In many cases the angle of contact is close to either \( 0 ° \) or \( 180 ° \), although it will be appreciated that if \( \theta \) were exactly zero, the liquid would spread. out in an infinitesimally thin layer to cover or "wet" the entire surface; and if it were exactly \( 180 ° \), the liquid, in the absence of other forces (such as its.

I'm aware that $$(2)\ \tan^2(\frac{1}{2}\theta)=\frac{1 - \cos(\theta)}{1 + \cos(\theta)} = \frac{\left(\frac{1 - \cos(\theta)}{2}\right)}{\left(\frac{1 + \cos(\theta)}{2}\right)}$$ And have been working towards solving the right-hand-side of the first identity listed into something along the lines of the right-hand-side of the second identity.

Sin Theta Formula bhukyarajeshkk Read Discuss The term Trigonometry is derived from Greek words i.e; trigonon and metron, which implies triangle and to measure respectively, θ. There are 6 Trigonometry ratios namely, Sine, Cosine, Tangent, Cotangent, Secant, and Cosecant.

If we let α = β = θ, then we have sin(θ + θ) = sinθcosθ + cosθsinθ sin(2θ) = 2sinθcosθ Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, cos(α + β) = cos α cos β − sin α sin β ,and letting α = β = θ, we have cos(θ + θ) = cosθcosθ − sinθsinθ cos(2θ) = cos2θ − sin2θ

Important formulas for tan theta tan (θ)=sin (θ)/cos (θ) tan (θ)=1/cot (θ) tan 2 (x)=sec 2 (x)-1 tan (-x)=-tan (x) tan (90 o -x)=cot (x) tan (x+π)=tan (x) tan (π-x)=-tan (x) tan (x+y)= \frac {tan (x)+tan (y)} {1-tan (x).tan (y)} tan (x-y)= \frac {tan (x)-tan (y)} {1+tan (x).tan (y)} tan (2x)=\frac {2tan (x)} {1-tan^2 (x)}

Sine, Cosine and Tangent (often shortened to sin, cos and tan) are each a ratio of sides of a right angled triangle: For a given angle θ each ratio stays the same. no matter how big or small the triangle is. To calculate them:

how is this possible? tan^2 is equal to sec^2 according to the calculations, they're just ignoring the one at the end of that original argument we're trying to simplify, like it wasn't there. If sin^2 + cos^2 =tan^2 = 1

Trigonometry 4sinθ cosθ = 2sinθ Linear equation y = 3x + 4 Arithmetic 699 ∗533

In trigonometrical ratios of angles (270° + θ) we will find the relation between all six trigonometrical ratios. Using the above proved results we will prove all six trigonometrical ratios of (180° - θ). Therefore, tan (270° + θ) = - cot θ, [since tan (90° + θ) = - cot θ] Therefore, cot (270° + θ) = - tan θ. 1. Find the value of.

\[ \sin 3 \theta = 3 \sin \theta - 4 \sin ^3 \theta \] \[ \cos 3\theta = 4 \cos ^ 3 \theta - 3 \cos \theta \] To prove the triple-angle identities, we can write \(\sin 3 \theta\) as \(\sin(2 \theta + \theta)\). Then we can use the sum formula and the double-angle identities to get the desired form:

2 Answers Harish Chandra Rajpoot Jun 29, 2018 Taking LHS as follows LH S = cotθ − tanθ cotθ + tanθ = cosθ sinθ − sinθ cosθ cosθ sinθ + sinθ cosθ = cos2θ −sin2θ cos2θ +sin2θ = cos2θ −(1 −cos2θ) 1 = cos2θ −1 + cos2θ = 2cos2θ − 1 = cos2θ = RH S proved. Answer link Rhys Jun 30, 2018 Shown below Explanation: Let t = tan( a 2) sina = 2t 1 + t2

You would need an expression to work with. For example: Given sinα = 3 5 and cosα = − 4 5, you could find sin2α by using the double angle identity. sin2α = 2sinαcosα. sin2α = 2(3 5)( − 4 5) = − 24 25. You could find cos2α by using any of: cos2α = cos2α −sin2α. cos2α = 1 −2sin2α. cos2α = 2cos2α − 1.