I'm having trouble with 3 other problems, here they are: (2) = squared, (3) = cubed and so on cot x + tan x --------------- = csc x sec x ____________________________________________________ 1 1 ----------- - ----------- = -2cot x csc x 1 + cos x 1 - cos x ____________________________________________________ sin(3) x + cos (3) x ------------------------ = 1 - sin x cos x sin x + cos x
Actually these are problems which were to be turned in yesterday but I didn't do these, I did the other ones but I couldn't figure these out.
Really... here is the first one: 1. (cot x + tan x) / sec x = csc x 2. ((cos x / sin x) + (sin x / cos x)) / sec x = csc x 3. ((cos x / sin x) / sec x) + ((sin x / cos x) / sec x) = csc x 4. (cos x / sin x) * (1 / sec x) + (sin x / cos x) * (1 / sec x) = csc x 5. (cos x / sin x) * (1 / (1/ cos x)) + (sin x / cos x) * (1 / (1/ cos x) = csc x 6. (cos x / sin x) * cos x + (sin x / cos x) * cos x = 1 / sin x 7. (cos^2 x / sin x) + sin x = 1 / sin x 8. (cos^2 x + sin^2 x) / sin x = 1 / sin x 9. 1 / sin x = 1 / sin x Which is the same as csc x = csc x Therefore, this identity is true. I think the other 2 requires sum and difference formulas, so someone else can do those for you.
sin x cos x cot x + tan x = ----- + ----- cos x sin x sin(2) x + cos(2) x writing fraction with a = --------------------------- common denominator cosx sinx 1 = ------------- because sin(2) x + cos(2) x = 1 cosx sinx 1 1 =--------- * ---------- cos x sin x = sec x csc x So: sec x csc x -------------- = csc x sec x
Mudbug's solution reminds me that #1 can be done several different ways. Another way of solving is to do this: Multiply both sides by sec x, this leaves: cot x + tan x = csc x sec x Use definitions to get: cos x / sin x + sin x / cos x = (1 / sin x) * (1 / cos x) Get a common denominator of sin x cos x, to get (cos^2 x + sin^2 x) / sin x cos x = 1 / (sin x cos x) cos^2 x + sin^2 x = 1, so 1 / sin x cos x = 1 / sin x cos x Which that might be easier to use than the one I used up above.
Let's see if I can make this post more readable. left hand side: Get a common denominator: = (1 - cosx - 1 - cos x) / (1 + cosx )( 1 -cosx ) Evaluate: = -2cosx / (1 - cos^2 x) Use sin^2 x + cos^2 x = 1: = -2cosx / ( sin^2 x + cos^2 x - cos^2 x ) Evaluate: = -2cosx /(sin^2x) Factor: = -2 * (cosx / sinx ) * (1 / sinx ) Evaluate: = -2 cotx cscx
First factor the numerator of the left hand side: sin^3 x + cos^3 x = (sin^2 x + cos^2 x )( sinx + cosx ) - (sin^2 x)(cosx ) - (cos^2 x)(sinx) since sin^2 x + cos^2 x = 1 = ( sinx + cosx ) - (sin^2 x)(cosx ) - (cos^2 x)(sinx) So the left hand side evaluates to: [( sinx + cosx ) - (sin^2 x)(cosx ) - (cos^2 x)(sinx)] / ( sinx + cosx ) = 1 - [(sin^2 x)(cosx ) + (cos^2 x)(sinx)] / ( sinx + cosx ) Factor numerator: = 1 - [(sinx)(cosx)( sinx + cosx )] / ( sinx + cosx ) = 1 - sinx cosx
not with my 1200 baud modem and my cheap parents who didn't want to pay for prodigy! i didn't enter the world of the internet until 1996 (with aol 2.0)... i still remember how there was no web browser integrated with aol 2.0, so i had to d/l it.... took 18 hours with my 9600 baud modem!
While we are at it, does anyone have a solution to this : find x,y,z where x^n + y^n = z^n where x,y,z are all nonzero and n > 2. thanks in advance!
ya know, if you guys know how to program, you could write some quicky source code with a big loop to get the answer... thats what i used to do back in school if there was a math problem i couldn't solve... PASCAL got me lots of extra credit!
