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.

29665 questions
8
votes
4 answers

What's the intuition behind the sin of an obtuse angle?

The definition of sin of an angle is opposite / hypotenuse. Obviously, this definition is easy to use for acute angles within a right triangle, but it's hard to see how it carries over for obtuse angles. I know there are trig identities and unit…
user1411469
  • 183
  • 1
  • 1
  • 3
8
votes
5 answers

Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$

Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$. I have no clue how to proceed and tried to prove that the whole equation becomes $0$ when $\sin\frac{\pi}{14}$ is placed in place of $x$ but couldn't do anything further. I…
user142971
8
votes
1 answer

Prove trig identity: $(\sin \theta + 1)(\sin \theta − 1) = −\cos^2 θ$

This is my attempt: $(\sin \theta+1)(\sin \theta-1) = \sin\theta^2 - \sin\theta + \sin\theta - 1$ $= \sin^2\theta - 1$ $= -\cos^2\theta$ Is it correct, and can it be improved? Thanks!
Learner
  • 719
8
votes
3 answers

Prove that $1 + \cos\alpha + \cos\beta + \cos\gamma = 0$

If $\alpha + \beta + \gamma = \pi $ and $\tan(\frac{-\alpha + \beta + \gamma}4)\tan(\frac{\alpha - \beta + \gamma}4)\tan(\frac{\alpha + \beta - \gamma}4) = 1$ Then prove that: $1 + \cos\alpha + \cos\beta + \cos\gamma = 0$. I have no idea how to go…
user2369284
  • 2,231
8
votes
1 answer

Trying to get the infinite product for $\sin x$

I start with the fact that the zeros of $\sin x$ are $ n\pi$, $n\in\mathbb{Z}$. Therefore, it should be possible to express it as an infinite product: $$\sin x = x (x-\pi)(x+\pi)(x-2\pi)(x+2\pi)\cdots$$ $$ = x\prod_{n=1}^\infty (x^2 - n^2\pi^2)…
kuch nahi
  • 6,789
8
votes
5 answers

$\cos x+\cos 3x+\cos 5x+\cos 7x=0$, Any quick methods?

How to solve the following equation by a quick method? \begin{eqnarray} \\\cos x+\cos 3x+\cos 5x+\cos 7x=0\\ \end{eqnarray} If I normally solve the equation, it takes so long time for me. I have typed it into a solution generator to see the steps.…
Casper
  • 1,039
8
votes
4 answers

If $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$, then what is the value of $\cos(\theta -\frac{\pi}{4})$?

Problem : If $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$, then what is the value of $\cos(\theta -\frac{\pi}{4})$? My approach : Solution: $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$ $\Rightarrow \tan(\pi \cos\theta) = \tan \{ \frac{\pi}{2}…
Sachin
  • 9,896
  • 16
  • 91
  • 182
8
votes
4 answers

Given $\sec \theta + \tan \theta = 5$ , Find $\csc \theta + \cot \theta $.

The question is to find the value of $ \csc \theta + \cot \theta $ if $\sec \theta + \tan \theta = 5$ . Here is what I did : $\sec \theta + \tan \theta = 5$ $\sec \theta = 5 - \tan \theta $ Squaring both sides , $$\sec^2 \theta = 25 +…
A Googler
  • 3,355
8
votes
4 answers

Why does the tangent of numbers very close to $\frac{\pi}{2}$ resemble the number of degrees in a radian?

Testing with my calculator in degree mode, I have found the following to be true: $$\tan \left(90 - \frac{1}{10^n}\right) \approx \frac{180}{\pi} \times 10^n, n \in \mathbb{N}$$ Why is this? What is the proof or explanation?
Lee Sleek
  • 1,682
8
votes
2 answers

Solving for $x$: $a \sin(bx + c) = \sin(x)$

Ok, I scoured the internet for more than a few months for this one (whenever I had the time). But just because of this question I've created my first Stack Exchange account. I have thrown this at all Computer Algebra System software I was accessible…
Andrew Karasev
  • 81
  • 5
8
votes
4 answers

Simplifying $\sin(2\tan^{-1} x)$

I've been working on this for a while. The answer in the book is $\frac{2x}{x^2 + 1}$ Here's my workings: $\sin(2\tan^{-1} x)$ Let $\alpha = \tan^{-1}x \Rightarrow \tan \alpha = x$ $\sin(2\alpha) = 2\sin\alpha\cos\alpha = 2\tan\alpha\cos^2\alpha =…
PeteUK
  • 1,570
8
votes
2 answers

Show that $\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x} = \frac{1}{2}\left(\tan 27x-\tan x\right)$

Show that $$\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x} = \frac{1}{2}\left(\tan 27x-\tan x\right)$$
kalpeshmpopat
  • 3,270
  • 4
  • 28
  • 42
8
votes
2 answers

Evaluating $\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$ for a triangle with sides $2$, $3$, $4$

What is $$\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$$ for a triangle with sides $2$, $3$, and $4$? One can use Heron's formula to get $\sin A$, etc, and use $\cos A = (b^2+c^2-a^2)/(2bc)$ to get the cosines. But that's lots…
user526427
  • 821
8
votes
5 answers

Proving $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$

Prove the identity $$8\cos^4 \theta -4\cos^3 \theta-8\cos^2 \theta+3\cos \theta +1=\cos4\theta-\cos3\theta$$ If $7\theta $ is a multiple of $2\pi,$ Show that $\cos4\theta=\cos3\theta$ and deduce, …
emil
  • 1,310
8
votes
1 answer

Solving $\displaystyle \cos(x-\alpha)\cos(x-\beta) = \cos{\alpha}\cos{\beta}+\sin^2{x}$

Solve $\displaystyle \cos(x-\alpha)\cos(x-\beta) = \cos{\alpha}\cos{\beta}+\sin^2{x}$. My attempt: $\displaystyle \cos(x-\alpha)\cos(x-\beta) = \cos{\alpha}\cos{\beta}+\sin^2{x} \Rightarrow \cos(x-\alpha)\cos(x-\beta)-\cos{\alpha}\cos{\beta} =…
Lyrebird
  • 1,123
1 2 3
…
99 100