Wednesday

October 1, 2014

October 1, 2014

Posted by **anonymous** on Sunday, May 23, 2010 at 5:56pm.

sin[arccos(-2/7)] ?

- precal -
**Damon**, Sunday, May 23, 2010 at 6:38pmcos is - in quadrants 2 and 3

adjacent is -2, hypotenuse is 7, what is opposite?

7^2 - 2^2 = 49 - 4 = 45 = 9 * 5

sqrt(9*5) = 3 sqrt (5)

so sin = +/- 3 sqrt(5) /7

- precal -
**anonymous**, Sunday, May 23, 2010 at 6:59pmThank you!

**Answer this Question**

**Related Questions**

Precal - Find the exact value of the expression. sin(arccos sqrt(5)/5)

Trigonometry - Write the trigonometric expression as an algebraic expression. 1...

Math - Calculate the limit: lim(((arccos x)^(n+1))/((arccos x)^n) with x tends ...

trig - solve for x in the following equations: 1) sin(arcsin x) = 1 2) 2arcsin x...

maths - Please can you help me with this question? Choose the option which is a ...

math - Evaluate. 1. sin^-1(-1/2) 2. cos^-1[(-root 3)/2] 3. arctan[(root3)/3] 4. ...

Pre-Calculus-check answers - State the period and phase shift of the function y...

Precal - Use a calculator to evaluate the expression. arcsin 0.45 arctan 15 ...

math - Evaluate. 1. sin^-1(-1/2) 2. cos^-1[(-root 3)/2] 3. arctan[(root3)/3] 4...

Precal - Verify the identity: sin^(1/2)x*cosx - sin^(5/2)*cosx = cos^3x sq root ...