# Calculus

Find the exact value of cot(arcsin(12/13))
and
cos(arcsin(1.7/2))

1. 👍 0
2. 👎 0
3. 👁 125
1. I got the second answer. Still confused on the first one.

1. 👍 0
2. 👎 0
posted by Emily
2. draw your triangle You have the opposite leg=12 and the hypotenuse=13

So, the adjacent leg is √(169-144) = 5

cot(x) = 5/12

1. 👍 0
2. 👎 0
posted by Steve
3. 300

1. 👍 0
2. 👎 0
posted by Sarah

## Similar Questions

1. ### math

Eliminate the parameter (What does that mean?) and write a rectangular equation for (could it be [t^2 + 3][2t]?) x= t^2 + 3 y = 2t Without a calculator (how can I do that?), determine the exact value of each expression. cos(Sin^-1

asked by Anonymous on August 3, 2007
2. ### calculus

Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x)

asked by Chelsea on March 10, 2011
3. ### Calc.

Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x)

asked by Chelsea on March 9, 2011
4. ### Calculus

Book works out integral of sqtr(a^2-x^2)dx as a^2/2*arcsin(x/a)+x/2*sqrt(a^2-x^2) by taking x=a sin theta, and advises to work it out using x=a cos theta. I did and got the first term as -a^2/2*arccos(x/a)and the same second term.

asked by ms on October 3, 2013
5. ### Trigonometry

Evaluate please: a.) arcsin(sin 13pi/16) b.) arccos(cos[-pi/18]) c.) arcsin(sin pi/18 d.) arcsec(-√2) e.) arctan(-√3/3)

asked by Liv on November 26, 2016
6. ### maths

Please can you help me with this question? Choose the option which is a false statement: A arctan(tan2/3pi))=-1/3pi B arccos(cos(3/4pi))=3/4pi C sin(arcsin(-1/2pi))=-1/2pi D arcsin(1/2squareroot3)=1/3pi E arcsin(sin(3/4pi))=1/4pi

asked by janet on May 26, 2011
7. ### Inverses

Problem: cos[arccos(-sqrt3/2)-arcsin(-1/2)] This is what I have: cos [(5pi/6)-(11pi/6)] cos (-6pi/6) or -pi giving an answer of -1 What I am wondering is should I use the (11pi/6) or should I use (-pi/6) because of the

asked by Dennis on April 21, 2008
8. ### Trigonometry

I need help with I just can't seem to get anywhere. this is as far as I have got: Solve for b arcsin(b)+ 2arctan(b)=pi arcsin(b)=pi-2arctan(b) b=sin(pi-2arctan(b)) Sub in Sin difference identity let 2U=(2arctan(b))

asked by Jen on July 27, 2007
9. ### trig

please help me with some questions I skipped on a review for our test coming up? solve 5-7 on the interval 0 greater than or equal to x less than or equal to 2pi. 5. sin x=sqrt3/2 6. cos x=-1/2 7. tan x=0 ---------------- 14. what

asked by BreAnne on February 22, 2012
10. ### math (repost)

Which of the following are inverse functions? 1. Arcsin x and sin x 2. cos^-1 x and cos x 3. csc x and sin x 4. e^x and ln x 5. x^2 and +/- sqrt x 6. x^3 and cubic root of x 7. cot x and tan x 8. sin x and cos x 9. log x/3 and 3^x

asked by Anonymous on April 16, 2009

More Similar Questions