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If y = sin x:

x can take on any value, so the domain is the set of real numbers.

y can take on values between -1 and 1 (including the extremes); so the range is -1 <= y <= 1.

If y = sin x:

x can take on any value, so the domain is the set of real numbers.

y can take on values between -1 and 1 (including the extremes); so the range is -1 <= y <= 1.

If y = sin x:

x can take on any value, so the domain is the set of real numbers.

y can take on values between -1 and 1 (including the extremes); so the range is -1 <= y <= 1.

If y = sin x:

x can take on any value, so the domain is the set of real numbers.

y can take on values between -1 and 1 (including the extremes); so the range is -1 <= y <= 1.

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15y ago

If y = sin x:

x can take on any value, so the domain is the set of real numbers.

y can take on values between -1 and 1 (including the extremes); so the range is -1 <= y <= 1.

This answer is:
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Q: Domains and ranges of sinx
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What jobs use domains and range and how do they use it?

Domains and ranges are commonly used in fields such as mathematics, computer science, economics, physics, and engineering. In mathematics, domains and ranges help define the inputs and outputs of functions, which are essential for solving equations and analyzing data. In computer science, domains and ranges are used in programming to determine the scope and limits of variables and functions. In economics, domains and ranges help model relationships between variables in economic systems. In physics and engineering, domains and ranges are crucial for understanding the behavior of physical systems and designing solutions based on specific input-output relationships.


Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?

2


How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?

(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx


How do you verify the identity sinx cscx 1?

sinx cscx = 1 is the same thing as sinx(1/sinx) = 1 which is the same as sinx/sinx = 1. This evaluates to 1=1, which is true.


Can you Show 1 over sinx cosx - cosx over sinx equals tanx?

From the Pythagorean identity, sin2x = 1-cos2x. LHS = 1/(sinx cosx) - cosx/sinx LHS = 1/(sinx cosx) - (cosx/sinx)(cosx/cosx) LHS = 1/(sinx cosx) - cos2x/(sinx cosx) LHS = (1- cos2x)/(sinx cosx) LHS = sin2x /(sinx cosx) [from Pythagorean identity] LHS = sin2x /(sinx cosx) LHS = sinx/cosx LHS = tanx [by definition] RHS = tanx LHS = RHS and so the identity is proven. Q.E.D.

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What jobs use domains and range and how do they use it?

Domains and ranges are commonly used in fields such as mathematics, computer science, economics, physics, and engineering. In mathematics, domains and ranges help define the inputs and outputs of functions, which are essential for solving equations and analyzing data. In computer science, domains and ranges are used in programming to determine the scope and limits of variables and functions. In economics, domains and ranges help model relationships between variables in economic systems. In physics and engineering, domains and ranges are crucial for understanding the behavior of physical systems and designing solutions based on specific input-output relationships.


Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?

2


How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?

(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx


How do you verify the identity sinx cscx 1?

sinx cscx = 1 is the same thing as sinx(1/sinx) = 1 which is the same as sinx/sinx = 1. This evaluates to 1=1, which is true.


Verify that Cos theta cot theta plus sin theta equals csc theta?

It's easiest to show all of the work (explanations/identities), and x represents theta. cosxcotx + sinx = cscx cosx times cosx/sinx + sinx = csc x (Quotient Identity) cosx2 /sinx + sinx = csc x (multiplied) 1-sinx2/sinx + sinx = csc x (Pythagorean Identity) 1/sinx - sinx2/sinx + sinx = csc x (seperate fraction) 1/sinx -sinx + sinx = csc x (canceled) 1/sinx = csc x (cancelled) csc x =csc x (Reciprocal Identity)


Can you Show 1 over sinx cosx - cosx over sinx equals tanx?

From the Pythagorean identity, sin2x = 1-cos2x. LHS = 1/(sinx cosx) - cosx/sinx LHS = 1/(sinx cosx) - (cosx/sinx)(cosx/cosx) LHS = 1/(sinx cosx) - cos2x/(sinx cosx) LHS = (1- cos2x)/(sinx cosx) LHS = sin2x /(sinx cosx) [from Pythagorean identity] LHS = sin2x /(sinx cosx) LHS = sinx/cosx LHS = tanx [by definition] RHS = tanx LHS = RHS and so the identity is proven. Q.E.D.


What is 1 sinx?

If you mean 1 - sinx = 0 then sinx = 1 (sin-1) x = 90


Integration of root sinx?

integration of (sinx)^1/2 is not possible.so integration of root sinx is impossible


What is the derivative of 1 divided by sinx?

y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx


How do you solve csc x-sin x equals cos x cot x?

cscx-sinx=(cosx)(cotx) 1/sinx-sinx=(cosx)(cosx/sinx) (1/sinx)-(sin^2x/sinx)=cos^2x/sinx cos^2x/sinx=cos^2x/sinx Therefore LS=RS You have to remember some trig identities when answering these questions. In this case, you need to recall that sin^2x+cos^2x=1. Also, always switch tanx cotx cscx secx in terms of sinx and cosx.


What is the derivative of sin x minus cos x?

d/dx(sinx-cosx)=cosx--sinx=cosx+sinx


What is sin 3x in terms of sin x?

given the identity sin(x+y)=sinx cosy + siny cosxsin2x = 2 sinx cosx andsin(2(x)+x) = sin 2x cos x + sinx cos 2xusing the last two identities givessin3x= 2 sinx cosx cosx + sinx cos2xfactoring the sinx we havesin3x = sinx(2cosx cosx+cos2x)which satisfies the requirement.However, we can simplify further since cos 2x = cosx cosx - sinx sinx (a well known identity)sin3x = sinx (2cosx cosx +cosx cosx - sinx sinx)so sin3x= sinx(3cosx cosx - sinx sinx)or sin 3x = 3.cos&Acirc;&sup2;x.sinx - sin&Acirc;&sup3;x* * * * *Good, but not good enough. The answer was required in terms of sin, not a mixture of sinx and cosx. Easily recitified, though, since cos&sup2;x = 1 - sin&sup2;xTherefore sin3x = 3*(1-sin&sup2;x)*sinx - sin&sup3;x= 3sinx - 3sin&sup3;x - sin&sup3;x= 3sinx - 4sin&sup3;x