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
4
votes
2 answers

Different representation of $\sin^2x$

I'm learning trigonometry, and I was just looking at the $y = \sin^2(x)$ graph. To me it looks the same as a $y = -\cos(x)$ shifted up. More specifically, it looks like $y = -0.5\cos(2x) + 0.5$ . Are these two functions the same? And this is my…
Archie Gertsman
  • 1,051
4
votes
4 answers

$\sin\alpha + \sin\beta + \sin\gamma = 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$ when $\alpha + \beta + \gamma = \pi$

Assume: $\alpha + \beta + \gamma = \pi$ (Say, angles of a triangle) Prove: $\sin\alpha + \sin\beta + \sin\gamma = 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$ There is already a solution on Math-SE, however I want to avoid…
Fine Man
  • 1,671
4
votes
5 answers

Proving $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$

Possible Duplicate: Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle Prove trigonometry identity? If $A$, $B$, and $C$ are to be taken as the angles of a triangle, then I beg someone to help me…
vini
  • 61
4
votes
1 answer

Where am I removing solutions in this equation

I have this equation : $\tan 2x = 3\cot x$ By rearranging I am getting the solutions: $37.8$, $142$, $218$ and $322$. However the mark scheme also has $90$, $270$. Hence I am wondering where I am getting rid of solutions. Here is my working:…
Cjen1
  • 259
4
votes
7 answers

Find the numerical value of $\sin 10^\circ \sin 50^\circ \sin 70^\circ$.

Prerequisite This problem is found in "Trigonometry" by I. M. Gelfand [in English]. It is asked in the section "Double the angle". So, assume that I know the sin/cos angle additions [i.e.: $\sin(A + B) = \sin A \cos B + \cos A \sin B$, etc.] as well…
Fine Man
  • 1,671
4
votes
3 answers

How can I Show that, $\arcsin\left(\frac{b}{c}\right)-\arcsin\left(\frac{a}{c}\right)=2\arcsin\left(\frac{b-a}{c\sqrt2}\right)$

Where $c^2=a^2+b^2$ is Pythagoras theorem. Sides a,b and c are of a right angle triangle. Show that, $$\arcsin\left(\frac{b}{c}\right)-\arcsin\left(\frac{a}{c}\right)=2\arcsin\left(\frac{b-a}{c\sqrt2}\right)$$ How do I go about proving this…
user334593
4
votes
2 answers

Prove that $(2\sqrt3+4)\sin x+4\cos x$ lies between $-2(2+\sqrt5)$ and $2(2+\sqrt5)$.

Prove that $(2\sqrt3+4)\sin x+4\cos x$ lies between $-2(2+\sqrt5)$ and $2(2+\sqrt5)$. Since we know that the minimum and maximum values of $a\cos x+b\sin x$ is $-\sqrt{a^2+b^2}$ and $\sqrt{a^2+b^2}$ I applied this formula to get the minimum and…
Vinod Kumar Punia
  • 5,648
  • 2
  • 41
  • 96
4
votes
3 answers

Prove that $\tan20^°\tan40^°\tan60^°\tan80^°=3$

Prove that $\tan20^°\tan40^°\tan60^°\tan80^°=3$ \begin{align} \tan20^°\tan40^°\tan60^°\tan80^°&=\frac{\sin20^°\sin40^°\sin60^°\sin80^°}{\cos20^°\cos40^°\cos60^°\cos80^°} \\ &=\frac{2^4(\sin20^°\sin40^°\sin60^°\sin80^°)^2}{\sin 160^°} \end{align} I…
Vinod Kumar Punia
  • 5,648
  • 2
  • 41
  • 96
4
votes
3 answers

Prove $\frac{1+\cos{(2A)}}{\sin{(2A)}}=\cot{A}$

I am sorry to ask so many of these questions in such as short time span. But how would I prove this following trigonometric identity. $$ \frac{1+\cos(2A)}{\sin(2A)}=\cot A $$ My work thus far is $$ \frac{1+\cos^2A-\sin^2A}{2\sin A\cos A} $$ I know…
Fernando Martinez
  • 389
4
votes
5 answers

Show that 3 sinusoidal phasors sum to zero

Prove that $\sin(\theta) + \sin(\theta+2\pi/3) + \sin(\theta+4\pi/3) = 0 $ for any angle $\theta$. This came up in the context of electricity: It is common in electrical power engineering to use three-phase circuits with sinusoidal currents out of…
crobar
  • 165
4
votes
2 answers

Calculate $\sin\frac{3\pi}{14}-\sin\frac{\pi}{14}-\sin\frac{5\pi}{14}$

I have interesting trigonometric expression for professionals in mathematical science. So, here it is: $$\sin\dfrac{3\pi}{14}-\sin\dfrac{\pi}{14}-\sin\dfrac{5\pi}{14};$$ Okay! I attempt calculate…
Yura
  • 117
4
votes
3 answers

Solving $E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$

$$E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$$ I got no idea how to find the solution to this. Can someone put me on the right track? Thank you very much.
Grozav Alex Ioan
  • 495
  • 5
  • 12
4
votes
2 answers

Solve for $x$; $\cos^2x-\sin^2x=\sin x; -\pi\lt x\leq\pi$

Solve for $x$; $\cos^2x-\sin^2x=\sin x; -\pi\lt x\leq\pi$ $$\cos^2x-\sin^2x=\sin$$ Edit $$1-\sin^2x-\sin^2x=\sin x$$ $$2\sin^2 x+\sin x-1=0$$ $\sin…
Austin Broussard
  • 1,195
4
votes
1 answer

Remembering exact sine cosine and tangent values?

There exists a common trick to remember exact sine cosine and tangent values. The trick is relatively long, so instead of reposting it, please refer to my answer on this page. Although I have used this trick for a while, I've never understood why it…
user26649
4
votes
6 answers

Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$

Question: Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$ My attempt: $$\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$$ $$ \frac{\frac {1}{\cos(x)} - \frac{1}{\sin(x)}}{\frac{\sin(x)}{\cos(x)}…
bigfocalchord
  • 5,597
1 2 3
…
99 100