In mathematics, the inverse trigonometric functions (occasionally also called arcus functions, antitrigonometric functions or cyclometric functions) are the inverse functions of the trigonometric functions (with suitably restricted domains). Specifically, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are used to obtain an angle from any of the angle's trigonometric ratios. Inverse trigonometric functions are widely used in engineering, navigation, physics, and geometry.
My understanding is that it probably means either ##+ \pi## or ## - \pi##, so that ##\pi## is either added to or subtracted from the first term, just like ##|x| = \pm x##.
Homework Statement
Hello!
Seems I have forgotten how to work with such values. Please, help me to find the correct approach to this type of exercises.
Homework Equations
The task is to find the domain of the function:
f(x) = arccsc(e2x)
The Attempt at a Solution
The answer in the book is [0...
Consider ##y=\cos{-x}=\cos x=\cosh ix##.
Thus, ##\pm x=\cos^{-1}y## and ##ix=\cosh^{-1}y##.
So ##\cosh^{-1}y=\pm i\cos^{-1}y##.
Renaming the variable ##y##, we have ##\cosh^{-1}x=\pm i\cos^{-1}x##.
Next, we evaluate the derivative of ##\cosh^{-1}x## by converting it to ##\cos^{-1}x## using...
Homework Statement
After having struggled yesterday with this as much as I could, I am posting this problem here-
if ##ax+bsec(tan^{-1}x)=c## and ##ay+bsec(tan^{-1}y)=c##, then prove that ##\frac{x+y}{1-xy}=\frac{2ac}{a^2-c^2}##.Homework EquationsThe Attempt at a Solution
My attempt-...
Homework Statement
The problem- if
$$\theta= tan^{-1}(\frac{a(a+b+c)}{bc})+tan^{-1}(\frac{b(a+b+c)}{ac})+tan^{-1}(\frac{c(a+b+c)}{ab})$$
, then find $$tan\theta$$
Homework EquationsThe Attempt at a Solution
I tried to use these as sides of a triangle and use their properties, but other than...
Homework Statement
Express $$ 2 arctan (\sqrt\frac{a-b}{a+b} tan (\theta/2))$$ in terms of inverse cosine
Homework Equations
I realize it amounts to find a smart substitution, but I can't find one.
The Attempt at a Solution
I tried ##b/a=tan \theta## , but I can't find any way to get rid of...
Hey guys,
I have a couple of questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.
Question:
For 1a, using inverse trigonometric derivative identities should work, right?
I got y' = 1/sinØ + 1/cosØ and multiplied by the common...
Homework Statement
Which of the following is the solution set of the equation
$$2\arccos(x)=\text{arccot}\left(\frac{2x^2-1}{2x\sqrt{1-x^2}}\right)$$
A)(0,1)
B)(-1,1)-{0}
C)(-1,0)
D)[-1,1]
Ans: A
Homework Equations
The Attempt at a Solution
I start by rewriting LHS in terms of ##\arctan##...
Homework Statement
The sum of the infinite terms of the series
\text{arccot}\left(1^2+\frac{3}{4}\right)+\text{arccot}\left(2^2+\frac{3}{4}\right)+\text{arccot}\left(3^2+\frac{3}{4}\right)+...
is equal to
A)arctan(1)
B)arctan(2)
C)arctan(3)
D)arctan(4)
Ans: B
Homework Equations
The Attempt at...
Homework Statement
If cos^{-1}\frac{x}{a}+cos^{-1}\frac{y}{b} = \alpha then show that \frac{x^2}{a^2}-\frac{2xy}{ab}cos \alpha + \frac{y^2}{b^2} = sin^2 \alpha
Homework Equations
The Attempt at a Solution
I assume inverse functions to be θ and β respectively. So in the second equation...
Homework Statement
cos(2tan-1(1/7)) equals-
a)sin(4cot-13)
b)sin(3cot-14)
c)cos(3cot-14)
d)sin(4cot-14)
Homework Equations
The Attempt at a Solution
I am completely stuck in this one. I tried using the formula 2tan^{-1}x=tan^{-1}(\frac{2x}{1-x^2}). Using this formula i got...
Homework Statement
tan-1[2x/(1-x2)]+cot-1[(1-x2)/2x]=2π/3
2. The attempt at a solution
tan-1[2x/(1-x2)]+cot-1[(1-x2)/2x]=2π/3
tan-1[2x/(1-x2)]=2π/6
take tan on both sides
2x/(1-x2) =sqrt(3)
quadratic equation so it should have 2 solutions(sqrt(3) and sqrt(1/3)).But this question...
This is a simple question but I really need to know why:
why is the alternate form of arccosh(x) = log(x + sqrt(x^2 - 1)) and not log(x - sqrt(x^2 - 1))
How do I justify the plus/minus sign?
I need to know this ASAP, thanks.