Verifying trig identities?
cotx - tanx = cosx/sinx - sinx/cosx = (cos^2x - sin^2x)/(cosx sinx) = cos(2x)/(cosx sinx) cotx + tanx = cosx/sinx + sinx/cosx = (cos^2x + sin^2x)/(cosx sinx) = 1/(cosx sinx) Therefore, (cotx - tanx )/(cotx + tanx ) = [cos(2x)/(cosx sinx)] / [ 1/(cosx sinx) ] = cos2x. The second question, 1-cos(2x)*sec(x) = (tan(x))^2 seems to be incorrect. Take x = pi/3. cos(2x) = cos(2*pi/3) = -0.5 secx = 1/cosx = 1/(cos pi/3) = 1/0.5 = 2 Therefore, 1-cos(2x)*sec(x) = 1 - (-0.5)(2) = 2 But tan^2x = tan^2(pi/3) = 3.
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
Prove the trigonometric identities?
First, express everything in terms of sine and cosine. a) I am assuming you meant: [ sin(x) + sin(2x) ] / [ 1 + cos(x) - cos(2x) ] = tan(x) LS = [ sin(x) + sin(2x) ] / [ 1 + cos(x) + cos(2x) ] RS = tan(x) = sin(x) / cos(x) There are three common forms of the expansion of cos(2x), but I will start using the one that only have cosines in it because the RS only have cosine in the denominator. cos(2x) = 2cos^2(x) - 1 LS = [ sin(x) + 2sin(x)cos(x) ] / [ 1 + cos(x) + 2 cos^2(x) - 1 ] LS = sin(x) [ 1 + 2cos(x) ] / [ cos(x) + 2cos^2(x) ] Factor cos(x) out of the denominator. LS = sin(x) [ 1 + 2cos(x) ] / cos(x) [ 1 + 2cos(x) ] Cancel common factors. LS = sin(x) / cos(x) = RS b) I am assuming you mean: 2tan(y/2) / [ 1 + tan^2(y/2) ] = sin y RS = sin(y) LS = 2tan(y/2) / [ 1+tan^2(y/2) ] There are two common forms for tangent half-angle. I will use the one that has sine in the numerator because the right side has sine in the numerator. For the tan^2 in the denominator, I will use the one with sine in the denominator for the same reason. LS = 2 sin(y) / [ 1 + cos(y) ] / [ 1 + (1-cos[y] / sin[y])^2 ] LS = 2sin(y) / [ (1 + cos[y] ) (1 + [ 1-cos(y) ] ^2 / sin^2[y] ) ] LS = 2sin(y) / [ (1 + cos[y] ) (sin^2[y] / sin^2[y] + [ 1 - 2cos(y) + cos^2(y) ] / sin^2[y]) ] LS = 2sin(y) / [ (1 + cos[y] ) (sin^2[y] + 1 - 2cos[y] + cos^2[y]) / sin^2(y) ] LS = 2sin^3(y) / [ (1 + cos[y]) (1 + 1 - 2cos[y]) ] LS = 2sin^3(y) / [ (1 + cos[y]) (2-2cos[y]) ] LS = 2 sin^3(y) / ( 2 - 2cos[y] + 2cos[y] - 2cos^2[y] ) LS = 2 sin^3(y) / (2 - 2cos^2[y]) LS = 2 sin^3(y) / 2(1 - cos^2[y]) Cancel the 2's on top and bottom. 1-cos^2(y) = sin^2(y) LS = sin^3(y) / (sin^2[y]) LS = sin(y) = RS c) I am not sure how to do this one.
HELP WITH THESE TRIG IDENTITIES IM BEGGING YOU SEE?
Okay, here are the questions. Make sure to show each step, and if you can explain them! 22.) 2cot2x=cot x-tan x 23.)(cos2x)/(1+sin2x)=tan(π/4-x) 24.)(cos x - sin x) / (cos x + sin x)= sec2x-tan2x 25.) (1-cos2x +sin 2x)/ (1+cos2x+sin2x)= tanx 26.) sin^2(x)+cos^4(x)=cos^2(x)+sin^4(x) 27.) tan x -cot x = (tan x -1)(cot x +1) 28.) cos x = sin x tan^2xcot^3x
Help me prove a trig identity? prove: tan2x = (2tanx)/ (1 - tan^2 x)?
tan(2x) = (2 tan x) / (1 - tan²x) Take the LHS. tan(2x) = Remember that tan x = sin x / cos x. sin(2x) / cos(2x) = Remember that sin 2A = 2 sin A cos A. 2 sin x cos x / cos(2x) = Remember that cos 2A = cos²A - sin²A. 2 sin x cos x / (cos²x - sin²x) = Divide the numerator and denominator by cos²x. (2 sin x cos x / cos²x) / [(cos²x - sin²x) / cos²x] = [2 sin x(1) / cos x] / [(cos²x / cos²x) - (sin²x / cos²x)] = [2(sin x / cos x)] / [1 - (sin²x / cos²x)] = Remember that tan x = sin x / cos x. (2 tan x) / [1 - (sin²x / cos²x)] = Remember that tan²x = sin²x / cos²x. (2 tan x) / (1 - tan²x) = RHS
Prove the identity (8 tan theta)/(1+tan^2 theta)=4 sin 2 theta?
I'll do a couple of them for you. i) Determine if the equation is an identity (cot(x) - tan(x)) / (sec(x)csc(x))=1. Yes or No? An equation is an identity if it holds true for all values of x (and y and z, and whatever other variables are in there) where both sides of the equation are defined. If something is not an identity, it's usually pretty easy to find a value of x for which it does not hold true, so let's start by assuming it's not, and finding proof. The way you prove something is not an identity is you find a value of x for which the left hand side does not equal the right hand side. Normally I'd try x = 0 or x = pi/2 first, because the trig values are so easy to compute, but unfortunately, tan(x) is not defined for x = pi/2, and cot(x) is not defined for x = 0, so we'll have to move on to my second choice, x = pi/4. When x = pi/4, we have: RHS = 1 ... (as always!) LHS = (cot(pi/4) - tan(pi/4)) / (sec(pi/4)csc(pi/4)) = (1 - 1) / (1/sqrt(2) * 1/sqrt(2)) = 0 Aha! So, RHS does not equal LHS for this value of x! The equation is not an identity. f) Factor and simplify. tan^3(x) + 8 = ? To do this, use the sum of two cubes factorisation: a^3 + b^3 = (a + b)(a^2 - ab + b^2) In this case a = tan(x), b = 2. We get: tan^3(x) + 2^3 = (tan(x) + 2)(tan^2(x) - 2tan(x) + 4) Hope that helps!
Proving Trigonometric Identity?
tan2x= 2tanx/1-tan^2x 2tanx/1-tan^2x= tan2x-------------- (1) & cos2x= cos^2x-sin^2x cos2x= cos^2x- 1+ cos^2x cos2x= 2cos^x-1 takinf reciprocals 1/cos2x= (1/2cos^2-1) (1/2cos^2-1)= 1/cos2x ------------------(2) adding (1) and (2) 2tanx/(1-tan^2 x) + 1/(2cos^2 x - 1) = tan2x + 1/cos2x 2tanx/(1-tan^2 x) + 1/(2cos^2 x - 1) = tan2x + sec2x 2tanx/(1-tan^2 x) + 1/(2cos^2 x - 1) = sin2x/cos2x + 1/cos2x 2tanx/(1-tan^2 x) + 1/(2cos^2 x - 1) = (sin2x+1)/ cos2x as sin2x= 2sinxcosx & 1= cos^2x + sin^2x & cos2x= cos^2x- sin^2x 2tanx/(1-tan^2 x) + 1/(2cos^2 x - 1) = (2sinxcosx +cos^2x + sin^2x)/ cos^2x- sin^2x 2tanx/(1-tan^2 x) + 1/(2cos^2 x - 1) = (cosx+sinx)^2/ (cosx+sinx)(cosx-sinx) 2tanx/(1-tan^2 x) + 1/(2cos^2 x - 1) =(cosx+sinx)/ (cosx-sinx) As Required
How would you prove the identity: [math] \frac{\cos^{4} x - \sin^{4} x} {\cos^{2} x} = 1 - \tan^{2} x [/math]
I am now going to solve the new question. I will first multiply all of it with [math]\cos^2x[/math], to get that [math]\cos^4x-\sin^4x=\cos^2x-\sin^2x[/math].Let's expand the left side to [math]\left(\cos^2x-\sin^2x\right)\left(\cos^2x+\sin^2x\right)[/math]; as the sum is 1, the identity is proved.OLD ANSWER (DUE TO MISUNDERSTANDING THE QUESTION)Attempting to prove: [math]\cos^4x-\frac{\sin^4x}{\cos^2x}=1-\tan^2x[/math]Let's attempt multiplying by cos^2 x, to get rid of the tangent (it's nasty)cos^6 x - sin^4 x = cos^2 x - sin^2 xLet y = sin^2 x, 1-y = cos^2 x (as the two sum to 1 by the fundamental theorem)(1-y)^3-y^2=(1-y)-y1-3y+y^2-y^3=1-2yy^3-y^2+y=01) y=0 which means sin^2 x = 0 or sin x = 0. Otherwhise:y^2-y+1: Δ=1-4 that gets to no other real solutions. Thus, this appears to have no other solutions when sin x is not 0. In particular, when sin^2 x is 1/2: 1/4 - 1/2 = 1 - 1 (false).
How do you prove sin^2x (1-tan^2x) =sin^2x-2sin^4x/cos^2x?
sin^2x(1-tan^2x)=sin^2x(1-sin^2x/cos^2x)= sin^2x*cos^2x/cos^2x - sin^4x/cos^4x= sin^2x*(1-sin^2x)/cos^2x - sin^4x/cos^4x= sin^2x/cos^2x - sin^4x/cos^4x-sin^4x/cos^4x= RHS
Verify that: tan^2x-sin^2x = (tan^2x)(sin^2x)?
Divide both sides by sin^2(x): tan^2 / sin^2 = (sin^2/cos^2) / sin^2 = 1/cos^2 = sec^2 sec^2(x) - 1 = tan^2(x) That's a trigonometric identity.