x)$ is an $8^{th}$ degree polynomial with rational coefficients. This $P(x)$ could in-turn be represented as a quadratic $Q(x)^2 + cQ(x) + d$, where ...
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