Let θ be an angle in quadrant III such that cosθ=(−3/4). Find the exact values of cscθ and cotθ.
cscθ = - 4 /√[ 4^2 - 3^2] = - 4 /√ [ 7] = - [ 4√ ( 7) / 7 ] = -(4/7) √ [ 7]
cotθ = 3 / √ [ 7] = (3/7) √ [ 7 ]
Yep...any fraction in the form 1/n will repeat unless n can be factored solely in terms of 2 or 5 [or both ]
BTW....the maximum length of any such repeat will be n - 1 digits
Be right with you in a sec.....
Length of diagonal = 4√3 cm
The short side is given by :
10 tan (27°) ≈ 5.095 = 5
The answer "cubed" would lead us back to the thing we took the cube root of....thus
2^3 = 8
∛8 =
∛[2 * 2 * 2 ] =
2
No equations are presented, Darkvoid......
2^4 * 3^8 = 9^n * 6^4
2^4 * 3^4 * 3^4 = (3^2)^n * 6^4
[ 2 * 3]^4 * 3^4 = (3^2)^n * 6^4
[ 6]^4 * 3^4 = (3^2)^n * 6^4 divide by 6^4
3^4 = (3)^(2n) solve for the exponents
4 = 2n → n = 2
No prob, DTA....!!!!!
