Questions tagged [trigonometry]

Questions about trigonometric functions (both geometric and circular), relationships between lengths and angles in triangles and other topics relating to measuring triangles.

Trigonometry is a branch of mathematics that studies relationships involving lengths and angles of triangles.

Trigonometry is most simply associated with planar right-angle triangles. The applicability to non-right-angle triangles exists, but, since any non-right-angle triangle (on a flat plane) can be bisected to create two right-angle triangles, most problems can be reduced to calculations on right-angle triangles. Thus the majority of applications relate to right-angle triangles.

One exception to this is spherical trigonometry, the study of triangles on spheres, surfaces of constant positive curvature, in elliptic geometry. Trigonometry on surfaces of negative curvature is part of hyperbolic geometry.

Trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables. Geometrically, these are identities involving certain functions of one or more angles.

See Wikipedia's list of trigonometric identities.

29665 questions
4
votes
2 answers

$\cot 30^\circ = \cot 20^\circ - \operatorname{cosec} 80^\circ$ (?)

Prove that, $\cot 30^\circ = \cot 20^\circ - \operatorname{cosec} 80^\circ$ I could not do that by trying heart and soul. Please solve it.
user884185
  • 49
  • 2
4
votes
2 answers

How do I bypass the limit for the tan function on the calculator?

So basically, my partner and I are creating problems for a project. She created one where we aren't sure what the correct answer is. Problem: $Find \ cos\frac{\theta}{2} \ if \ tan \ \theta = \frac{3}{4}; \pi < \theta<\frac{3\pi}{2}$ I'll show you…
4
votes
1 answer

Is it permitted to use $\tan 90^\circ$ to prove an equation/identity even if it is not defined? defined?

Domain of tangent function limits to all real numbers except odd integral multiples of $\frac{\pi}{2}$. I noticed when proving an equality that using $\tan 90^\circ$ yielded the exact same result. Here what I did: For a triangle $\mathrm{ABC}$,…
Eyy boss
  • 213
4
votes
5 answers

How to express $2 \cos X = \sin X$ in terms of $\sin X$?

The Question was: Express $2\cos{X} = \sin{X}$ in terms of $\sin{X}$ only. I have had dealings with similar problems but for some reason, due to I believe a minor oversight, I am terribly vexed.
4
votes
3 answers

$\arccos(1/2)$ products

I'm finding points of intersection between two functions, $y = \tan(x)$ and $y = 2\sin(x)$, which are bound between $-\pi/3$ and $\pi/3$. Solving for the intersections points, I end up performing $x = \arccos\left(\frac12\right)$ I know this…
lime
  • 143
4
votes
2 answers

For natural $n$, prove $\prod_{k=1}^n \tan\left(\frac{k\pi}{2n+1}\right) = 2^n \prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}$

Prove that, for a natural number $n$, $$\prod_{k=1}^n \tan\left(\frac{k\pi}{2n+1}\right) = 2^n \prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}$$ This follows from a continued fraction identity for which, I think, there is a lengthy…
DVD
  • 1,137
4
votes
1 answer

solve ;$\sqrt{\frac{1-4\cos^2 4x}{8\cos (2x-2\pi/3)}}=\cos (2x-\pi/6)$

This is one more of my unsolved trigonometry questions: solve ;$$\sqrt{\frac{1-4\cos^2 4x}{8\cos (2x-2\pi/3)}}=\cos (2x-\pi/6)$$ My Try provided that $\cos (2x-2\pi/3)\neq 0$ and squaring both sides $$1-4\cos^2 4x=8\cos…
4
votes
2 answers

Write $\cos(5x)$ as a function of $\cos(x)$, answer with a polynomial.

Write $\cos(5x)$ as a function of $\cos(x)$, answer with a polynomial. $$e^{i\cdot5x} = \cos(5x) + i\sin(5x)$$ $$\cos(5x) + i\sin(5x) = (\cos(x) + i\sin(x))^5$$ and using the binomial theorem I think the answer is the real part of the expansion…
user766526
4
votes
3 answers

Evaluating $\left| \frac{\tan40^\circ + \tan100^\circ + \tan160^\circ}{\tan20^\circ\tan40^\circ\tan80^\circ} \right| $

How do I find the value of the following expression? $$ \left| \frac{\tan40^\circ + \tan100^\circ + \tan160^\circ}{\tan20^\circ\tan40^\circ\tan80^\circ} \right| $$ I tried writing the numerator as $\tan 40^\circ - \tan80^\circ -\tan20^\circ,$ but…
4
votes
2 answers

How were amplitudes of the $\cos$ and $\sin$ chosen?

I don't understand why we use $\displaystyle\sqrt{1^2+\left(\frac{1}{2}\right)^2}$ in the below transformation. Can someone help to explain? from $$f(x)=\frac{3}{5}-\frac{3}{5}e^t\left(\cos(2t)+\frac{1}{2}\sin(2t)\right)$$ transform…
Cable W
  • 143
4
votes
1 answer

eye vision problem

Imagine that the smallest letter that ken can read on the Snellen Eye chart is 3 inches tall. What is Ken's vision, using 20/XX notation? I have a question about eye vision, but I have no idea how can i start? am i finding the XX? and is XX…
Rex Rau
  • 85
4
votes
4 answers

Find the complete solution set of the equation ${\sin ^{ -1}}( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} ) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is

The complete solution set of the equation ${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is (A)[-1,0] (B)[0,1] (C)$[-1,\frac{1}{\sqrt{2}}]$ (D)$[\frac{1}{\sqrt{2}},1]$ My approach is…
4
votes
3 answers

How to find the solutions $x$ of $ 2\sin{11^{\circ}}\sin{71^{\circ}}\sin{(x^{\circ}+30^{\circ})}=\sin{2013^{\circ}}\sin{210^{\circ}}$

Let $$2\sin{11^{\circ}}\sin{71^{\circ}}\sin{(x^{\circ}+30^{\circ})}=\sin{2013^{\circ}}\sin{210^{\circ}}$$ where $90^{\circ}
math110
  • 93,304
4
votes
1 answer

Time Average of Cosine squared function

I've carried out the steps for the time average for $\cos^2x$ for limits $0$ to $T$. I've gotten : $\frac{1}{T}\left[\frac{1}{2}[T+\frac{1}{4}\sin2T\right]$ I'm trying to find the average over a long period of time. Would it be correct to set $T =0$…
4
votes
2 answers

Simple Trig Question

Hi I am not sure I am solving this trig question correctly: $\tan\left(\sin^{-1}\left(\dfrac{2\sqrt{x}}{1+x}\right)\right) = ?$ I drew a right triangle and set an angle equal to $\sin^{-1}\left(\dfrac{2\sqrt{x}}{1+x}\right)$ so that I could find the…
Matt
  • 1,123