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There is no sensible or useful simplification.

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โˆ™ 2012-04-23 14:46:55
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Algebra

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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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Q: What is sinx plus cosx plus cotx simplified?
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Related questions

How do you express the term 1-csc squared x all over sin x cot x to cos x?

(1 - csc2x)/(sinx*cotx) = -cot2x/sinxcotx = -cotx/sinx = -(cosx/sinx)/sinx = -cosx/sin2x = -cosx/(1-cos2x) = cosx/(cos2x - 1)


How does secx plus 1 divided by cotx equal 1 plus sinx divided by cosx?

secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = 1 + sinx/cosx, andsin/cos = tanx, therefore1/cosx + tanx = 1 + tanx, therefore1/cosx = 1, therfore1 = cosx.So, therfore, it is not neccesarily true.But if you meansecx plus 1 divided by cotx equals (1 plus sinx) divided by cosx(this is probably what you mean) Let's start over!secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = (1+sinx)/cosx therefore1/cosx + tanx = 1/cosx + sinx/cosxsinx/cosx = tanx therfore1/cosx + tanx = 1/cosx + tanxDo you think this is correct? Subtract both sides by 1/cosx + tanx:0 = 0So, therefore, this is correct!(BTW, I'm in Grade 6! :P)


Tan plus cot divided by tan equals csc squared?

(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.


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

2


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 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 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.


How do you simplify cosx plus sinx tanx?

to simplify Cosx=Sinx Tanx you should remember your fundamental and pythagorean identities.. Cosx + Sinx Tanx Cosx + Sinx (Sinx/Cosx) <---------- From Tanx= Sinx/Cosx Cosx + Sin2x/ Cos x <------------- do the LCD Cosx (Cosx/Cosx) + Sin2x/Cosx (Cos2x+Sin2x)/Cosx 1/Cosx <--------- From Sin2x + Cos2x =1 or Secx <-------- answer Comment if you have questions...:))


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.


Find the critical numbers of sinx plus cosx?

f(x)=sinx+cosx take the derivative f'(x)=cosx-sinx critical number when x=pi/4


What is the derivative of sin x minus cos x?

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


Integral of cos2x log cosx-sinx coax plus since?

it is not possible to get the Integral of cos2x log cosx-sinx coax plus since there are no symbols given in the equation.


How do you take the derivative of a trig function?

Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x


What is the derivative of sin x to the e to the xth power?

y = (sinx)^(e^x) ln(y) = ln((sinx)^(e^x)) ln(y) = (e^x)ln(sinx) (1/y)dy = (e^x)(1/sinx)(cosx)+ln(sinx)(e^x)dx (1/y)dy = (e^x)(cotx)+ln(sinx)(e^x)dx dy = ((sinx)^(e^x))((cotx)(e^x)+ln(sinx)(e^x))dx dy = ((e^x)(sinx)^(e^x))(cotx+ln(sinx))dx


Sinx plus cosx equals 0?

x = 3pi/4


Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?

(1 + tanx)/sinxMultiply by sinx/sinxsinx + tanxsinxDivide by sin2x (1/sin2x) = cscxcscx + tan(x)csc(x)tanx = sinx/cosx and cscx = 1/sinxcscx + (sinx/cosx)(1/sinx)sinx cancels outcscx + 1/cosx1/cosx = secxcscx + secx


What is the derivative of -cosx?

-sinx


How do you prove the following equation the quantity of sin theta divided by 1 minus cos theta minus the quantity 1 plus cos theta divided by sin theta equals 0?

You will have to bear with the angle being represented by x because this browser will not allow characters from other alphabets!sin^2x + cos^2x = 1=> sin^2x = 1 - cos^x = (1 + cosx)(1 - cosx)Divide both sides by sinx (assuming that sinx is not zero).=> sinx = (1 + cosx)(1 - cosx)/sinxDivide both sides by (1 - cosx)=> sinx/(1 - cosx) = (1 + cosx)/sinx=> sinx/(1 - cosx) - (1 + cosx)/sinx = 0


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.cos²x.sinx - sin³x* * * * *Good, but not good enough. The answer was required in terms of sin, not a mixture of sinx and cosx. Easily recitified, though, since cos²x = 1 - sin²xTherefore sin3x = 3*(1-sin²x)*sinx - sin³x= 3sinx - 3sin³x - sin³x= 3sinx - 4sin³x


How do you integrate sinx divide sinx plus cosx?

1/2(x-ln(sin(x)+cos(x)))


Cos x plus sin x equals 0?

cosx + sinx = 0 when sinx = -cosx. By dividing both sides by cosx you get: sinx/cosx = -1 tanx = -1 The values where tanx = -1 are 3pi/4, 7pi/4, etc. Those are equivalent to 135 degrees, 315 degrees, etc.


How do you break 1 sinx divided 1-cosx?

0


What is the derivative of -Cos x?

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


How do you prove that the derivative of csc x is equals to -csc x cot x?

d/dx cscx = d/dx 1/sinx = d/dx (sinx)-1= -(sinx)-2 cosx = -cosx/sin2x = -1/sinx.cosx/sinx = -cscx cotx I suggest you copy this out onto paper so it is more clear. The / signs make it harder to see what is happening compared to horizontal divide lines.


Cosx - sinx?

Given: = cosx - sinxJust isolate sx= sx(co-in)