Mathematics Stack Exchange
Mathematics Stack Exchange

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.

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How to use $ S=\frac{\sin\frac{n+1}{2}\alpha\sin(x+\frac{n}{2}\alpha)}{\sin\frac{\alpha}{2}}$

I am struggling to understand how to use the following: A general formula for the sum $$ S=\sin x + \sin(x+\alpha) +\sin(x+2\alpha) +\dots+\sin(x+n\alpha)$$ is $$…
mikoyan
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How to solve $\tan x=x^3$

I was playing with functions I just learned, like sin, cos and tan, then I saw that the graph of $ y = \tan x$ and $y = x^3$ are pretty similar, that's how I thought of the equation $\tan x = x^3$. No matter how hard I try, I can't find a way to…
Tsuruta Junichiro
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Solving $\sin x=\cos 2x$ by expressing left side as $\cos (\pi/2-x)$ doesn't give correct solution

I can solve the equation in different ways but I'm not getting the right answer when solving as below: $$ \sin x=\cos 2x \\ \cos(\pi/2-x)=\cos 2x \\ \pi/2-x = \pm2x+2\pi k$$ $$x = -\pi /2+2\pi k\\ x = \pi /6-(2\pi k)/3$$ However the second solution…
Lenny White
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Why range of $\sin(x) - \cos(x)$ is not [-2,2]

If I want to calculate range of $$f(x)=\sin(x) - \cos(x) $$ Watching solution I got to know that we have to change this in a single trigonometric ratio (that is whole equation in form of sine or cosine) And then range will be $[-\sqrt2,\sqrt2]$ But…
OpenLearner
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Simplfying a trigonometric expression

I would like to simplify: $$\frac{\cos^2(80)+5\sin^2(80)-3}{\cos(50)}$$ By using the fact that $\sin^2(\theta) + \cos^2(\theta) = 1$, $$\frac{\cos^2(80)+5\sin^2(80)-3}{\cos(50)} = \frac{\cos^2(80)+5(1-\cos^2(80))-3}{\cos(50)} =…
user1107963
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If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$

If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$ My approach is as follow $\sqrt 3 \sin C = 2\sec A - \tan A \Rightarrow \sqrt 3 \sin C = \sec A + \sec A - \tan A$ $\sqrt 3 \sin C =…
Samar Imam Zaidi
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How to factorize $\frac{\cos(3x)-\cos(x)}{\tan(2x)-\tan(x)}$?

How to factorize $\dfrac{\cos(3x)-\cos(x)}{\tan(2x)-\tan(x)}$? Which trigonometric identities to use? I'm stuck when it comes to $\tan(2x)+\tan(x)$. I don't know which identity to use to turn it into the product. I was thinking of just transforming…
development mentor
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Show that $\frac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\frac{3\pi}{4}+\alpha\right)$

Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$ I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For the…
yinivem462
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Gelfands Trigonometry $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$

Trying Prove the identity $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$ using the figure provided in Gelfands trigonometry. What I have so far $\sin(\alpha - \beta) = \frac{CD}{AC} = \frac{PQ}{AC} = \frac{BQ}{AC} -…
jonathan parlett
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Show that $\cos^2\left(x\right)\left(2+\cos 2x\right)-2 \cos^2\left(x+\frac{\sin(2x)}{4}\right)\ge0$

Show that for all $x\in\mathbb{R},$ \begin{equation} \cos^2\left(x\right)\left(2+\cos 2x\right)-2 \cos^2\left(x+\frac{\sin 2x}{4}\right)\ge0 \end{equation} I have checked this numerically and this is correct but I cannot show analytically. I have…
JamesV
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Proving (or disproving) that the sine and cosine of integers are always unique

Can it be proven that $ \forall x, y \in \mathbb{Z} \left( \sin(x) = \sin(y) \iff x = y\right)$ ? Or disproven, of course. And likewise with cosine? Since sine and cosine have periods of $2\pi$, for $x$ and $y$ radians to have the same location on…
scott_fakename
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$\sin(25°)+\cos(115°)$?

What is the value of $\sin(25°)+\cos(115°)$? Using $\cos(90°+\theta)=-\sin(\theta)$, we get, $$\sin(25°)+\cos(115°)=\sin(25°)-\sin(25°)=0$$ But when I searched the same on Google, it showed $-0.45816155531$ as result on their calculator. Which…
user1071283
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Minimum of $\begin{aligned}\frac{a^2+b^2}{c^2}\end{aligned}$ in $\Delta ABC$

In $\triangle ABC$, $\sin B=-\cos C$. Find the minimum of $\begin{aligned}\frac{a^2+b^2}{c^2}\end{aligned}$. According to the law of sines, $\begin{aligned}\frac{a^2+b^2}{c^2}=\frac{\sin^2A+\sin^2B}{\sin^2C}\end{aligned}$. Solution $1$ Let $\sin…
user1034536
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Find the value of $\cos^{12}\theta + 3\cos^{10}\theta + 3\cos^{8}\theta + \cos^6\theta + 2\cos^4\theta + 2\cos^2\theta - 2$

We are given that $\sin\theta + \sin^3\theta + \sin^2\theta = 1$ Find the value of $\cos^{12}\theta + 3\cos^{10}\theta + 3\cos^{8}\theta + \cos^6\theta + 2\cos^4\theta + 2\cos^2\theta - 2$ Now, I was able to establish the following from the first…
Gerard
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Proof of cos(A-B) and geometric intuition

In the above figure, $a,b$ are unit vectors. The angle between them is $A-B$. It is easy to see the green highlighted part is: $$\cos A \cos B$$ This means the magenta part must be $$\sin A \sin B$$ Is there a way to prove this elegantly? My…
across
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