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Sin 2x = 2tan(x) /(1 tan 2 (x)) So sin 2A = 2tan(A) /(1 tan 2 (A)) Putting the value of tan A = 4/3, we get Sin 2A = 2(4/3) /(1 (4/3) 2) Sin2A = (8/3)/(25/9) Sin 2A = 24/25 Answer sin 2A = 24/25 1tan^2x=sec^2x Change to sines and cosines then simplify 1tan^2x=1(sin^2x)/cos^2x =(cos^2xsin^2x)/cos^2x but cos^2xsin^2x=1 we have1tan^2x=1/cos^2x=sec^2x Trigonometry ScienceDividing both numerator and denominator by \(\cos ^{2}\)X, we get \(\cos 2X = \frac{1\tan ^{2}X}{1\tan ^{2}X} Since, \tan X = \frac{\sin X}{\cos X} \) Hence, \(\cos 2X = \frac{1\tan ^{2}X}{1\tan ^{2}X} \) Solved Examples Now that we have seen the formula of Cos 2X, let us try some examples to deepen our understanding Q Prove that,

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1 tan^2x formula- As far as I know, so far the $\tan2x$ can be converted to $\frac{2\tan x}{1\tan^2x}$ using the double angle formula and the $\tan 2x$ can be further be substituted to the following given item ($\tan(xa·)\tan(xb)$) however the problem now is that there are no ways to simplify this to my knowledge2tan 2x = 1tan x fromwhich tan2 x =1 Takingthesquarerootthengives tanx =1 or

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=(1tan^2x/22tan x/2)/(1tan^2 x/2) =(1tan x/2)^2/(1tan^2 x/2)(2) AnswerThe lowercase form of the same letter denotes the opposite side ofTan adjacent q= adjacent cot opposite q= Unit circle definition For this definition q is any angle sin 1 y q==y 1 csc y q= cos 1 x q==x 1 sec x q= tan y x q= cot x y q= Facts and Properties Domain The domain is all the values of q that can be plugged into the function sinq, q can be any angle cosq, q can be any angle tanq, 1,0,1,2, 2 qpnn

 Now rewrite this as sin x/cos x times 1/cos x However, the sin x divided by cos square x is equal to the tangent of x Also, the 1/cos x is equal to the secant of x So the derivative of sec x is equal to the tan x multiplied by sec x So, we have f (x) = 1/cos x = u/v By quotient rule, f' (x) = (vu' – uv') / v2 General solution of tan⁡( 2 x) tan⁡( x) = 1 x, and got the solution as ( 2 n 1) π 6 ( π 2 − x) So, 2 x = n π π 2 − x So, 3 x = ( 2 n 1) π 2 ( 2 x), and got the solution as ( 6 n ± 1) π 6 I can see that my solution has odd multiples of π / 2, which should be discarded, but I thought of it only after checking the solutionLet's see tan(AB)= (tanAtanB)/(1tanA*tanB) Now putting B=A , we get, tan(AA)= (tanAtanA)/(1tanA*tanA) tan2A= 2tanA/(1tan^2A) Hope, now it's fine

Answer (1 of 4) We use the following identities or definitions sec^2 x = 1/ ( cos^2 x) tan x = sin x / cos x & sin^2 x cos^2 x = 1 Now left hand side = sec^2 x tan^2 x =1/(cos^2 x) (sin^2 x)/(cos^2 x) =(1 sin^2 x)/ (cos^2 x) = cos^2 x / cos^2 x =1 = RHSWe know the identity sin 2 (x)cos 2 (x)=1 —— (i) Dividing throughout the equation by cos 2 (x) We get sin 2 (x)/cos 2 (x) cos 2 (x)/cos 2 (x) = 1/cos 2 (x) We know that sin 2 (x)/cos 2 (x)= tan 2 (x), and cos 2 (x)/cos 2 (x) = 1 So the equation (i) after substituting becomes tan 2 (x) 1= 1/cos 2 (xFormula $\cos{2\theta}$ $\,=\,$ $\dfrac{1\tan^2{\theta}}{1\tan^2{\theta}}$ A mathematical identity that expresses the expansion of cosine of double angle in terms of tan squared of angle is called the cosine of double angle identity in tangent Introduction Let

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D is the differential operator, int is the integration operator, C is the constant of integration Identities tan x = sin x/cos x equation 1 cot x = cos x/sin x equation 2 sec x = 1/cos x equation 3 csc x = 1/sin x equation 4Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!The points labelled 1, Sec(θ), Csc(θ) represent the length of the line segment from the origin to that point Sin(θ), Tan(θ), and 1 are the heights to the line starting from the axis, while Cos(θ), 1, and Cot(θ) are lengths along the axis starting from the origin

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 Half Angle Formula Cosine Using a similar process, with the same substitution of `theta=alpha/2` (so 2θ = α) we subsitute into the identity cos 2θ = 2cos 2 θ − 1 (see cosine of a double angle) We obtain `cos alpha=2\ cos^2(alpha/2)1`Solve for x tan (2x)=1 tan (2x) = 1 tan ( 2 x) = 1 Take the inverse tangent of both sides of the equation to extract x x from inside the tangent 2x = arctan(1) 2 x = arctan ( 1) The exact value of arctan(1) arctan ( 1) is π 4 π 4 2x = π 4 2 x = π 4 Divide each term by 2Solve for x tan(2x)tan(x)=1 Divide each term by and simplify Tap for more steps Divide each term in by Cancel the common factor of Tap for more steps Cancel the common factor Replace with in the formula for period Solve the equation Tap for more steps The absolute value is the distance between a number and zero The

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Free derivative calculator differentiate functions with all the steps Type in any function derivative to get the solution, steps and graphCos 2x = cos2 x – (1 – cos2 x) cos 2x = cos 2 x – 1 cos 2 x cos 2x = 2cos 2 x – 1 Third doubleangle identity for cosine Summary of DoubleAngles • Sine sin 2x = 2 sin x cos x • Cosine cos 2x = cos2 x – sin2 x = 1 – 2 sin2 x = 2 cos2 x – 1 • Tangent tan 2x = 2 tan x/1 tan2 x = 2 cot x/ cot2 x 1 = 2/cot x – tan xThe trigonometric formulas like Sin2x, Cos 2x, Tan 2x are popular as double angle formulae, because they have double angles in their trigonometric functions For solving many problems we may use these widely The Sin 2x formula is \(Sin 2x = 2 sin x cos x\) Where x is the angle Source enwikipediaorg Derivation of the Formula

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Sin(θ), Tan(θ), and 1 are the heights to the line starting from the xaxis, while Cos(θ), 1, and Cot(θ) are lengths along the xaxis starting from the origin In this section, an uppercase letter denotes a vertex of a triangle and the measure of the corresponding angle;The range of cscx is the same as that of secx, for the same reasons (except that now we are dealing with the multiplicative inverse of sine of x, not cosine of x)Therefore the range of cscx is cscx ‚ 1 or cscx • ¡1 The period of cscx is the same as that of sinx, which is 2Since sinx is an odd function, cscx is also an odd function Finally, at all of the points where cscx isSome common Identities and formulas generally used in finding Trigonometric ratios are stated below Double or Triple angle identities 1) sin 2x = 2sin x cos x 2) cos2x = cos²x – sin²x = 1 – 2sin²x = 2cos²x – 1 3) tan 2x = 2 tan x / (1tan ²x) 4) sin 3x =

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 tan(2x) To prove tan(2x) = (2 tan x) / (1 – tan²x) Proof First let us start from LHS tan(2x) We know that tan x = sin x / cos x sin(2x) / cos(2x) We know that sin 2A = 2 sin A cos A 2 sin x cos x / cos(2x) Also cos 2A = cos²A – sin²A 2 sin x cos x / (cos²x – sin²x) = Divide the numerator and denominator by cos²x 1tan^2x/1 tan^2x formula2tan 2x = 1tan x fromwhich tan2 x =1 Takingthesquarerootthengives tanx =1 orView M2 Formuladocx from AMA 1110 at The Hong Kong Polytechnic University M2 Formula 2 Trigonometric 2 2 sin xcos x =1 1 sinx cosx= sin 2 x 2 2 sec x=1 tan x 2 2 cose c x=1cot x 1 3Tan(xy) = (tan x tan y)/ (1−tan x •tan y) sin(x–y) = sin(x)cos(y)–cos(x)sin(y) cos(x–y) = cos(x)cos(y) sin(x)sin(y) tan(x−y) = (tan x–tan y)/ (1tan x • tan y) Double Angle Identities sin(2x) = 2sin(x) • cos(x) = 2tan x/(1tan 2 x) cos(2x) = cos 2 (x)–sin 2 (x) = (1tan 2 x)/(1tan 2

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 1tan^2(x) = 1 (sin 2 x)/(cos 2 x) = cos 2 x sin 2 x/cos 2 x = cos 2x/cos 2 x is a posibly 'simplified' version in that it has been boiled down to only cosines7 rows We will use the following trigonometric formula to prove the formula for tan 2x tan (a b)Answer (1 of 8) Yes, there is We know that cos 2x = cos^2 x sin^2 x As cos^2 x sin^2 x = 1, we can write, cos 2x = 1 sin^2 x sin^2 x = 1 2 sin^2 x or 1 cos 2x = 2 sin^2 x Cheers!!!

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Tan2x = 2tanx/(1tan^2x) How!Free online tangent calculator tan(x) calculator This website uses cookies to improve your experience, analyze traffic and display adsDerive Double Angle Formulae for Tan 2 Theta \(Tan 2x =\frac{2tan x}{1tan^{2}x} \) let's recall the addition formula \(tan(ab) =\frac{ tan a tan b }{1 tan a tanb}\) So, for this let a = b , it becomes \(tan(aa) =\frac{ tan a tan a }{1 tan a tana}\) \(Tan 2a =\frac{2tan a}{1tan^{2}a} \) Practice Example for tan 2 theta Question

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Legend x and y are independent variables, ;Tan(x y) = (tan x tan y) / (1 tan x tan y) sin(2x) = 2 sin x cos x cos(2x) = cos ^2 (x) sin ^2 (x) = 2 cos ^2 (x) 1 = 1 2 sin ^2 (x) tan(2x) = 2 tan(x) / (1 #2 tan x/(1tan^2x)=(2sin x/cos x)/(1(sin^2x/cos^2x)# #=2 sin x cos x/(cos^2xsin^2x)# #=(sin 2x)/(cos 2x)=tan 2x# Proofs for #sin 2x = 2 sin x cos x and cos 2x = 1 2 sin^2x# Use Area of a #triangle# ABC = 1/2(base)(altitude) = 1/2 bc sin A Here, it is the #triangle# ABC of a unit circle, with center at A, B and C on the circle and #angle# A = 2x Here, AB = AC = 1, BC =

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Cosine 2x or Cos 2x formula is also one such trigonometric formula, which is also known as double angle formula It is called a double angle formula because it has a double angle in it This is the reason why it is driven by the expressions for trigonometric functions of the sum and difference of two numbers (angles) and related expressions Inverse Trigonometric Formulas Trigonometry is a part of geometry, where we learn about the relationships between angles and sides of a rightangled triangleIn Class 11 and 12 Maths syllabus, you will come across a list of trigonometry formulas, based on the functions and ratios such as, sin, cos and tanSimilarly, we have learned about inverse trigonometry concepts alsoV t e In trigonometry, tangent halfangle formulas relate the tangent of half of an angle to trigonometric functions of the entire angle The tangent of half an angle is the stereographic projection of the circle onto a line Among these formulas are the following tan ⁡ 1 2 ( η ± θ ) = tan ⁡ 1 2 η ± tan ⁡ 1 2 θ 1 ∓ tan ⁡ 1 2

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Integral of tan^2 (x) \square! The solution you were presented is surely wrong, or at least would require further explanation of context It does not make sense to have a solution outside the domain of definition of the involved expressionsDouble angle formulas We can prove the double angle identities using the sum formulas for sine and cosine From these formulas, we also have the following identities sin ⁡ 2 x = 1 2 ( 1 − cos ⁡ 2 x) cos ⁡ 2 x = 1 2 ( 1 cos ⁡ 2 x) sin ⁡ x cos ⁡ x = 1 2 ( sin ⁡ 2 x) tan ⁡ 2 x = 1 − cos ⁡ 2 x 1 cos ⁡ 2 x

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