Trigonometry Definition and 662 Threads

Attempt : ##\begin{align}\tan(x+100^{\circ})&=\tan(x+50^{\circ})\tan x\tan(x-50^{\circ})\\ \Rightarrow \dfrac{\sin(x+100^{\circ})\cos x}{\cos(x+100^{\circ})\sin x}&=\dfrac{\sin(x+50^{\circ})\cos (x-50^{\circ})}{\cos(x+50^{\circ})\sin (x-50^{\circ})}\quad{\text{(Using tan x=sin x/cos x)}}\\...

I like optimization problems a lot because they apply math to make the right decisions. However, I often come up with problems that are too hard for me to solve. Hopefully, this one will be simple but interesting enough for someone here to help me to crack it. Imagine there is a building or...

Teens who solved 2,000-year-old math puzzle expand on their work in publication https://www.cbsnews.com/news/teens-pythagorean-theorem-proofs-published-60-minutes/

Greetings, I would like to gain some insight when it comes to dealing with this problem. Personally, I wasn't able to solve it. I had to look for the solution in the book. I can just tell you that it's a telescopic series, the rest would be too much. You don't have to post the whole solution...

I would like to know the "why" of trigo of non acute angles in a unit triangle. why is it equals to the reference angle? how did it even come about? For example, sin 150 degrees. why is it equal to sin 30? i understand sin 30 because there is a right angle triangle of opposite and hypoteneuse...

Question; My take, I have, then using sine rule; ##\dfrac{x}{x+y} = \dfrac{\sin 20^{\circ}}{\sin 80^{\circ}}## ##\dfrac{x}{x+y} =0.347## ##x=3.47## then ##y=6.53##. then, ##BD^2=3.47^2+10^2-(2×3.47×10×\cos 20^{\circ})## ##BD= 6.842## ... ##10^2=3.47^2+6.842^2-(2×3.47×6.842 ×\cos...

I was able to solve with a rather longer way; there could be a more straightforward approach; My steps are along these lines; ##\sinh^{-1} x = 2 \ln (2+ \sqrt{3})## ##\sinh^{-1} x = \ln (7+ 4\sqrt{3})## ##x = \sinh[ \ln (7+ 4\sqrt{3})]## ##x = \dfrac {e^{\ln (7+ 4 \sqrt{3})} - e^{-[\ln 7+ 4...

So I am studying precalculus along with some basic calculus (I am not very patient but I feel relatively confident about my precalculus knowledge). Do you think there’s any real use of memorizing all identities for tangent and cotangent?

My question is on the highlighted part (circled in red); Why is it wrong to pre-multiply each term by ##e^x##? to realize , ##5e^{2x} -2-9e^x=0## as opposed to factorising by ##e^{-x} ## ? The other steps to required solution ##x=\ln 2## is quite clear and straightforward to me.

Thanks a lot in advance!

I have, Using ##\ cosh 2x = 2 \cosh^2 x - 1## ##\cosh x = 2 \cosh^2\dfrac{x}{2} -1## Therefore, ##\cosh x -1 = 2 \cosh^2\dfrac{x}{2} -1 - 1## ##\cosh x -1 = 2 \cosh^2\dfrac{x}{2} -2## ##=2\left[ \cosh^2 \dfrac{x}{2}...

Four points lie on the surface of a sphere. Given the six distances between the points, calculate the radius of the sphere. This is (allegedly) an advanced high school level problem. However, it is a remembered problem, so it is possibly misremembered (i.e. there might have been some “bice...

Hi all, I am starting with the following equation: ##2\cot\left(\frac{\theta}{2}\right) = \cot\left(\frac{k_{1}}{2}\right) - \cot\left(\frac{k_{2}}{2}\right)## with the following definitions: ##k_{1} = \frac{K}{2} + ik, k_{2} = \frac{K}{2}-ik, \theta = \pi(I_{2}-I_{1}) + iNk##, where...

I'm not sure how to go about solving this mathematically? In just using what seems obvious, I know the angle pi would work, because pi = -1, and 2pi = 1. However, as far as manipulating the equations in a way where it can solve itself without me having to look at a chart where cos for both x...

In my approach i have the following lines ##\ln (x + \sqrt{x^2+1}) = 2\ln (2+\sqrt 3)## ##\ln (x + \sqrt{x^2+1} = \ln (2+\sqrt 3)^2## ##⇒x+ \sqrt{x^2+1} =(2+\sqrt 3)^2## ##\sqrt{x^2+1}=-x +7+4\sqrt{3}## ##x^2+1 = x^2-14x-8\sqrt 3 x + 56\sqrt 3 +97## ##1 = -14x-8\sqrt 3 x + 56\sqrt 3 +97##...

My take, ##5 \cos 0 = 10 \cos θ## ##\cos θ = 0.5## ##⇒θ = 60^0## and ##X= 10 \cos (90^0-θ)=\cos 30^0= 8.66## (to two decimal places). ...insight welcome

I was trying to show that ##sin(x-y) = sin(x)cos(y)-cos(x)sin(y)## using Pythagoras' theorem and ##cos(x-y)=cos(x)cos(y)+sin(x)sin(y)##. I have: $$sin^2(x-y)=1-cos^2(x-y)$$ $$sin^2(x-y)=1-(cos(x)cos(y)+sin(x)sin(y))^2$$...

I proceeded as follows $$\int\frac{2(\sqrt3-1)(cosx-sinx)}{2(\sqrt3+2sin2x)}dx$$ $$\int\frac{(cos(\pi/6)-sin(\pi/6))(cosx-sinx)}{(sin(\pi/3)+sin2x)}dx$$ $$\frac{1}{2}\int\frac{cos(\pi/6-x)-sin(\pi/6+x)}{sin(\pi/6+x)cos(\pi/6-x)}dx$$ $$\frac{1}{2}\int cosec(\pi/6+x)-sec(\pi/6-x)dx$$ [FONT=times...

Problem Statement : I draw a picture of the given problem alongside. P is the location of the man and Q that of his friend at a height ##h## above. If the sun is at a position ##\text{S}_1## at 6 pm, at what time is the sun at position ##\text{S}_2##? Attempt : If ##\text{S}_2Q## is inclined to...

Can't attempt to solve the task. I'd appreciate it a lot if you could help!

Was solving a problem in mathematics and came across the following integration. Unable to move further. Can somebody provide answer for the following ( a and b are constants ).

In circuit analysis, everything seems to work out when you set i*cos = sin. But thats not a legitimate equation, so why does that work? Is there a proof that this is a real equation?

The problem is based on a similar thread. In fact, the first question is extremely similar. However, the second question is the one I consider more interesting but I posted the first one too for context. If this was just 1 pulley and two masses, then equilibrium is only possible if both masses...

In my working i have, ... ##\cos C = 2\cos^2 \dfrac{1}{2} C -1## ##c^2= a^2+b^2-2ab(2\cos^2 \dfrac{1}{2} C-1)## ##c^2= a^2+b^2+2ab(1-2\cos^2 \dfrac{1}{2} C)## ##c^2= (a+b)^2 (1-2\cos^2 \dfrac{1}{2} C)## Now from here, ##k^2 =2## but text gives different solution. I am still checking...

I let, ## 4\tan^{-1}\left[\dfrac{1}{5}\right]- \tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{4}## ##\tan^{-1}\left[\dfrac{1}{5}\right]- \dfrac{1}{4}\tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{16}## Then i let, ##\tan^{-1}\left[\dfrac{1}{5}\right] = α ...

I let ##\tan θ = 2x+1## and ##\tan β = 2x-1## ##θ + β = \tan^{-1} \left[\dfrac{(2x+1)+(2x-1)}{1- (2x+1)(2x-1)}\right]## ... ##θ + β = \tan^{-1} \left[\dfrac{4x}{1- 2x^2+1}\right]## ##θ + β = \tan^{-1} \left[\dfrac{4x}{2(1-x^2)}\right]## then ##\tan^{-1} \left[\dfrac{4x}{2(1-x^2)}\right]=...

In my approach (using a right angled triangle) i let, ##\cos^{-1} x = C## ⇒##\cos C = \sqrt{1-y^2}## and ##\cos^{-1} y= A## ⇒ ##\cos A= \sqrt{1-x^2}## Also, ##A+C = \dfrac{π}{2}## and ##\cos \dfrac{π}{2}= 0## ##xy - \sqrt{(y^2) ⋅(x^2)}=xy-xy=0## It follows that, ##\cos^{-1} [xy -...

In my approach i have, ##α + β = \tan^{-1} \left[ \dfrac{\dfrac{a}{a+1} + \dfrac{1}{2a+1}}{1-\dfrac{a}{a+1} ⋅\dfrac{1}{2a+1}}\right]## ... ##α + β = \tan^{-1} \left[ \dfrac{2a^2+3a+1}{(a+1)(2a+1)}\right] \div \left[\dfrac{2a^2+2a+1}{(a+1)(2a+1)}\right]## ##α + β = \tan^{-1}...

Ok in my approach i have, ##2 \tan^{-1} \left(\dfrac{1}{5}\right)= \sin^{-1} \left(\dfrac{3}{5}\right) - \cos^{-1} \left(\dfrac{63}{65}\right)##Consider the rhs, Let ##\sin^{-1} \left(\dfrac{3}{5}\right)= m## then ##\tan m =\dfrac{3}{4}## also let ##\cos^{-1} \left(\dfrac{63}{65}\right)=...

Hi. I have two trigonometric equations whose graphs I am trying to understand. Here are the equations: 1. a sin(x) - b cos(y) = y; a = 2, b = 2 2. a sin(x) + b cos(y) = 1; a = 1, b = 1 My question is why the graphs are the way they are. What should I do to understand them? Can anyone...

Is there an existing ray trace program that can trace planar light rays through this monocentric, model lens? Parameter values are given above. Input ray angles are all zero. Does some program give the output ray angle values at the second surface? How about for any arbitrary ray incoming to...

How do I do this from here without using the derivatives of sin or cos ?

For part (a), We know that ##\cos (-θ)=\cos (θ)## and ##\sin (-θ)=-\sin (θ)## ##\cos (A-B)=\cos A\cos (-B) -\sin A\sin(-B)## ##\cos (A-B)=\cos A\cos (B) +\sin A\sin(B)## ##\cos (A-B)=\cos A\cos B+\sin A\sin B## For part (b) ... ##f(θ)=\cos 60^0- \sin (θ+30^0)\sin (θ-30^0)## ##f(θ)=\cos...

My take: I got ##BC=10.25## cm, using cosine rule...no issue there. For part (b) ##BK=3cm## using sine rule i.e ##\sin 30^0 =\dfrac{BK}{6}## Thus it follows that ##∠BDK=48.59^0## ...⇒##∠ADB=131.4^0## correct...any other approach? Also: ##∠ADB=48.59^0## when BD is on the other side of the...

I just need to know how to find Θ in sin2Θ=0.51 I know I can use Θ = arcsin(0.51) but what about sin2Θ = 0.51

Hi all! I've never been studied the identities and such of secant, cosecant and cotangent. Yet I think, it would be useful to have them in my toolbox. Thus I'm asking, if anyone would know a reasonable book or other kind of material (paper or pdf) about trigonometry that has brief theory...

My take; ##x^2=\dfrac{(1+\sin θ)^2}{cos^2θ}=\dfrac{(1+\sin θ)^2}{1-\sin ^2θ}=\dfrac{1+\sin θ}{1-\sin θ}## we know that, ##x=\dfrac{1+\sin θ}{\cos θ}## ##⇒1+\sin θ=x\cos θ## therefore, ##x^2=\dfrac{x\cos θ}{1-\sin θ}##...

my notebook says that we can rewrite the integral $$\int {75\sin^3⁡(x) \cos^2⁡(x)dx}$$ as $$\int {75 \cos^2(x)\sin(x)dx} - \int {75\sin(x)\cos^4(x)dx}$$ however, i have literally no idea how it got to this point, and i unfortunately can't really provide an "attempt at a solution" for this...

text solution here; I was solving this today...got stuck and wanted to consult here...but i eventually found the solution...any insight/alternative approach is welcome... My approach; ... ##\sin^2y+ cos^2 y= 2a^2-2a \sin x - 2a\cos x+1## It follows that,##2a(\sin x + \cos x)=2a^2## ##\sin...

I'm doing self-study out of a free .PDF book entitled, Trigonometry, by Richard W. Beveridge (©June 18, 2014). The problem I'm interested in is as follows: "A woman standing on a hill sees a building that she knows is 55 feet tall. The angle of depression to the bottom of the building is 27°...

The official solution says ±25.4°, but I'm having trouble reproducing it. Here is my solution: 1) The components of the velocity of firework F with respect to the ground G in the moment of explosion are the following (Notice, I'm using sin, because the statement says 30.0° from vertical.)...

Refreshing on trig. today...a good day it is...ok find the text problem here; With maths i realize one has to keep on refreshing at all times... my target is to solve 5 questions from a collection of 10 textbooks i.e 50 questions on a day-day basis...motivation from late Erdos...

I’m teaching myself algebra right now so I’m not at that point, but I was wondering when I finish algebra what should I study next? Trig or Geometry?

I am looking for good textbooks in physics, algebra, and trigonometry textbooks that are up to date and a good read. I heard that Feynman’s Lectures was really good. Is it still up to date enough? Any opinions?

I can solve this by using the double-angle formula but the teacher expects another method not involving the double-angle formula. Is there a way to solve this without using double-angle formula? Thanks

This is the problem. The question is simple i just need some clarification as indicated on the part highlighted below in red. Now from my understanding tangent repeats on a cycle of ##π## radians...why do we have 2 the part circled in red below i.e ##2##? This is the part that i need clarity...

Some textbooks I found online ( open source ) College Trigonometry 3rd Corrected Edition - STITZ ZEAGER OPEN SOURCE MATHEMATICS Precalculus 3rd, Corrected Edition - Lakeland Community College, Lorain County Community College A First Course in Linear Algebra - Robert A. Beezer Cheers.

I just simply used the formula to solve. Note the "x" represents multiplication in this case 0.5 x a x c Sin B This is based on the conditions given in the textbook I am using which quotes "Use this formula to find the area of any triangle when you know 2 sides and an angle between them" So I...

This is the question... My attempt on part (i), ##b=\dfrac {16π}{2π}=8## ##11=a sin 32π+c## ##c=11## ##5=-a\frac {\sqrt 3}{2} +11## ##10=-a\sqrt 3+22## ##12=a\sqrt 3## ##a=\dfrac {12}{\sqrt 3}## Is this correct? Thanks...

My interest is on finding the value of ##A## only. From my calculations, ##A=1##and not ##2## as indicated on textbook solution. In my working we have; i.e ##4=A +3.## The values of ##B##and ##C## are correct though. Kindly advise. Find the question and textbook solution.