Cos tester

Author: u | 2025-04-24

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This function is defined in header file.[Mathematics] cos x = cos(x) [In C++ Programming]cos() prototype (As of C++ 11 standard)double cos(double x);float cos(float x);long double cos(long double x);double cos(T x); // Here, T is an integral type.cos() ParametersThe cos() function takes a single mandatory argument in radians.cos() Return valueThe cos() function returns the value in the range of [-1, 1]. The returned value is either in double, float, or long double.Note: To learn more about float and double in C++, visit C++ float and double.Example 1: How cos() works in C++?#include #include using namespace std;int main(){ double x = 0.5, result; result = cos(x); cout When you run the program, the output will be:cos(x) = 0.877583cos(x) = 0.906308Example 2: cos() function with integral type#include #include using namespace std;int main(){ int x = 1; double result; // result is in double result = cos(x); cout When you run the program, the output will be:cos(x) = 0.540302Also Read:C++ acos()

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Cell loss priority (CLP) bit QoS to PPP over X (PPPoX) sessions. The map is accepted only if you do not specify the set atm-clp command. For an example using the set atm-clp command to configure egress marking, please refer to Example 2: Configuring Egress Marking. set cos To set the Layer 2 class of service (CoS) value of an outgoing packet, use the set cos command in policy-map class configuration mode. To disable this setting, use the no form of this command. [no]setcoscos-value Syntax Description cos-value Specifies the IEEE 802.1Q CoS value of an outgoing packet ranging from 0 to 7 Command Default Either IP Precedence or MPLS EXP bits are copied from the encapsulated datagram. Command Modes policy-map (config-pmap) Usage Guidelines You can use the set cos command to propagate service-class information to a Layer 2 switched network. Although a Layer 2 switch may not be able to parse embedded Layer 3 information (such as DSCP), it might be able to provide differentiated service based on CoS value. Switches can leverage Layer 2 header information, including the marking of a CoS value. Traditionally the set cos command had meaning only in service policies that are attached in the egress direction of an interface because routers discard Layer 2 information from received frames. With the introduction of features like EoMPLS and EVC, the setting of CoS on ingress has meaning, such that you can preserve Layer 2 information throughout the routed network. set cos-inner To set the Layer 2 CoS value in the inner VLAN tag of a QinQ packet, use the set cos-inner command in policy-map class configuration mode. To disable this setting, use the no form of this command. [no]setcos-innercos-value Syntax Description cos-value Specifies a IEEE 802.1q CoS value ranging from 0-7 Command Default Either IP Precedence

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Theorem. Thus,[latex]\begin{array}{llllll} {a}^{2}={\left(x-c\right)}^{2}+{y}^{2}\hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \text{ }={\left(b\mathrm{cos}\,\theta -c\right)}^{2}+{\left(b\mathrm{sin}\,\theta \right)}^{2}\hfill & \hfill & \hfill & \hfill & \hfill & \text{Substitute }\left(b\mathrm{cos}\,\theta \right)\text{ for}\,x\,\,\text{and }\left(b\mathrm{sin}\,\theta \right)\,\text{for }y.\hfill \\ \text{ }=\left({b}^{2}{\mathrm{cos}}^{2}\theta -2bc\mathrm{cos}\,\theta +{c}^{2}\right)+{b}^{2}{\mathrm{sin}}^{2}\theta \hfill & \hfill & \hfill & \hfill & \hfill & \text{Expand the perfect square}.\hfill \\ \text{ }={b}^{2}{\mathrm{cos}}^{2}\theta +{b}^{2}{\mathrm{sin}}^{2}\theta +{c}^{2}-2bc\mathrm{cos}\,\theta \hfill & \hfill & \hfill & \hfill & \hfill & \text{Group terms noting that }{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1.\hfill \\ \text{ }={b}^{2}\left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)+{c}^{2}-2bc\mathrm{cos}\,\theta \hfill & \hfill & \hfill & \hfill & \hfill & \text{Factor out }{b}^{2}.\hfill \\ {a}^{2}={b}^{2}+{c}^{2}-2bc\mathrm{cos}\,\theta \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \end{array}[/latex]The formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion.Keep in mind that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, try to draw a diagram of the situation. As more information emerges, the diagram may have to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve.Law of CosinesThe Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle. For triangles labeled as in (Figure), with angles[latex]\,\alpha ,\beta ,[/latex] and[latex]\,\gamma ,[/latex] and opposite corresponding sides[latex]\,a,b,[/latex] and[latex]\,c,\,[/latex]respectively, the Law of Cosines is given as three equations.[latex]\begin{array}{l}{a}^{2}={b}^{2}+{c}^{2}-2bc\,\,\mathrm{cos}\,\alpha \\ {b}^{2}={a}^{2}+{c}^{2}-2ac\,\,\mathrm{cos}\,\beta \\ {c}^{2}={a}^{2}+{b}^{2}-2ab\,\,\mathrm{cos}\,\gamma \end{array}[/latex]Figure 3.To solve for a missing side measurement, the corresponding opposite angle measure is needed.When solving for an angle, the corresponding opposite side measure is needed. We can use another version of the Law of Cosines to solve for an angle.[latex]\begin{array}{l}\hfill \\ \begin{array}{l}\begin{array}{l}\hfill \\ \mathrm{cos}\text{ }\alpha =\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\hfill \end{array}\hfill \\ \mathrm{cos}\text{ }\beta =\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}\hfill \\ \mathrm{cos}\text{ }\gamma =\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\hfill \end{array}\hfill \end{array}[/latex]Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle.Sketch the triangle. Identify the measures of the known sides and angles. Use. Contribute to cos-onboarding/cos-tester development by creating an account on GitHub. View the profiles of people named Cos Tester. Join Facebook to connect with Cos Tester and others you may know. Facebook gives people the power to share

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= cos(β)0sin(β) 010 sin(β)0cos(β) * v Some might notice a missing sign in the rotation matrix because β is defined positive in a clockwise direction when looking down range. This gives v' = [sin(ε)*sin(β), cos(ε), sin(ε)*cos(β)] With this new vector and the definition of the vector dot product, A·B = |A||B|cos(c) where A·B = Ax*Bx + Ay*By + Az*Bz |A| = [Ax*Ax + Ay*Ay + Az*Az]1/2 |B| = [Bx*Bx + By*By + Bz*Bz]1/2 and c is the angle between vectors A and B we can find the new elevation and azimuth angles. The vector in the X-Y plane, v'' (parallel to the "ground") is just the X and Y components of v' or v'' = [sin(ε)*sin(β), cos(ε), 0] Then the new elevation angle, ε', be found by solving v'·v'' = |v'||v''|cos(ε') and v'·v'' = sin2(ε)*sin2(β) + cos2(ε) or v'·v'' = |v'||v''|cos(ε') = [sin2(ε)*cos2(β) + cos2(ε)]1/2 cos(ε') Equating v'·v'' and solving for cos(ε') we have cos(ε') = [sin2(ε)*sin2(β) + cos2(ε)]1/2 or ε' = cos-1[sin2(ε)*sin2(β) + cos2(ε)]1/2 Note that |v'| = 1.0 since v is a unit vector (length 1.0 by definition) and v' is just a rotated copy of v. To find the new azimuth angle, the angle between v'' and the Y axis, we use the dot product of v'' and a unit vector in the Y direction, y = [0,1,0] y·v'' = |v''||y|cos(α') But like |v| and |v'| the length of |y| is 1 so y·v'' = cos(ε)*1 = |v''|cos(α') and as before |v''| = [sin2(ε)*sin2(β) + cos2(ε)]1/2 Plugging this into the previous equation and solving for α' gives α' = cos-1(cos(ε)/[sin2(ε)*sin2(β) + cos2(ε)]1/2) The two equations for ε' and α' give the new elevation and azimuth angles. To correct for canted drop and windage, one only need to take the difference between the initial elevation and the canted

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M as$$ M = {\text{Def}}_{0} + 2 \cdot d \cdot {\text{Ha}} \to {\text{Ha}} = M/\left( {2 \cdot d} \right) $$ (19) Considering stage 1 beam enlargement and substituting Ha, we have$$ {\text{Def}}_{1h} = {\text{Def}}_{1v} = {\text{Def}}_{1} = {\text{Def}}_{0} + 2 \cdot ({\text{OPL}}^{\prime } - d) \cdot M/\left( {2 \cdot d} \right) $$ (20) $$ {\text{Def}}_{1} = {\text{Def}}_{0} + \left( {1/\cos \left( \Theta \right) - 1} \right) \cdot M $$ (21) Stage 2 beam enlargement affects only the vertical axis, thus$$ {\text{Def}}_{2v} = {\text{Def}}_{1} /\cos \left( \Theta \right) $$ (22) where Def2h remains unchanged. We can now calculate a new Def for the two stages combined by calculating the area of the ellipse, then normalizing it to an equal circular area, resulting in$$ {\text{Def}} = \frac{1}{{\sqrt {\cos \left( \Theta \right)} }} \cdot {\text{Def}}_{1} = \frac{1}{{\sqrt {\cos \left( \Theta \right)} }} \cdot \left( {{\text{Def}}_{0} + \left( {\frac{1}{\cos \left( \Theta \right)} - 1} \right) \cdot M } \right) $$ (23) The beam speed is extracted from Fig. 2 as$$ v = {\text{d}}h/{\text{d}}\Theta = \frac{{d \left( {d \cdot \tan \left( \Theta \right) } \right)}}{d\Theta } = \frac{d}{{{\text{cos}}^{2} \left( \Theta \right)}} $$ (24) Using the Ev in Eqs. 2, 22 and 23 and applying Lambert’s cosine law, this yields (Fig. 3)$$ {\text{Ev}} = \frac{{P_{{\text{l}}} }}{{v \cdot {\text{Def}} \cdot L_{h} }} = \frac{{P_{{\text{l}}} \cdot \cos (\Theta )}}{{\frac{2 \cdot \pi \cdot R \cdot f \cdot d}{{\cos^{2} (\Theta )}} \cdot {\text{Def}} \cdot L_{h} }} = P_{{\text{l}}} \cdot \frac{{\cos^{3} (\Theta )}}{{L_{h} \cdot v_{0} \cdot \left( {\frac{{{\text{Def}}_{0} + \left( {\frac{1}{\cos (\Theta )} - 1} \right) \cdot M}}{{\sqrt {\cos (\Theta )} }}} \right)}} $$ (25) $$ {\text{Ev}} = \frac{{P_{{\text{l}}} }}{{L_{h} \cdot V_{0} }} \cdot \frac{{\cos^{3} (\Theta ) \cdot \sqrt {\cos (\Theta )} }}{{\left( {{\text{Def}}_{0} + \left( {\frac{1}{\cos (\Theta )} - 1} \right) \cdot M} \right)}} $$ (26) Fig. 3Def and Ev

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In trigonometry, the Cosine Rule says that the square of the length of any side of a given triangle is equal to the sum of the squares of the length of the other sides minus twice the product of the other two sides multiplied by the cosine of angle included between them. Cosine rule is also called law of cosines or Cosine Formula.Suppose, a, b and c are lengths of the side of a triangle ABC, then;a2 = b2 + c2 – 2bc cos ∠xb2 = a2 + c2 – 2ac cos ∠yc2 = a2 + b2 – 2ab cos ∠zwhere ∠x, ∠y and ∠z are the angles between the sides of the triangle. The cosine rule relates to the lengths of the sides of a triangle with any of its angles being a cosine angle. With the help of this rule, we can calculate the length of the side of a triangle or can find the measure of the angle between the sides.What are the Laws of Cosine?As per the diagram, Cosine rules to find the length of the sides a, b & c of the triangle ABC is given by;a2 = b2 + c2 – 2bc cos xb2 = a2 + c2 – 2ac cos yc2 = a2 + b2 – 2ab cos zSimilarly, to find the angles x, y and z, these formulae can be re-written as :cos x = (b2 + c2 -a2)/2bccos y = (a2 + c2 -b2)/2accos z = (a2 + b2 – c2)/2abAlso, read:Cosine FunctionInverse CosineThe law of cosine states that for any given triangle say ABC, with sides a, b and c, we have;c2 = a2 + b2 – 2ab cos CNow let us prove this law.Suppose a triangle ABC is given to us here. From the vertex of angle B, we draw a perpendicular touching the side AC at point D. This is the height of the triangle denoted by h.Now in triangle BCD, as per the trigonometry ratio, we know;cos C = CD/a [cos θ = Base/Hypotenuse]or we can write;CD = a cos C ………… (1)Subtracting equation 1 from side

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Karaoke Version CDG Video Karaoke Pop Coldplay A Sky Full of Stars Formats included: The CDG format (also called CD+G or MP3+G) is suitable for most karaoke machines. It includes an MP3 file and synchronized lyrics (Karaoke Version only sells digital files (MP3+G) and you will NOT receive a CD). This universal format works with almost any device (Windows, Mac, iPhone, iPad, Android, Connected TVs...) Your purchase allows you to download your video in all of these formats as often as you like. About With backing vocals (with or without vocals in the KFN version) Same as the original tempo: 125 BPM In the same key as the original: E♭m Duration: 04:29 - Preview at: 01:41 Release date: 2014 Genres: Pop, Dance, Alternative, In English Original songwriters: Guy Berryman, Jonathan Mark Buckland, William Champion, Chris Martin, Avicii All the content on our website is entirely reproduced by our musicians in studio. We do not use any parts of the original recordings and do not make use of AI stem separation technology in any way. Lyrics A Sky Full of Stars 'Cos you're a sky 'Cos you're a sky full of stars I'm gonna give you my heart 'Cos you're a sky 'Cos you're a sky full of stars 'Cos you light up the path I don't care go on and tear me apart I don't care if you do ooh ooh ooh ooh 'Cos in a sky 'Cos in a sky full of stars I think I saw you ooh ooh ooh ooh ooh ooh 'Cos you're a sky 'Cos you're a sky full of stars I wanna die in your arms ah ah ah ah ah 'Cos you get lighter the more it gets dark I'm gonna give you my heart oh I don't care go on and tear me apart I don't care if you do ooh ooh ooh ooh ooh 'Cos in a sky 'Cos in a sky full of stars I think I see you ooh ooh ooh ooh ooh ooh I think I see you ooh ooh ooh ooh ooh ooh 'Cos you're a sky. Contribute to cos-onboarding/cos-tester development by creating an account on GitHub.

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Described in this report is detected and blocked by FortiGuard Antivirus as:PDF/Agent.A6DC!tr.dldrW32/Agent.7BBA!trW64/UACMe.O!trW64/ValleyRat.A!tr.spyFortiGate, FortiMail, FortiClient, and FortiEDR support the FortiGuard AntiVirus service. The FortiGuard AntiVirus engine is part of each of these solutions. As a result, customers who have these products with up-to-date protections are protected.The FortiGuard CDR (content disarm and reconstruction) service, which runs on both FortiGate and FortiMail, can disarm the malicious macros in the document.We also suggest that organizations go through Fortinet’s free NSE training module: FCF Fortinet Certified Fundamentals. This module is designed to help end users learn how to identify and protect themselves from phishing attacks.FortiGuard IP Reputation and Anti-Botnet Security Service proactively block these attacks by aggregating malicious source IP data from the Fortinet distributed network of threat sensors, CERTs, MITRE, cooperative competitors, and other global sources that collaborate to provide up-to-date threat intelligence about hostile sources.If you believe this or any other cybersecurity threat has impacted your organization, please contact our Global FortiGuard Incident Response Team.IOCsIP43[.]137[.]42[.]254206[.]238[.]221[.]60206[.]238[.]221[.]240124[.]156[.]100[.]172206[.]238[.]221[.]244Domain1234[.]360sdgg[.]com9001[.]360sdgg[.]com9002[.]360sdgg[.]com9003[.]360sdgg[.]com9005[.]360sdgg[.]com9006[.]360sdgg[.]com9007[.]360sdgg[.]com9009[.]360sdgg[.]com9010[.]360sdgg[.]comffggssa-1329400280[.]cos[.]ap-guangzhou[.]myqcloud[.]comfuued5-1329400280[.]cos[.]ap-guangzhou[.]myqcloud[.]com0107-1333855056[.]cos[.]ap-guangzhou[.]myqcloud[.]comrgghrt1140120-1336065333[.]cos[.]ap-guangzhou[.]myqcloud[.]comhei-1333855056[.]cos[.]ap-guangzhou[.]myqcloud[.]comchakan202501-1329400280[.]cos[.]ap-guangzhou[.]myqcloud[.]comwrwyrdujtw114117-1336065333[.]cos[.]ap-guangzhou[.]myqcloud[.]comfdsjg114-1336065333[.]cos[.]ap-guangzhou[.]myqcloud[.]comsjujfde-1329400280[.]cos[.]ap-guangzhou[.]myqcloud[.]comhtrfe4-1329400280[.]cos[.]ap-guangzhou[.]myqcloud[.]com0611-1333855056[.]cos[.]ap-guangzhou[.]myqcloud[.]comtwzfw[.]vipPhishing 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(20c34b5f0983021414b168913c3da267caf298d8f0f5e3ec0ce97db5f4f48316 Corrupt)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B on both the sides, we get;b – CD = b – a cos CorDA = b – a cos CAgain, in triangle BCD, as per the trigonometry ratio, we know;sin C = BD/a [sin θ = Perpendicular/Hypotenuse]or we can write;BD = a sin C ……….(2)Now using Pythagoras theorem in triangle ADB, we get;c2 = BD2 + DA2 [Hypotenuse2 = Perpendicular2 + Base2 ]Substituting the value of DA and BD from equation 1 and 2, we get;c2 = (a sin C)2 + (b – a cos C)2c2 = a2 sin2C + b2 – 2ab cos C + a2 cos2 Cc2 = a2 (sin2C + cos2 C) + b2 – 2ab cos CBy trigonometric identities, we know;sin2θ+ cos2θ = 1Therefore,c2 = a2 + b2 – 2ab cos CHence, proved.Cosine FormulaThe formula to find the sides of the triangle using cosine rule is given below:\(\begin{array}{l}a = \sqrt{b^2 + c^2 – 2~b~c~ cos x}\end{array} \)\(\begin{array}{l}b = \sqrt{a^2 + c^2 – 2~a~c~ cos y}\end{array} \)\(\begin{array}{l}c = \sqrt{a^2 + b^2 – 2~a~b~cos z}\end{array} \)Sine FormulaAs per sine law, a / Sin A= b/ Sin B= c / Sin CWhere a,b and c are the sides of a triangle and A, B and C are the respective angles. Also, we can write:a: b: c = Sin A: Sin B: Sin CSolved ExampleFind the length of x in the following figure.Solution: By applying the Cosine rule, we get:x2 = 222 +282 – 2 x 22 x 28 cos 97x2 = 1418.143x = √ 1418.143Learn more about different Math topics with BYJU’S – The Learning AppFrequently Asked Questions – FAQsQ1 What is the cosine rule for a triangle?According to the Cosine Rule, the square of the length of any one side of a triangle is equal to the sum of the squares of the length of the other two sides subtracted by twice their product multiplied by the cosine of their included angle. Mathematically it is given as:a2 = b2 + c2 – 2bc cos xQ2 When can we use the cosine rule?We can use the cosine rule,Either, when all three sides of the triangle are given and. Contribute to cos-onboarding/cos-tester development by creating an account on GitHub. View the profiles of people named Cos Tester. Join Facebook to connect with Cos Tester and others you may know. Facebook gives people the power to share

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WEAPONGUI ANIMATE.Parent = nil for _,v in next, Humanoid:GetPlayingAnimationTracks() do v:Stop(); end if Character:FindFirstChildOfClass("Humanoid") == nil then Humanoid = IT("Humanoid",Character) end SINE = SINE + CHANGE local TORSOVELOCITY = (RootPart.Velocity * VT(1, 0, 1)).magnitude local TORSOVERTICALVELOCITY = RootPart.Velocity.y local HITFLOOR = Raycast(RootPart.Position, (CF(RootPart.Position, RootPart.Position + VT(0, -1, 0))).lookVector, 4, Character) local WALKSPEEDVALUE = 4 if ANIM == "Walk" and TORSOVELOCITY > 1 and PLAYSONG == true then RootJoint.C1 = Clerp(RootJoint.C1, ROOTC0 * CF(0, 0, -0.15 * COS(SINE / (WALKSPEEDVALUE / 2))) * ANGLES(RAD(0), RAD(0) - RootPart.RotVelocity.Y / 75, RAD(0)), 2 * (Humanoid.WalkSpeed / 16) / Animation_Speed) Neck.C1 = Clerp(Neck.C1, CF(0, -0.5, 0) * ANGLES(RAD(-90), RAD(0), RAD(180)) * ANGLES(RAD(2.5 * SIN(SINE / (WALKSPEEDVALUE / 2))), RAD(0), RAD(0) - Head.RotVelocity.Y / 30), 0.2 * (Humanoid.WalkSpeed / 16) / Animation_Speed) RightHip.C1 = Clerp(RightHip.C1, CF(0.5, 0.875 - 0.125 * SIN(SINE / WALKSPEEDVALUE) - 0.15 * COS(SINE / WALKSPEEDVALUE*2), -0.125 * COS(SINE / WALKSPEEDVALUE) +0.2- 0.2 * COS(SINE / WALKSPEEDVALUE)) * ANGLES(RAD(0), RAD(90), RAD(0)) * ANGLES(RAD(0) - RightLeg.RotVelocity.Y / 75, RAD(0), RAD(65 * COS(SINE / WALKSPEEDVALUE))), 0.5 * (Humanoid.WalkSpeed / 16) / Animation_Speed) LeftHip.C1 = Clerp(LeftHip.C1, CF(-0.5, 0.875 + 0.125 * SIN(SINE / WALKSPEEDVALUE) - 0.15 * COS(SINE / WALKSPEEDVALUE*2), 0.125 * COS(SINE / WALKSPEEDVALUE) +0.2+ 0.2 * COS(SINE / WALKSPEEDVALUE)) * ANGLES(RAD(0), RAD(-90), RAD(0)) * ANGLES(RAD(0) + LeftLeg.RotVelocity.Y / 75, RAD(0), RAD(65 * COS(SINE / WALKSPEEDVALUE))), 0.5 * (Humanoid.WalkSpeed / 16) / Animation_Speed) elseif (ANIM ~= "Walk") or (TORSOVELOCITY RootJoint.C1 = Clerp(RootJoint.C1, ROOTC0 * CF(0, 0, 0) * ANGLES(RAD(0), RAD(0), RAD(0)), 0.2 / Animation_Speed) Neck.C1 = Clerp(Neck.C1, CF(0, -0.5, 0) * ANGLES(RAD(-90), RAD(0), RAD(180)) * ANGLES(RAD(0), RAD(0), RAD(0)), 0.2 / Animation_Speed) RightHip.C1 = Clerp(RightHip.C1, CF(0.5, 1, 0) * ANGLES(RAD(0), RAD(90), RAD(0)) * ANGLES(RAD(0), RAD(0), RAD(0)), 1 / Animation_Speed) LeftHip.C1 = Clerp(LeftHip.C1, CF(-0.5, 1, 0) * ANGLES(RAD(0), RAD(-90), RAD(0)) * ANGLES(RAD(0), RAD(0), RAD(0)), 1 /