#Solve tan(pi/2+theta)

[devphilamaths]

Solve tan(pi/2+theta) | tan(pi/2 +x) | tan pi/2 + x formula, Find Exact value tan pi by 2 + x

 Find the Value of tan(pi/2+theta), tan(pi/2 +x)tan pi/2 + x formula Exact Value of tan pi/2 + x Find the value of tan pi by 2 +x

[rowanstree-blog]

Day #167

          Today I started with a math class on the graphs of the trigonometric functions, sine, cosine and tangent. I started the lesson of with a video on the graph of the sinusoidal function, y=sin(x). I started with a question asking what the domain of the sinusoidal function was. The domain is all sets of valuable inputs for the function. I used the unit circle and a y, theta graph to wrote and plot the sinusoidal equation. Once I did all of that, I concluded that the domain of this function is (-1,1) including -1 and 1. I then watched a video on how to construct the graph of y=tan(x). I concluded that the graph would have a vertical asymptote every five radians and approach positive snd negative infinity while doing so. Considering the tangent of theta is the slope of the terminal ray, I just had to take the sin of theta over the cosine of theta to find out the tangent of several pi radians. I then watched a video on intersection points of y=sin(x) and y=cos(x). In this video, I wrote terms of the table in terms of radians for both the cosine of theta and the sine of theta on the unit circle and graphed accordingly. After that, I watched a video on basic trigonometric entities involving symmetry. In the lesson, I drew several terminal rays in all four quadrants of a graph in a unit circle. I then wrote down which sine and cosine functions were equivalent and then followed it up with a video on tangent identities which is the proportion of a sin function over its equivalent cosine function. Next, I watched a video on sine and cosine identities involving symmetry in which I took random sine and cosine functions and determined if they were equal to each other. I then watched a video on tangent identities periodicity. In this video, I did pretty much the same thing except I determined if its slope, or tangent was equal to 1/2. After all of that, I took an Algebra 1 class on the Quadratic equation sample problems as a refresher. I then took a French lesson on Duolingo followed by a Chemistry class on how to fill in an ICE table according to the equilibrium constant expression of a reaction. I stands for initial concentration which should be given in order to calculate E which stands for equilibrium concentration which can be found after adding or subtracting I and C which stands for change in concentration. First, you must write all of your initial concentrations If your equilibrium constant is larger than or equal to 10^4, then your products are favored. If Kc is less than or equal to 10^-2, then your reactants are favored. This information can help you figure out which side is equal to zero which will later determine how you fill in your ICE table. Depending on which side is favored, you must add or subtract the molecules stoichiometric coefficient of x. Add, subtract and solve for x to find out your equilibrium constant. After that, I took a space science class on a simulation video of Earth’s tilt and rotation which determines its season. This information can contribute to the reasoning behind why the equator doesn't have seasons. It is because it is never turned away from the sun, like us hence our four seasons as we turn towards and away from the sun. I then took an environmental science class on various studies launched and questions asked on the transgenic contamination of Mexican maize. Even though there have been both false positives and false negatives in order to protect and expose the firms causing this transgenic contamination. It is true that transgenic contamination is becoming more and more abundant and the effects are still unknown. I then did a quick exercise followed by an hour long lunch. After that, I took a women’s history class on Mary Harris “Mother” Jones who was a labor organizer who played an outspoken role in the world that was mislead by her appearance. She was a passionate and resilient woman that fought for what she believed in and due to that, became known as “the most dangerous woman in America.” I then took an economics class on Allocatice efficiency which is when the marginal cost is equal to the marginal benefit and I did an example that showed this both literally and graphically. Next, I took a grammar class on commas and introductory elements which are followed by a comma and are placed in the beginning of a sentence. After that, I read some of “The Beauty Myth” on how markets rely so much on feminine appearance more that the actual product to “appeal” to the reader while they are actually making them more self conscious.

[mathematicianadda]

Simplify $\tan^{-1} ( \frac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}} )$ with trigonometric substitution https://ift.tt/eA8V8J

I will explain my approach, help me with the last step please! $$ \tan^{-1} {\left(\frac {x - \sqrt {1-x^2}}{x + \sqrt {1-x^2}}\right)}$$

substituting x = $\sin \theta$ (as learnt from book) and solving 1-$\sin^2 \theta$ = $\cos^2 \theta$ $$ \tan^{-1} {\left(\frac {\sin \theta - |\cos \theta|}{\sin \theta + |\cos \theta| }\right)}$$

For solving modulus, it was important to determine range of $\theta$ , therefore I defined it (as it is my variable,i can define it my way) for [-$\pi$/2 , $\pi$/2] so that sine covers all values from $-1$ to $1$ (as , $ -1 \le x \le 1 \,$ , from domain ) and $\cos \theta$ is positive , and hence $|\cos \theta| = \cos \theta$.

$$ \tan^{-1} {\left(\frac {\sin \theta - \cos \theta}{\sin \theta + \cos \theta }\right)}$$ = dividing by $\cos \theta$ $$ \tan^{-1} {\left(\frac {\tan \theta - 1}{\tan\theta + 1 }\right)}$$

= by formula of $\tan (\theta - \pi/4)$ $$ \tan^{-1}( \tan{\left(\theta - \pi/4\right)})$$

That's where I am stuck ,as according to the identity,$\quad$ $tan^{-1} ( \tan \alpha) = \alpha$ $\quad$ only when $\, -\pi/2 <\alpha < \pi/2$ . But here $$ -3\pi/4 \le \,(\theta-\pi/4) \, \le \pi/4 $$ Therefore, I am not going to get ($ \,\theta - \pi/4 $) out of the expression. What i get will be based on that graph of $\bf {\tan^{-1} (\tan x)}$ . $$ (\theta - \pi/4) +\pi \,$$ for $\,-3\pi/4 \le \, (\theta -\pi/4) \, < -\pi/2 \,\,$ and

$$\theta -\pi/4$$ for $\,-\pi/2 < \, (\theta -\pi/4) \, \le \pi/4 \,\,$

My teacher just cancelled arctan and tan and wrote $\theta - \pi/4$ and he didn't even include that modulus function over $\cos \theta$.

So what will be the exact answer because if everyone decide $\theta$ as per they like then there will not be a finite answer. Everyone will have their own answers and in each answer they have multiple cases as I just discussed above.

So please help me, very hopefully I signed up in stackexchange!

Found Solution :-

I was confused because I was thinking that there can be many solutions differing person to person, but even if you choose any value of $\theta$ , you are going to get two solutions which are in the asked question above. The problem resolves when we write $\theta$ in terms of $sin^{-1} x$ as then we would not simply write like $$ \theta = \sin^{-1} x $$ we would write an equation,$$ \sin^{-1} x = \sin^{-1} (\sin \theta)$$, now if $\theta$ is not in range of $-\pi/2$ and $\,\pi/2$ , then there would be some constant in $\pi$ (like , $\pi/4 , 2\pi$ etc. we would have to add or subtract according to the graph of 'sin inverse sin' and when we would put that value of $\theta$ , we would end with the solutions as answered by people. (I write the answer in this edit to help anyone who will reach here after searching web , thanks to everyone for answers)

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[mathematicianadda]

If $a^2>b^2$ prove that $\int\limits_0^{\pi} \dfrac{dx}{(a+b\cos x)^3}=\dfrac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$.

Problem: If $a^2>b^2$ prove that $\int\limits_0^\pi \dfrac{dx}{(a+b\cos x)^3} = \dfrac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$.

My effort: If we choose $$x=\tan\frac{\theta}{2}\Longrightarrow d\theta=\frac{2}{x^2+1} \, dx\;\;,\;\;\cos\theta=\frac{1-x^2}{1+x^2}$$ then the integral becomes critical. What is the simplest way to solve?

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