$$0 &lt; \cos (\theta)&lt;\frac {\sin (\theta)}{\theta}&lt;\frac {1}{\cos(\theta)} $$ for $\theta\in(0,\pi/2)$.
121759
How can I find the following product using elementary trigonometry?     Suppose $0 \lt x \lt \frac{\pi}{2}$   is an angle measured in radians. Use the trigonometric circle and   show that $\cos(x) \le \frac{\sin(x)}{x} \le \frac{1}{\cos(x)}$.   I have been trying to solve this question. I can't figure out whether or not the solution requires a trigonometric circle or if it can be done using another method.
53574
If angle $\alpha$ is reflex, and $\cos \alpha = -\frac{9}{41}$, without using a calculator, evaluate  (a) $\sin \alpha$ (b) $\tan \alpha$ (c) $\cos (\alpha - 180\deg)$     .
93877
Show that $$\sum_{n=0}^\infty r^n e^{i n \theta} = \frac{1- r\cos(\theta)+i r \sin(\theta)}{1+r^2-2r\cos(\theta)},$$ where $0\leq r &lt;1$.  Using this, prove that $\sum_{n=0}^\infty r^n \cos(n\theta)$ and $\sum_{n=0}^\infty r^n \sin(n\theta)$ are convergent.
14883
How do you prove the following trigonometric identity: $$ \sin^2\theta+\cos^2\theta=1$$  I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
42032
The number $\operatorname{cis} 75 + \operatorname{cis} 83 + \operatorname{cis} 91 +\dots+ \operatorname{cis} 147$ is expressed in the form $r\operatorname{cis}(\theta)$, where $0\leq \theta&lt; 360$.    Find $\theta$ in degrees   I'm having major trouble with this problem.
17614
Prove that $x - \frac{x^3}{3!} &lt; \sin(x) &lt; x$ for all $x&gt;0$   This should be fairly straightforward but the proof seems to be alluding me.    I want to show $x - \frac{x^3}{3!} &lt; \sin(x) &lt; x$ for all $x&gt;0$.  I recognize this shouldn't be too difficult but perhaps finals have fried my brain.
114279
Prove that for $ 0&lt;x&lt; \pi $, $$ \quad S_n(x) = \sin x + \frac{ \sin3x }{3} + ... + \frac{ \sin((2n-1)x) }{2n-1} &gt;0 \quad \forall n = 1,2,... $$  Having trouble with this problem. This is an olympiad-style question, so an answer that doesn't use calculus or analysis would be preferred. A possible approach is induction, but for this we need to find a function in terms of $n$ and $x$ so that we can actually use the inductive step. If anyone has any ideas they would be appreciated.  If you really want to go down the calculus route (at this point I don't mind), then $ S_n' (x) = \cos x + \cos 3x + ... +\cos((2n-1)x) $ , which you can find a closed form for, but I don't know how useful that is.
77661
If $I_n=\int _0^{\pi }\sin^{2n}\theta \:d\theta $, show that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, and hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $  Hence calculate $\int _0^{\pi }\:\sin^4t\cos^6t\:dt$  I knew how to prove that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, but I am not very good at English, what does it mean Hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $ do we need to prove this part as well or is it just a hint to use? and for the other calculation to find $\int _0^{\pi }\:\sin^4t\cos^6t\:dt$ is there something in the first part that I can do to help me solve this question because otherwise it becomes very long and it's part of the question.
87534
I need to calculate $\cos(\pi/20)$ with an error less than $10^{−5}$ of Taylor/Maclaurin series, What I did is like that:  $$|R_n(X)|=\left|\frac{(-1)^nC}{(n+1)!}\right|=\left|\frac{C}{(n+1)!}\right|_{0&lt;C&lt;\frac{\pi}{20}}&lt;\frac{\frac{\pi}{20}}{(n+1)!}=\frac{\pi}{20(n+1)!}&lt;10^{-5}$$  The statement is true when n=7, is my answer correct?
110923
$$\int^{ \pi /2}_{0} \ln (\sin x)\  dx$$  The answer is $- \frac{\pi}{2} \ln 2$.  I have changed it into $$\frac{1}{2} \int^{1}_{0} t d \ln t^{2}$$  But I didn't get the answer with it.
101742
I need solve this integral, and I tried various methods of solving and did not get it. The integral is:   $$\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{1-2t\cos\theta +t^2}d\theta,$$  where $t$ is a positive integer.
46536
I need to show that $\dfrac{2}{\pi}&lt;\dfrac{\sin(x)}{x}&lt;1$ for $0&lt;x&lt;\dfrac{\pi}{2}$.  I know I need to use the mean value theorem, would I just say that since $f$ is continuous in the interval (call it I).  We know $\exists c\in I$ such that $f'(c) = \dfrac{f\left(\dfrac{\pi}{2}\right)-f(0)}{\dfrac{\pi}{2}}$. I don't know where this is going though.  Also, would I need to show $\lim_{x \to 0}\dfrac{\sin(x)}{x} = 1$? Is there a way to do this without L'H Rule?
90876
If you take a linear combination of the cosine and sine function, then the result is , but with a new amplitude and phase shift. $$a \cos(\theta) + b \sin(\theta) = A \cos(\theta + \theta_0)$$ This is one of those facts which I've always felt should be more obvious to me than it is. I can't help feeling that I might be missing out on some good "conceptual reason" for this to be true. I did a bit of thinking about it today and I came up with the following explanation which I'm not particularly satisfied with. I feel like there should be some perspective on this which renders the fact obvious.      Multiplying the polar equation $r = 2a \cos(\theta) + 2b \sin(\theta)$ by $r$ yields $x^2+y^2 = 2a x + 2by$. Rearranging terms and adding $a^2 + b^2$ to both sides gives $x^2 - 2ax + a^2 +y^2 -2by + b^2 = a^2 + b^2$ which is identical to   $$(x-a)^2 + (y-b)^2 = a^2 + b^2$$   which we recognize as the equation for the circle with centre at $(a,b)$ and passing through the origin. This hints that $r = 2a \cos(\theta) + 2b \sin(\theta)$ should define a parametrization of this circle and, indeed, this holds. This circle can be rotated about the origin so as to have its centre on the $x$-axis. Such a circle will have the simpler equation   $$r = D \cos(\theta) $$   where $D=2(a^2+b^2)$ is the diameter of the circle. In polar coordinates, performing this rotation amounts to introducing a phase $\theta_0$. In fact $\theta_0$ is an argument for the point $(a,b)$.    Note I'm not really looking for rigorous proofs of this fact, more like interpretations, really.
59035
$$L~=~ \frac{1}{2} \dot{q}  \sin^2{q} $$  Is it zero or not defined?
42761
How to find bounds of $$I=\int_0^1\frac{\sin x}{\sqrt x}$$ and $$J=\int_0^1\frac{\cos x}{\sqrt x}$$     A) $I&gt;2/3$ and $J&gt;2$   B) $I&lt;2/3$ and $J&gt;2$   C) $I&gt;2/3$ and $J&lt;2$   D) $I&lt;2/3$ and $J&lt;2$
77032
Given that $x =r\cos \theta$ and $y = r\sin\theta$, determine $\frac{\partial r}{\partial x}$ and $\frac{\partial \theta}{\partial x}$   Attempt:  I thought I'd calculate $\frac{\partial x}{\partial r}$, treat it as a fraction and say $\frac{\partial r}{\partial x} = 1/\frac{\partial x}{\partial r}$, however that doesn't work.  Then I thought I'd set $r = \frac{x}{\cos \theta}$ and differentiate about $x$, but this path leads to the same result above.  I'm not sure how $y = r\sin \theta$ comes into play here.  Official Answer:  $$ \frac{\partial r}{\partial x} = \cos \theta, \quad \frac{\partial \theta}{\partial x} = \frac{- \sin\theta }{r}$$
96608
$f(r,\theta)=(r\cos\theta,r\sin\theta)$ Then  $1.$ $Df(r,\theta)$ is always nonzero for any $(r,\theta)\in\mathbb{R}^2,r\ne 0$  $2.$ $f$ is injective on $\{(r,\theta):r\ne 0\}$  $3.$ On any neighborhood of $(r,\theta):r\ne 0$, $f$ is one one.  $4$ $Df(r,\theta)=r^2I\forall (r,\theta):r\ne 0$  I have calculated that $Df(r,\theta)$ is the matrix $$\begin{bmatrix} \cos\theta &amp;-r \sin\theta \\ \sin\theta &amp; r\cos\theta \end{bmatrix}$$ so its determinant is r which is nonzero.so by inverse mapping theorem $3$ is true, $1$ is true also, I don't know about $2$ and $4$, could any one tell me how to prove or disprove , for $4$ I have taken $(r,\theta)=(r,\pi)$ and saw the contradiction.