I was given 20 problems for extra credit and I cannot firgure 6 of them out. I have about over 30 trig identities right here in my face and still can't figure these out. If anybody could help me please do. Here are the problems: Verify each identity and compare graphs of the two to support. 1. (1 + tan^3 x)/(1 + tan x) = 1 - tan x + tan^2 x 2. (sin x - 2 + (1/sin x))/(sin x - (1/sin x)) = (sin x - 1)/(sin x +1) 3. csc x = (cot x + tan x)/sec x 4. (2 sin x cot x + sin x - 4 cot x-2)/ (2 cot x + 1) = sin x - 2 5. (sin x/1 - sin x) - (cos x/1 - sin x) = (1 - cot x)/(csc x -1) 6. ((1/tan x) + cot x)/((1/tan x) + tan x = 2/sec^2 x
When are they due? If no one can get them, I'll ask my roommate for you. She's a math/psychology major with a 4.0. I, on the other hand, routinely cry in math classes! I'll check back later and see how you're doing! Good luck!
Okay, just eyeballing this one led me to think that I could do it, and this is what I came up with: 1) csc x = (cot x + tan x) / sec x - given 2) 1/sin x = (cos x/sin x + sin x/cos x) / 1/cos x - definition of csc x, cot x, tan x, and sec x 3) 1/sin x = (cos x/sin x + sin x/cos x) * cos x - definition of multiplication of fractions 4) 1/sin x = cos x^2/ sin x + sin x - after multiplying the right side by cos x 5) sin x * (1/sin x) = (cos x^2/ sin x + sin x) * sin x - multiply both sides by sin x to get rid of fraction 6) cos x^2 + sin x^2 = 1 - rearranging and simplified after multiplying both sides by sin x 7) 1 = 1 because cos x^2 + sin x^2 = 1 is well- known identity of trig...therefore identity is true. It's been over 10 years since I had trig, but I can tell you to try to get everything in terms of sin and cos...i.e. get rid of tan, cot, sec, and csc. Good luck on the others
1. (1 + tan^3 x)/(1 + tan x) = 1 - tan x + tan^2 x separate (1+tan^3x) into (1-tanx+tan^2 x)(1+ tanx) (1-tanx+tan^2 x)(1+ tanx)/(1+tanx) = 1 - tan x + tan^2x 1 - tan x + tan^2 x = 1 - tan x + tan^2 x
2. (sin x - 2 + (1/sin x))/(sin x - (1/sin x)) = (sin x - 1)/(sin x +1) make sin x LCD for both numerator and denomenator. sin^2 x - 2 sinx + 1 ________________ sin x ________________ = (sin x - 1)/(sin x +1) sin^2 x - 1 ________________ sin x Cancel sin x sin^2 x - 2 sinx + 1 _______________ = (sin x - 1)/(sin x +1) sin^2 x - 1 (sin x-1)^2 ________________= (sin x - 1)/(sin x +1) (sinx + 1)(sin x - 1) (sin x - 1)/(sin x +1) = (sin x - 1)/(sin x +1)
6. ((1/tan x) + cot x)/((1/tan x) + tan x = 2/sec^2 x Answer: ((1/tan x) + cot x)/((1/tan x) + tan x --- Evaluating cot x = 1/tan x) ((1/tan x)+(1/tan x)) / (( 1/tan x) + tan x) --- perform algebra (2/tanx) / ((tan^2 x + 1)/tanx) --- Cancel out tan x 2/(tan^2 x + 1) --- Expanding tan x = sin x /cos x 2/ ((sin x/cos x)^2 + 1) --- expanding the denominator 2/((sin^2 x / cos ^2 x ) + 1) --- more algebra 2/ (( sin^2 x + cos^2 x)/ cos ^2 x) --- evaluating sin^2 x + cos^2 x = 1 2/(1/cos^2 x) --- 2/ (1/cos x)^2 -- evaluating 1/ cos x = sec x 2/sec^2 x
4. (2 sin x cot x + sin x - 4 cot x-2)/ (2 cot x + 1) = sin x - 2 (2 cos + sin - 4 cos/sin - 2)/( 2 cos/sin + 1) = (2 cos + sin - 4 cos/sin - 2) _____________________ = (2cos + sin)/sin 2cos x sin x + sin^2 -4cos -2sin __________________________= 2cos + sin (2cos + sin)(-2+sin) ______________ = 2cos + sin sin x - 2 = sin x - 2
5. (sin x/1 - sin x) - (cos x/1 - sin x) = (1 - cot x)/(csc x -1) 1-sin x is LCD (sin x - cos x)(1-sinx) = multiply by 1/sin x/1/sin x on left side. 1 - cot x _______ = csc - 1 My trick is to get rid of the fractions inside a fraction by finding LCD. Get rid of all division signs but one. After that look to factor. See what needs to be done to one side of numerator to equal other sides numerator.
I'll go ahead and give you #3 assuming nobody's done it and it seems a quickie before my 2:30 meeting : csc x = (cot x + tan x) / sec x 1/sin x = ((cos x/sin x) + (sin x/cos x))/ (1/cos x) 1/sin x = ((cos x/sin x) + (sin x/cos x)) * cos x 1/sin x = = (cos^2x/sin x) + sin x 1 = cos^2x + sin^2x 1 = 1
5. (sin x/1 - sin x) - (cos x/1 - sin x) = (1 - cot x)/(csc x -1) s -- by the notation of a/x - b/x = a- b/x sin x - cos x --------------- 1 - sin x -- sin x= 1/csc x sin x - cos x -------------- 1 - 1/csc x -- expanding the denominator (sinx - cos x) ----------------- (csc x - 1 ) ------- csc x --more algebra (sinx - cos x) csc x ----------------- csc x - 1 ) -- by definition csc x = 1/sinx 1 - cot x --------- csc x -1
Sorry about that Manny, I took the one I knew I could finish before my meeting and just did it. Never even read any of the others. lol. Oh well, I guess all have been solved. Those were pretty easy Lil Pun...
OK, I have a couple more problems which I need help, these are some HW problem which I got stuck on, here they are : Give exact values: A. Given tan a = 24/7, a in Quadrant I and sin B = -8/17, B in Quadrant III. Find sin(a + B) B. Given cos a = 8/17, a in Quadrant IV and sin B = -24/25, B in Quadrant III. Find tan(a + B) Write as a product of two funcitons: A. cos 2a + cos 5a B. sin 2B - sin 6B
Oh yeah, there is one more I need help with as well as the 4 above this post: Give exact value if possible, otherwise round to the nearst ten-thousandth cos(2 tan^-1 1)
Sorry to butt in, but you're not going to claim the extra credit for these problems are you? Or did the criteria for the extra credit include "it's O.K. to use other people to work on your problems"?
Actually the extra credit ones were already turned in, these are just homework porblems that I am having trouble on now.
