Trigonometric Definition and 1000 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)}}\\...

what is the difference? It seems like in T, you choose the RHS first, but in f, you choose the LHS first. Is this the only difference? Because the Fourier coefficients of f is derived in a standard way, right? As in, couldn't I derive the coefficients for T in the same way as I did for f? In...

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.

I am looking for a closed form solution to an integral of the form: $$ \int_0^\infty \frac{e^{-Du^2t}u \sin{ux}}{u^2+h^2} du $$ D, t, and h are positive and x is unrestricted. I have tried everything, integration by parts, substitution, even complex integration with residue analysis. I've...

I came across the mentioned equation aftet doing a integral for an area related problem.Doing the maclaurin series expansion for the inverse sine function,I considered the first two terms(as the latter terms involved higher power of the argument divided by factorial of higher numbers),doing so...

Returning if I have to show the effort, I came to this: \frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}. =...

##8 \sin^4u = 3-4\cos 2u+\cos 4u## ##8 \sinh^4u = 3-4(1+2\sinh^2 u)+ \cosh ( 2u+2u)## ##8 \sin^4u = 3-4-8\sinh^2 u+ \cosh 2u \cosh 2u + \sinh 2u \sinh 2u## ##8 \sinh^4u = 3-4+1-8\sinh^2 u+ 4\sinh^2u +4\sinh^4 u + 4\sinh^2 u + 4\sinh^4 u## ##8 \sinh^4u = -8\sinh^2 u+ 8\sinh^2u +8\sinh^4 u##...

Last night I tried to calculate from an automatically generated Wolfram Alpha problem set: $$\int{\frac{1}{\sqrt{x^2+4}}}dx$$ I answered $$\ln({\frac{\sqrt{x^2+4}}{2}+\frac{x}{2}})+C$$ The answer sheet gave: $$\ln({\sqrt{x^2+4}+x})+C$$ I couldn't see where I had gone wrong, so I tried...

I'm trying to use the following trigonometric identity: $$ a \cos ( \omega t ) + b \sin ( \omega t ) = \sqrt{a^2+b^2} \cos ( \omega t - \phi )$$ Where ##\phi = \tan^{-1} \left( \frac{b}{a} \right)## for the following equation: $$ x(t) = -\frac{g}{ \omega^2} \cos ( \omega t) + \frac{v_o}{...

Summary: Hey, I'm getting confused with this question and don't think I'm doing it right, I was wondering if anyone could help me Tides vary so the high tide and low tide height of the water is different every day. At certain times of the year, such as a Spring tide, the water can be very deep...

R is the triangle which area is enclosed by the line x=2, y=0 and y=x. Let us try the substitution ##u = \frac{x+y}{2}, v=\frac{x-y}{2}, \rightarrow x=2u-y , y= x-2v \rightarrow x= 2u-x + 2v \therefore x= u +v## ## y=x-2v \rightarrow y=2u-y-2v, \therefore y=u- v## The sketch of triangle is as...

So I came across a question regarding sine rule. This is to find a missing angle. The question goes as follows Sin7x= Sin24 * 6.4 I tried two methods to solve this. Method 1: x= (Sin24 * 6.4)/ Sin 7 =21.35 I basically divided both sides by Sin 7 Method 2: Sin x= (Sin 24 * 6.4)/ 7 x=...

Summary:: I have a series of three equations that transform three angles of a system (J1, J2, J3), into three spatial x, y, z coordinates. I want to invert them to find the angles from the coordinates. Reference: https://www.physicsforums.com/forums/general-math.73/post-thread I have a series...

Can you prove the following? [sec(x)]^6 - [tan(x)]^6 = 1 + 3*[tan(x)]^2*[sec(x)]^2 [sin(x)]^2*tan(x) + [cos(x)]^2*cot(x) + 2*sin(x)*cos(x) = tan(x) + cot(x) If not, the following free math tutoring video shows you the method:

I’m stuck on how to begin. I’ve tried to factor out sin theta from both of the terms on the left hand side but that led to nowhere. Could I have a hint on how to continue? Than you!

What are the step-by-step in solving these problems?

I have been doing the resolutions of vectors on x and y-axis with making triangles and reference angles in all quadrants. But I want to calculate now how to find something like ##\sin 235## without the help of reference angles. I know we don’t need to. Calculator and Taylor theorem is handy here...

Solve the equation, $$cos ∅+ \sqrt 3⋅ sin ∅=1$$ in the interval, $$0≤∅≤2π$$

How can i prove that 6cos(x+45) cos(x-45) is equal to 3cosx?

Can a situation at angle ##\theta## happen, where: ##d sin \theta = 3 \lambda## ##b sin \theta = 2 \lambda## ##d/b = 3/2##

eg ##\cos (\sin x)## Asking this question out of curiosity.

Given : The equation ##\sin m\theta + \sin n\theta = 0##. Attempt : Using the formula for ##\text{sin C + sin D}## (see Relevant Equation 3 above), the given equation simplifies to \begin{equation*} 2 \sin \frac{(m+n)\theta}{2} \cos \frac{(m-n)\theta}{2} = 0 \end{equation*} This implies the...

Find the limit of csc(2x) as x tends to pi/2 from the right side. I decided to graph the function. Based on the graph, I stated the answer to be positive infinity. According to the textbook, the answer is negative infinity. Why is negative infinity the right answer? Thanks

Find the limit of cot (x) as x tends to pi from the left side. Seeking a hint or two. Does the graph of the given function help in terms of finding the limit?

Prove $\sin^2(x+a)+\sin^2(x+b)-2\cos (a-b)\sin (x+a)\sin (x+b)$ is a constant function of $x$.

Let $x$ be a real number such that $\dfrac{\sin^4 x}{20}+\dfrac{\cos^4 x}{21}=\dfrac{1}{41}$. If the value of $\dfrac{\sin^6 x}{20^3}+\dfrac{\cos^6 x}{21^3}$ can be expressed as $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers, find the remainder when $m+n$ is divided by 1000.

Hi! I am so confused about the given and what is being asked, I don't know how to solve it. This topic is solving situational problems involving trigonometric identities. Your help would be a big one for me :) Thank you so much in advance!

Prove that $\cos(\sin x))+\cos(\cos x))<\dfrac{\pi}{2}$.

Simplify $\left(\tan \dfrac{2\pi}{7}-4\sin \dfrac{\pi}{7}\right)\left(\tan \dfrac{3\pi}{7}-4\sin \dfrac{2\pi}{7}\right)\left(\tan \dfrac{6\pi}{7}-4\sin \dfrac{3\pi}{7}\right)$.

Attempt : I could not progress far, but the following is what I could do. $$\begin{align*} \mathbf{\text{LHS}} & = (\tan A+\tan B+\tan C)(\cot A+\cot B+\cot C) \\ & = 3+\tan A \cot B+\tan B \cot A+\tan A \cot C+\tan C \cot A+\tan B \cot C+\tan C \cot B\\ & = 3+\frac{\tan^2A+\tan^2B}{\tan A \tan...

What are the units of the trigonometric functions sinus, cosinus etc? If I take say Sin(0.5), what would the units of the output be?

Considering the measure of angles in radians, that are real numbers, the concept of of trigonometric function spreads to all real numbers. Any real number can be considered as an angle of the first circumference and a ##\mathbb{K}## number of circumferences. We can consider the function...

Create one equation of a reciprocal trigonometric function that has the following: Domain: ##x\neq \frac{5\pi}{6}+\frac{\pi}{3}n## Range: ##y\le1## or ##y\ge9## I think the solution has to be in the form of ##y=4sec( )+5## OR ##y=4csc( )+5##, but I am not sure on what to include...

cot^2θ+5cosecθ=4 cot^2θ+5cosecθ-4=0 cosec^2θ+5cosec-4-1=0 cosec^2θ+5cosec-5=0 Let u=cosecθ u^2+5u-5=0 Solve using the quadratic formula; u=(-5± 3√5)/2 u=(-5+ 3√5)/2=0.8541... Substitute cosecθ=u Therefore, cosecθ=0.8541 1/sinθ=0.8541 sinθ=1/0.8541=1.170... which is not true since sin x cannot be...

1. The factor theorem states that (x-a) is a factor of f(x) if f(a)=0 Therefore, suppose (x+1) is a factor: f(-1)=3(-1)^3-4(-1)^2-5(-1)+2 f(-1)=0 So, (x+1) is a factor. 3x^3-4x^2-5x+2=(x+1)(3x^2+...) Expand the RHS = 3x^3+3x^2 Leaving a remainder of -7x^2-5x+2 3x^3-4x^2-5x+2=(x+1)(3x^2-7x+...)...

a. I have just plotted the graph using desmos and attached an image here. Clearly, there are two values of x that satisfy the equation in the range. Do I need to add anything to this statement, I feel the response is a little brief for the question? b. Using the trigonometric identities; tan...

Question 1; a. sin θ=√3/2 θ=arcsin √3/2 θ=π/3 rad sin √3/2=60 degrees 60 degrees *π/180=π/3 rad. To find the other solutions in the range, sin θ=sin(π-θ) π-π/3=2π/3 The solutions are π/3 and 2π/3 in the range 0 ≤θ ≤2 π b. cos2θ=0.5 2θ=arccos 0.5 2θ=π/3 rad Divide both sides by 2; θ=π/6 rad...

Question 1; a. sin θ=√3/2 θ=arcsin √3/2 θ=π/3 rad sin √3/2=60 degrees 60 degrees *π/180=π/3 rad. To find the other solutions in the range, sin θ=sin(π-θ) π-π/3=2π/3 The solutions are π/3 and 2π/3 in the range 0 ≤θ ≤2 π b. cos2θ=0.5 2θ=arccos 0.5 2θ=π/3 rad Divide both sides by 2; θ=π/6 rad...

Show that for all $0<x<\dfrac{\pi}{2}$, the following inequality holds: $\left(1+\dfrac{1}{\sin x}\right)\left(1+\dfrac{1}{\cos x}\right)\ge 5\left[1+x^4\left(\dfrac{\pi}{2}-x\right)^4\right]$

Prove that $\sin^7 x=\dfrac{35\sin x-21\sin 3x+7\sin 5x-\sin 7x}{64}$.

Is tan^2 (x) the same as tan(x)^2? Note: I could have used any trig function. I know that tan^2 (x) means (tan x)^2. What does tan (x)^2 mean? Is it proper notation?

Evaluate $\dfrac{1}{1-\cos \dfrac{\pi}{9}}+\dfrac{1}{1-\cos \dfrac{5\pi}{9}}+\dfrac{1}{1-\cos \dfrac{7\pi}{9}}$.

Prove that if $[a,\,b]\subset \left(0,\,\dfrac{\pi}{2}\right)$, $\displaystyle \int_a^b \sin x\,dx>\sqrt{b^2+1}-\sqrt{1^2+1}$.

First, I try to define the function in the figure above: ##V(t)=100\left[sin(120\{pi}\right]##. Then, I use the fact that absolute value function is an even function, so only Fourier series only contain cosine terms. In other words, ##b_n = 0## Next, I want to determine Fourier coefficient...

Evaluate $\dfrac{\sin^2 \dfrac{\pi}{7}}{\sin^4 \dfrac{2\pi}{7}}+\dfrac{\sin^2 \dfrac{2\pi}{7}}{\sin^4 \dfrac{3\pi}{7}}+\dfrac{\sin^2 \dfrac{3\pi}{7}}{\sin^4 \dfrac{\pi}{7}}$ without the help of a calculator.

(Sinx-2cosx)/ (cotx - sinx) Substitute tan instead of cot (Tanx(sinx-2cosx)/(1-sinx) What do I do from here I don't think what I did there is correct That's why I didn't expand the tan to sin/cos

Let: equation 1 : sin A + sin B = 1 equation 2 : cos A + cos B = 0 Squaring both sides of equation 1 and 2 then add the result gives me: cos (A - B) = -1/2 Then how to proceed? Thanks

So far I've got the real part and imaginary part of this complex number. Assume: ##z=\sin (x+iy)##, then 1. Real part: ##\sin x \cosh y## 2. Imaginary part: ##\cos x \sinh y## If I use the absolute value formula, I got ##|z|=\sqrt{\sin^2 {x}.\cosh^2 {y}+\cos^2 {x}.\sinh^2 {y} }## How to...

Hi, K₁cos(θt+φ)=K₁cos(θt)cos(φ)-K₁sin(θt)sin(φ)=K₁K₂cos(θt)-K₁K₃sin(θt) Let's assume φ=30° , K₁=5 5cos(θt+30°) = 5cos(θt)cos(30°)-5sin(θt)sin(30°) = (5)0.866cos(θt)-(5)0.5sin(θt) = 4.33cos(θt)-2.5sin(θt) If only the final result, 4.33cos(θt)-2.5sin(θt), is given, how do I find the original...

I can’t get the angle, answer given is x=56.33 , x=9.545. (All steps before the equation are correct.)