Help verifying trig identities?
Do not convert everything to sinθ and cosθ for either of these identities. ________________ 1) Verify the identity. [(secθ - tanθ)² + 1] / (secθ cscθ - tanθ cscθ) = 2 tan θ Let's start with the left hand side. Left Hand Side = [(secθ - tanθ)² + 1] / (secθ cscθ - tanθ cscθ) = [sec²θ - 2tanθ secθ + tan²θ + 1] / [cscθ(secθ - tanθ)] = [sec²θ + (tan²θ + 1) - 2tanθ secθ] / [cscθ(secθ - tanθ)] = [2sec²θ - 2tanθ secθ] / [cscθ(secθ - tanθ)] = [2secθ(secθ - tanθ)] / [cscθ(secθ - tanθ)] = 2secθ / cscθ = 2sinθ / cosθ = 2 tanθ = Right Hand Side ___________________________ 2) Verify the identity. 1 / (secθ - tanθ) = secθ + tanθ Let's work with the left hand side. Left Hand Side = 1 / (secθ - tanθ) = (secθ + tanθ) / [(secθ - tanθ)(secθ + tanθ)] = (secθ + tanθ) / (sec²θ - tan²θ) = (secθ + tanθ) / 1 = secθ + tanθ = Right Hand Side
Verifying trig identities?
sin x ( 1 + cot^2 x) ==> use the Pyth ID that csc^2 x = 1 + cot^2 x sin x ( csc^2 x ) sin x (1/sin^2 x) sinx / sin^2 x 1/ sin x csc x QED! ______________________________________... 1 + sin A = cos^2 A / ( 1 - sin A) work with left side: plug in Pyth ID that cos^2 A + sin^2 A = 1 ==> cos^2 A = 1 - sin^2 A cos^2 A / ( 1 - sin A) (1- sin^2 A) / (1 - sin A) ==> factor the difference of two squares (1- sin A) (1+sin A) / (1-sin A) 1 + sin A QED!
Verifying trig identities?
I don't see how you got: (cos^2 / sin^2) - (1/sin^2) (1 - cos) / (1 + cos) multiply the top and bottom by (1 - cos) (1 - cos)^2 / (1 - cos^2) (1 - 2cos + cos^2) / (sin ^2) 1/sin^2 - 2cos/sin^2 + cos^2/sin^2 csc^2 - 2(cos/sin)(1/sin) + cot^2 csc^2 - 2(cot)(csc) + cot^2 (csc - cot)^2 --------------------------------------... For #2... Get a common denominator (tan - sec)(tan + sec) [1(tan + sec) + 1(tan - sec)] / [(tan + sec)(tan - sec)] (2tan) / (tan^2 - sec^2) (2tan) / [(-1)(sec^2 - tan^2)] 2 tan / (-1)(1) -2tan
Help verifying the trig identity?
I don't have theta on my keyboard, so I'll use ß (beta) instead. You have: (cotß/cosß) + (secß/cotß). Using the ordinary rules for adding fractions, we get: (cot²ß + cosß.secß) / cosß.cotß. Now cosß.secß ≡ 1 and (1 + cot²ß) ≡ cosec²ß, so now we have: cosec²ß / cosß.cotß = (cosec²ß / cosß).(1/cotß), which reduces to: (cosec²ß / cosß).(sinß/cosß) = (1 / sin²ß). (1/cosß).(sinß/cosß), and that reduces to: (1/sinß).(1/cos²ß) = cosecß.sec²ß, as required.
Help!! on verifying trig identity?
The unique thing about hyperbolic functions is that they mirror almost exactly their regular counterparts, with one exception: cosh²x -sinh²x = 1 Now, take coshx - sinhxtanhx and turn it into coshx - sinhxsinhx/coshx Then this becomes (cosh²x -sinh²x)/coshx. So, along with the tanhx changed, your problem becomes: =[sinhx/coshx]/ (cosh²x -sinh²x)/coshx =[sinhx/coshx]*coshx/( cosh²x -sinh²x) =[sinhx/coshx]*coshx/1 ==> because cosh²x -sinh²x = 1 =sinhx
I need help verifying my trig identities, i suck at it!!!?
(1+cosx + 1 - cos x)/((1-cosx)(1+cosx)) = 2 /(1-cos^2x) = 2/ sin^2 x = 2 csc^2 x sin a (sin a/cos a + cos a/sin a) =sin a (sin^2 a + cos^2 a)/(sin a cos a) =sin a/(sin a cos a) =1/cos a = sec a
Verifying Trig Identities Please Help!!!?
So your probably thinking that this is for homework...well its not, I have a test tomorrow and I need to understand a few trig identities that i've been trying to figure out for 3 days. Thanks for your help! θ= theta/variable (sec^4 θ) - (sec^2 θ) = (tan^4 θ) + (tan^2 θ) 3sin^2 θ + 4cos^2 θ = 3 + cos^2θ 9sec^2 θ - 5tan^2 θ = 5 + 4sec^2 θ
Help Verify a Trig. Identity?
4tanx = [4tanx - 4tan^3(x)] / [1 - 6tan^2(x) + tan^4(x)] This is not an identity, so you shouldn't be able to prove it. Here's how you can see that it is not true: suppose x is 45 degrees. Then tan x = 1 left hand side = 4 But the numerator of the right hand side is (4-4)=0 Maybe you made a mistake typing the problem? As written, the problem only involves tan(x), no multiple angles are involved.
Need a bit of help verifying a Trig. Identity..?
The solution of the problem lies in the fact that Sin²θ + Cos²θ = 1 but the resolution path you have chosen is probably not the best one. Actually, (2aSinθCosθ)² is zero only when θ is a multiple of pi/2. That is 0, pi/2, pi, 3*(pi/2), 2*pi, etc. You can use the sin and cos of 2θ to solve your problem. Or otherwise you develop the equation and then regroup. One solution is the following: Let A = (2aSinθCosθ)² + a²(Sin²θ - Cos²θ)² We know that 2sinθcosθ = sin (2θ) And cos²θ - sin²θ = cos (2θ) Then A is written like: A = (a sin(2θ))² + a²(- cos (2θ))² = a²sin²(2θ)+a²(cos (2θ))² If we factor a² then A is: A = a² [sin²(2θ)+cos² (2θ) ] and since sin²x + cos²x = 1 for any x Then A = a² Which needed to be proved