• MATH

sec2(π2−x)−1=cot2(x)sec2(π2-x)-1=cot2(x) Verfiy the identity.

First, sec(x)=1cos(x)sec(x)=1cos(x) and cot(x)=cos(x)sin(x).cot(x)=cos(x)sin(x). Second, cos(π2−x)=sin(x)cos(π2-x)=sin(x) and sin(π2−x)=cos(x).sin(π2-x)=cos(x). Therefore the left part is

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• MATH

sin(t)csc(π2−t)=tan(t)sin(t)csc(π2-t)=tan(t) Verfiy the identity.

Verify the identity: sin(t)csc(π2−t)=tan(t)sin(t)csc(π2-t)=tan(t) Simplify the left side of the equation by using the cofunction identity csc((pi/2)-t)=sec(t). sin(t)sec(t)=tan(t)sin(t)sec(t)=tan(t) Simplify the left side of the…

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• MATH

sec2(y)−cot2(π2−y)=1sec2(y)-cot2(π2-y)=1 Verfiy the identity.

First, sec(y)=1cos(y)sec(y)=1cos(y) and cot(y)=cos(y)sin(y).cot(y)=cos(y)sin(y). Second, cos(π2−y)=sin(y)cos(π2-y)=sin(y) and sin(π2−y)=cos(y).sin(π2-y)=cos(y). Therefore sec2(y)−cot2(π2−y)=1cos2(y)−sin2(y)cos2(y)=sec2(y)-cot2(π2-y)=1cos2(y)-sin2(y)cos2(y)=…

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• MATH

cos2(β)+cos2(π2−β)=1cos2(β)+cos2(π2-β)=1 Verfiy the identity.

cos(π2−b)=sin(b)cos(π2-b)=sin(b) for all b. So the initial identity is equivalent to cos2(b)+sin2(b)=1,cos2(b)+sin2(b)=1, which is true.

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• MATH

1−cos(θ)1+cos(θ)−−−−−−−−−√=1−cos(θ)|sin(θ)|1-cos(θ)1+cos(θ)=1-cos(θ)|sin(θ)| Verfiy the identity.

By squaring both sides you are assuming that the equality is true, but this is what was to be established. Better is to multiply left side by sqrt(1-cos)/sqrt(1-cos) resulting in…

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• MATH

1+sin(θ)1−sin(θ)−−−−−−−−−√=1+sin(θ)|cos(θ)|1+sin(θ)1-sin(θ)=1+sin(θ)|cos(θ)| Verfiy the identity.

Verify 1+sin(θ)1−sin(θ)−−−−−−−−−√=1+sin(θ)|cos(θ)|1+sin(θ)1-sin(θ)=1+sin(θ)|cos(θ)| Working from the left side, we show that it is equivalent to the right side: 1+sin(θ)1−sin(θ)−−−−−−−−−√1+sin(θ)1-sin(θ)…

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• MATH

cos(x)−cos(y)sin(x)+sin(y)+sin(x)−sin(y)cos(x)+cos(y)=0cos(x)-cos(y)sin(x)+sin(y)+sin(x)-sin(y)cos(x)+cos(y)=0 Verfiy the…

We’ll use the formula (a−b)⋅(a+b)=a2−b2.(a-b)⋅(a+b)=a2-b2. Multiply equation by (sin(x)+sin(y))⋅(cos(x)+cos(y))(sin(x)+sin(y))⋅(cos(x)+cos(y)) (the product of the denominators):

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• MATH

tan(x)+cot(y)tan(x)cot(y)=tan(y)+cot(x)tan(x)+cot(y)tan(x)cot(y)=tan(y)+cot(x) Verfiy the identity.

Verify the identity: tan(x)+cot(y)tan(x)cot(y)=tan(y)+cot(x)tan(x)+cot(y)tan(x)cot(y)=tan(y)+cot(x) Rewrite the left side of the equation as two fractions. tan(x)tan(x)cot(y)+cot(y)tan(x)cot(y)=tan(y)+cot(x)tan(x)tan(x)cot(y)+cot(y)tan(x)cot(y)=tan(y)+cot(x)…

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• MATH

tan(x)+tan(y)1−tan(x)tan(y)=cot(x)+cot(y)cot(x)cot(y)−1tan(x)+tan(y)1-tan(x)tan(y)=cot(x)+cot(y)cot(x)cot(y)-1 Verfiy the…

Verify the identity: tan(x)+tan(y)1−tan(x)tan(y)=cot(x)+cot(y)cot(x)cot(y)−1tan(x)+tan(y)1-tan(x)tan(y)=cot(x)+cot(y)cot(x)cot(y)-1 Divide every term on the left side of the equation by tan(x)tan(y)…

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• MATH

(1+sin(y))[1+sin(−y)]=cos2(y)(1+sin(y))[1+sin(-y)]=cos2(y) Verfiy the identity.

Verify (1+sin(y))(1-sin(y))=cos^2(y): Working from the left side we show that it is equivalent to the right side: (1+sin(y))(1-sin(y)) =1-sin^2(y) =cos^2(y) using the pythagorean identity

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• MATH

csc(−x)sec(−x)=−cot(x)csc(-x)sec(-x)=-cot(x) Verfiy the identity.

Verify the identity: csc(−x)sec(−x)=−cot(x)csc(-x)sec(-x)=-cot(x) Simplify the left side of the equation using the following even/odd identities: csc(-x)=-csc(x) and sec(-x)=sec(x) −csc(x)sec(x)=−cot(x)-csc(x)sec(x)=-cot(x)…

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• MATH

tan(x)cot(x)cos(x)=sec(x)tan(x)cot(x)cos(x)=sec(x) Verfiy the identity.

Verify the identity: tan(x)cot(x)cos(x)=sec(x)tan(x)cot(x)cos(x)=sec(x) Simplify the numerator on the left side of the equation. Since tan(x) and cot(x) are reciprocals their product is 1. 1cos(x)=sec(x)1cos(x)=sec(x) Simplify…

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• MATH

cos[(π2)−x]sin[(π2)−x]=tan(x)cos[(π2)-x]sin[(π2)-x]=tan(x) Verfiy the identity.

Verify the identity: cos[(π2)−x]sin[(π2)−x]=tanxcos[(π2)-x]sin[(π2)-x]=tanx Simplify the left side of the equation using the cofunction identity. sin(x)cos(x)=tanxsin(x)cos(x)=tanx Simplify the left side of the equation using…

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• MATH

tan(π2−θ)tan(θ)=1tan(π2-θ)tan(θ)=1 Verfiy the identity.

By definition, tan(t)=sin(t)/cos(t). Also, sin(π2−t)=cos(t)sin(π2-t)=cos(t) and cos(π2−t)=sin(t).cos(π2-t)=sin(t). Therefore tan(π2−t)=cos(t)sin(t)tan(π2-t)=cos(t)sin(t) and tan(π2−t)⋅tan(t)=1,tan(π2-t)⋅tan(t)=1, QED.

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• MATH

cos(x)−cos(x)1−tan(x)=sin(x)cos(x)sin(x)−cos(x)cos(x)-cos(x)1-tan(x)=sin(x)cos(x)sin(x)-cos(x) Verify the identity.

Verify the identity. cos(x)−[cos(x)1−tan(x)]=sinxcosxsinx−cosxcos(x)-[cos(x)1-tan(x)]=sinxcosxsinx-cosx cos(x)(1−tan(x))−cos(x)1−tan(x)=sin(x)cos(x)sin(x)−cos(x)cos(x)(1-tan(x))-cos(x)1-tan(x)=sin(x)cos(x)sin(x)-cos(x)…

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• MATH

1cos(x)+1+1cos(x)−1=−2csc(x)cot(x)1cos(x)+1+1cos(x)-1=-2csc(x)cot(x) Verify the identity.

Verify the identity: 1cos(x)+1+1cos(x)−1=−2csc(x)cot(x)1cos(x)+1+1cos(x)-1=-2csc(x)cot(x)cos(x)−1+cos(x)+1cos2(x)−1=−2csc(x)cot(x)cos(x)-1+cos(x)+1cos2(x)-1=-2csc(x)cot(x) 2cos(x)cos2(x)−1=−2csc(x)cot(x)2cos(x)cos2(x)-1=-2csc(x)cot(x) Use the pythagorean identity…

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• MATH

cos(θ)cot(θ)1−sin(θ)−1=csc(θ)cos(θ)cot(θ)1-sin(θ)-1=csc(θ) Verfiy the identity.

Verify the identity cos(θ)cot(θ)1−sin(θ)−1=csc(θ)cos(θ)cot(θ)1-sin(θ)-1=csc(θ) cos(θ)cot(θ)1−sin(θ)−1=csc(θ)cos(θ)cot(θ)1-sin(θ)-1=csc(θ) Rewrite cot(θ)cot(θ) as a quotient….

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• MATH

1+sin(θ)cos(θ)+cos(θ)1+sin(θ)=2sec(θ)1+sin(θ)cos(θ)+cos(θ)1+sin(θ)=2sec(θ) Verfiy the identity.

Transform the left part: 1+sin(θ)cos(θ)+cos(θ)1+sin(θ)=1+sin(θ)cos(θ)+cos(θ)1+sin(θ)= (1+sin(θ))2+cos2(θ)cos(θ)⋅(1+sin(θ))=(1+sin(θ))2+cos2(θ)cos(θ)⋅(1+sin(θ))=

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• MATH

1sin(x)−1csc(x)=csc(x)−sin(x)1sin(x)-1csc(x)=csc(x)-sin(x) Verfiy the identity.

Verify the identity: 1sin(x)−1csc(x)=csc(x)−sin(x)1sin(x)-1csc(x)=csc(x)-sin(x) Simplify the left side of the equation using the following reciprocal identities: 1/sin(x)=csc(x) and 1/csc(x)=sin(x)….

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• MATH

1tan(x)+1cot(x)=tan(x)+cot(x)1tan(x)+1cot(x)=tan(x)+cot(x) Verfiy the identity.

cot(x)=1/tan(x) and tan(x)=1/cot(x). So the left side is cot(x)+tan(x) and the right is tan(x)+cot(x), which are the same.

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• MATH

sec(x)(csc(x)−2sin(x))=cot(x)−tan(x)sec(x)(csc(x)-2sin(x))=cot(x)-tan(x) Verify the identity.

Transform the left side: sec(x)⋅(csc(x)−2sin(x))=(1cos(x))⋅(1−2sin2(x)sin(x)=cos2(x)−sin2(x)cos(x)⋅sin(x).sec(x)⋅(csc(x)-2sin(x))=(1cos(x))⋅(1-2sin2(x)sin(x)=cos2(x)-sin2(x)cos(x)⋅sin(x).The right side is…

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• MATH

sec(x)−cos(x)=sin(x)tan(x)sec(x)-cos(x)=sin(x)tan(x) Verfiy the identity.

Verify the identity: sec(x)−cos(x)=sin(x)tan(x)sec(x)-cos(x)=sin(x)tan(x) The reciprocal of sec(x)=1/cos(x). sec(x)−cos(x)=sin(x)tan(x)sec(x)-cos(x)=sin(x)tan(x) 1cos(x)−cos(x)=sin(x)tan(x)1cos(x)-cos(x)=sin(x)tan(x) 1−cos2(x)cos(x)=sin(x)tan(x)1-cos2(x)cos(x)=sin(x)tan(x)Use the…

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• MATH

sec(θ)−11−cos(θ)=sec(θ)sec(θ)-11-cos(θ)=sec(θ) Verfiy the identity.

Verify the identity: sec(θ)−11−cos(θ)=sec(θ)sec(θ)-11-cos(θ)=sec(θ) Use the reciprocal identity cos(θ)=1sec(θ)cos(θ)=1sec(θ) sec(θ)−11−cos(θ)=sec(θ)sec(θ)-11-cos(θ)=sec(θ)…

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• MATH

cot(x)sec(x)=csc(x)−sin(x)cot(x)sec(x)=csc(x)-sin(x) Verfiy the identity.

Verify cot(x)/sec(x)=csc(x)-sin(x): Working from the left side, we show that it is equivalent to the right side: cot(x)/sec(x) =cos(x)sin(x)1cos(x)=cos(x)sin(x)1cos(x) =cos2(x)sin(x)=cos2(x)sin(x)…

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• MATH

sec6(x)(sec(x)tan(x))−sec4(x)(sec(x)tan(x))=sec5(x)tan3(x)sec6(x)(sec(x)tan(x))-sec4(x)(sec(x)tan(x))=sec5(x)tan3(x)Verify the identity.

Verify the identity. sec6(x)(sec(x)tan(x))−sec4(x)(sec(x)tan(x))=sec5(x)tan3(x)sec6(x)(sec(x)tan(x))-sec4(x)(sec(x)tan(x))=sec5(x)tan3(x)Factor out the GCF sec4(x)(sec(x)tan(x))sec4(x)(sec(x)tan(x)) sec4(x)(sec(x)tan(x))[sec2(x)−1]=sec5(x)tan3(x)sec4(x)(sec(x)tan(x))[sec2(x)-1]=sec5(x)tan3(x) Use the…

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• MATH

sin12xcosx−sin52xcosx=cos3(x)sin(x)−−−−−√sin12xcosx-sin52xcosx=cos3(x)sin(x) Verify the identity.

Verify the identity. sin12(x)cos(x)−sin52(x)cos(x)=cos3(x)sin(x)−−−−−√sin12(x)cos(x)-sin52(x)cos(x)=cos3(x)sin(x) Factor out the GCF sin12(x)cos(x).sin12(x)cos(x). sin12(x)cos(x)[1−sin2(x)]=cos3(x)sin(x)−−−−−√sin12(x)cos(x)[1-sin2(x)]=cos3(x)sin(x) Use the pythagorean…

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• MATH

1tan(β)+tan(β)=sec2(β)tan(β)1tan(β)+tan(β)=sec2(β)tan(β) Verfiy the identity.

Verify the identity: 1tan(β)+tan(β)=sec2(β)tan(β)1tan(β)+tan(β)=sec2(β)tan(β) 1+tan2(β)tan(β)=sec2(β)tan(β)1+tan2(β)tan(β)=sec2(β)tan(β) Use the pythagorean identity 1+tan2(β)=sec2(β).1+tan2(β)=sec2(β)….

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• MATH

cot2(t)csc(t)=1−sin2(t)sin(t)cot2(t)csc(t)=1-sin2(t)sin(t) Verfiy the identity.

cot(t)=cos(t)sin(t),cot(t)=cos(t)sin(t), csc(t)=1sin(t).csc(t)=1sin(t). So cot2(t)csc(t)=cos2(t)sin2(t)⋅⎛⎝11sin(t)⎞⎠=cos2(t)sin2(t)⋅sin(t)=cos2(t)sin(t),cot2(t)csc(t)=cos2(t)sin2(t)⋅(11sin(t))=cos2(t)sin2(t)⋅sin(t)=cos2(t)sin(t),which is the same as the right part…

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• MATH

cot3(t)csc(t)=cos(t)(csc2(t)−1)cot3(t)csc(t)=cos(t)(csc2(t)-1) Verfiy the identity.

You need to remember that 1csct=sint1csct=sint , hence, replacing sintsint for 1csct1csct to the left side, yields: sint⋅cot3t=(cost)⋅(csc2t−1)sint⋅cot3t=(cost)⋅(csc2t-1)You need to replace cos3tsin3tcos3tsin3t…

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• MATH

tan2(θ)sec(θ)=sin(θ)tan(θ)tan2(θ)sec(θ)=sin(θ)tan(θ) Verfiy the identity.

L.H.S tan2(θ)sec(θ)=tan2(θ)⋅cos(θ)=tan(θ)⋅cos(θ)⋅{tan(θ)}=tan2(θ)sec(θ)=tan2(θ)⋅cos(θ)=tan(θ)⋅cos(θ)⋅{tan(θ)}= tan(θ)⋅cos(θ)⋅{sin(θ)cos(θ)}=tan(θ)⋅sin(θ)=tan(θ)⋅cos(θ)⋅{sin(θ)cos(θ)}=tan(θ)⋅sin(θ)= R.H.S

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• MATH

sin2(α)−sin4(α)=cos2(α)−cos4(α)sin2(α)-sin4(α)=cos2(α)-cos4(α) Verfiy the identity.

Verify: sin2(α)−sin4(α)=cos2(α)−cos4(α)sin2(α)-sin4(α)=cos2(α)-cos4(α) Use the pythagorean identity sin2(α)+cos2(α)=1sin2(α)+cos2(α)=1 if sin2(α)sin2(α) is isolated the pythagorean identity is…

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• MATH

cos2(β)−sin2(β)=1−2sin2(β)cos2(β)-sin2(β)=1-2sin2(β) Verfiy the identity.

Verify cos2(β)−sin2(β)=1−2sin2(β)cos2(β)-sin2(β)=1-2sin2(β) Working from the left side, we show that it is equivalent to the right side: cos2(β)−sin2(β)cos2(β)-sin2(β) =(1−sin2(β))−sin2(β)=(1-sin2(β))-sin2(β)…

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• MATH

cos2(β)−sin2(β)=2cos2(β)−1cos2(β)-sin2(β)=2cos2(β)-1 Verfiy the identity.

Verify the identity: cos2(β)−sin2(β)=2cos2(β)−1cos2(β)-sin2(β)=2cos2(β)-1 Use the pythagorean identity sin2(β)+cos2(β)=1.sin2(β)+cos2(β)=1. If sin2(β)sin2(β) is isolated the pythagorean identity would be…

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• MATH

(1+sin(α))(1−sin(α))=cos2(α)(1+sin(α))(1-sin(α))=cos2(α) Verfiy the identity.

First, (1−x)⋅(1+x)=1−x2.(1-x)⋅(1+x)=1-x2. So the left part is 1−sin2(α),1-sin2(α),which is obviously equal to cos2(α).cos2(α).

1 educator answer

• MATH

cos(x)+sin(x)tan(x)=sec(x)cos(x)+sin(x)tan(x)=sec(x) Verfiy the identity.

Verify cos(x)+sin(x)tan(x)=sec(x)cos(x)+sin(x)tan(x)=sec(x) : cos(x)+sin(x)tan(x)cos(x)+sin(x)tan(x) =cos(x)+sin(x)⋅sin(x)cos(x)=cos(x)+sin(x)⋅sin(x)cos(x) =cos(x)+sin2(x)cos(x)=cos(x)+sin2(x)cos(x) =cos2(x)+sin2(x)cos(x)=cos2(x)+sin2(x)cos(x) =1cos(x)=1cos(x) =sec(x)=sec(x) as required.

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• MATH

cot2(y)(sec2(y)−1)=1cot2(y)(sec2(y)-1)=1 Verfiy the identity.

Verify the identity: cot2(y)[sec2(y)−1]=1cot2(y)[sec2(y)-1]=1 Use the pythagorean identity 1+tan2(y)=sec2(y).1+tan2(y)=sec2(y). If tan2(y)tan2(y) is isolated the identity would be tan2(y)=sec2(y)−1.tan2(y)=sec2(y)-1. cot2(y)[sec2(y)−1]=1cot2(y)[sec2(y)-1]=1…

1 educator answer

• MATH

sec(y)cos(y)=1sec(y)cos(y)=1 Verfiy the identity.

By definition, sec(y)=1cos(y).sec(y)=1cos(y). Therefore sec(y)⋅cos(y)=[1cos(y)]⋅cos(y)=1sec(y)⋅cos(y)=[1cos(y)]⋅cos(y)=1 , QED.

1 educator answer

• MATH

earctan(x)=x3+1−−−−−√earctan(x)=x3+1 Use Newton’s method to find all roots of the equation correct to…

You need to remember how Newton’s method is used, to evaluate the roots of a transcendent equation. xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn) , where xnxn is the approximate solution to the equation…

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• MATH

4(e−x2)sin(x)=x2−x+14(e-x2)sin(x)=x2-x+1 Use Newton’s method to find all roots of the equation…

4(e−x2)sin(x)=x2−x+14(e-x2)sin(x)=x2-x+1 f(x)=4(e−x2)sin(x)−x2+x−1=0f(x)=4(e-x2)sin(x)-x2+x-1=0 To solve equation using Newton’s method apply the formula, xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn)…

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• MATH

cos(x2−x)=x4cos(x2-x)=x4 Use Newton’s method to find all roots of the equation correct to eight…

cos(x2−x)=x4cos(x2-x)=x4 Set the left side equal to zero. 0=x4−cos(x2−x)0=x4-cos(x2-x) To solve using Newton’s method, apply the formula: xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn) Let the function of the given equation…

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• MATH

xx2+1=1−x−−−−−√xx2+1=1-x Use Newton’s method to find all roots of the equation correct to…

xx2+1=1−x−−−−−√xx2+1=1-x f(x)=xx2+1−1−x−−−−−√=0f(x)=xx2+1-1-x=0 f'(x)=(x2+1)−x(2x)(x2+1)2−(12)(1−x)−12(−1)f′(x)=(x2+1)-x(2x)(x2+1)2-(12)(1-x)-12(-1) f'(x)=1−x2(x2+1)2+121−x−−−−−√f′(x)=1-x2(x2+1)2+121-x See the attached graph. From the graph the curve…

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• MATH

x5−3×4+x3−x2−x+6=0x5-3×4+x3-x2-x+6=0 Use Newton’s method to find all roots of the equation…

x5−3×4+x3−x2−x+6=0x5-3×4+x3-x2-x+6=0 To solve this, using Newton’s method, apply the formula:  xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn) Let the function be:  f(x)=x5−3×4+x3−x2−x+6f(x)=x5-3×4+x3-x2-x+6 Then, take its…

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• MATH

x6−x5−6×4−x2+x+10=0x6-x5-6×4-x2+x+10=0 Use Newton’s method to find all roots of the equation…

f(x)=x6−x5−6×4−x2+x+10f(x)=x6-x5-6×4-x2+x+10 f'(x)=6×5−5×4−24×3−2x+1f′(x)=6×5-5×4-24×3-2x+1 xn+1=xn−(xn)6−(xn)5−6(xn)4−(xn)2+xn+106(xn)5−5(xn)4−24(xn)3−2(xn)+1xn+1=xn-(xn)6-(xn)5-6(xn)4-(xn)2+xn+106(xn)5-5(xn)4-24(xn)3-2(xn)+1 See the attached graph. From the graph the…

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• MATH

sin(x)=x2−2sin(x)=x2-2 Use Newton’s method to find all roots of the equation correct to six…

sin(x)=x2−2sin(x)=x2-2 f(x)=x2−2−sin(x)=0f(x)=x2-2-sin(x)=0 f'(x)=2x−cos(x)f′(x)=2x-cos(x) See the attached graph. From the graph, the curve of the function intersects the x-axis at ≈≈ -1.05 and 1.70. These can be used as initial…

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• MATH

x3=tan−1(x)x3=tan-1(x) Use Newton’s method to find all roots of the equation correct to six…

x3=tan−1(x)x3=tan-1(x) f(x)=x3−tan−1(x)=0f(x)=x3-tan-1(x)=0 f'(x)=3×2−(1×2+1)f′(x)=3×2-(1×2+1) To solve using Newton’s method apply the formula, xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn) Plug in f(x) and f'(x) in the formula…

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• MATH

1x=1+x31x=1+x3 Use Newton’s method to find all roots of the equation correct to six decimal…

1x=1+x31x=1+x3 Set the left side equal to zero. 0=1+x3−1×0=1+x3-1x To solve this using Newton’s method, apply the formula: xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn)Let the function be: f(x)=1+x3−1xf(x)=1+x3-1x Take…

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• MATH

(x−2)2=ln(x)(x-2)2=ln(x) Use Newton’s method to find all roots of the equation correct to six…

(x−2)2=ln(x)(x-2)2=ln(x) f(x)=(x−2)2−ln(x)=0f(x)=(x-2)2-ln(x)=0 f'(x)=2(x−2)−1xf′(x)=2(x-2)-1x See the attached graph. The curve of the function intersects the x-axis at x ≈≈ 1.4 and 3. These can be used as initial approximates….

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• MATH

x+1−−−−−√=x2−xx+1=x2-x Use Newton’s method to find all roots of the equation correct to six…

x+1−−−−−√=x2−xx+1=x2-x Set the left side equal to zero. 0=x2−x−x−1−−−−−√0=x2-x-x-1 To solve using Newton’s method, apply the formula: xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn) Let the function f(x) be:…

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• MATH

3cos(x)=x+13cos(x)=x+1 Use Newton’s method to find all roots of the equation correct to six…

3cos(x)=x+13cos(x)=x+1 f(x)=x+1−3cos(x)=0f(x)=x+1-3cos(x)=0 To solve using Newton’s method apply the formula, xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn) f'(x)=1+3sin(x)f′(x)=1+3sin(x) Plug in f(x) and f'(x) in the formula,…

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• MATH

100−−−√100100100 Use Newton’s method to approximate the given number correct to eight…

Newton’s method for the equation f(x)=0f(x)=0 is given by xn+1=xn−f(xn)f'(xn)xn+1=xn-f(xn)f′(xn) where xnxn is the nth iteration. In this case, we need to consider the number (100)1100(100)1100 as…

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