Formation of Equations, Expansion of Trigonometric functions

1. Allied Mathematics II
SWATHI SUNDARI.S
Assistant Professor of Mathematics
DEPARTMENT OF MATHEMATICSC(SF)
V.V.VANNIAPERUMAL COLLEGE FOR WOMEN
VIRUDHUNAGAR - 626001.
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2. Formation of Equations
Definition
Let f(x) and g(x) be two polynomials. We say that g(x) divides f(x) if there
exists another polynomial h(x) such that f(x) = h(x)g(x).
Definition
A real or complex number a is called a root of the polynomial f(x) if f(a) = 0.
Remark
a is a root of f(x) if and only if x − a divides f(x) and we say that x − a is a
factor of f(x).
Definition
If r is a positive integer such that (x − a)r
divides f(x) and (x − a)r+1
does not
divide f(x), then we say α is a root of multiplicity r.
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3. Theorem
Fundamental Theorem of Algebra: Every polynomial of degree n > 0 has
atleast one real or imaginary root.
Theorem
Every polynomial of degree n > 0 has exactly n roots where the roots are
counted according to multiplicity r.
Theorem
For any equation with real coefficients the complex roots occur in conjugate
pairs. (i.e) If α + iβ is a root then α − iβ is also a root.
Theorem
Let f(x) be a polynomial with rational coefficients . Let α, β be rational
numbers where β > 0 and β is not a perfect square. Then if α +
√
β is a root
of f(x) = 0 then α −
√
β is also a root.
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4. Expansion of Sin θ,Cos θ and Tan θ
Theorem
When θ is expressed in radians
(i) sin θ = θ -
θ3
3! +
θ5
5! - ....
(ii) cos θ = 1 -
θ2
2! +
θ4
4! - ....
(iii) tan θ = θ +
θ3
3 +
2θ15
15 + ....
Proof :
(i) Let f(θ)= sin θ . By Taylor’s expansion of f about the origin is given by
f(θ) = f(0) + f (0) θ
1! + f (0)θ2
2! + f (0)θ3
3! + ... (1)
Now, f(θ)= sin θ ⇒f(0)= 0; f’(θ)= cos θ ⇒f’(0)= 1;
f (θ)= -sin θ ⇒f”(0)= 0; f”’(θ)= -cos θ ⇒f”’(0)= −1.
Therefore, From (1), sin θ = θ -
θ3
3! +
θ5
5! - ....
(ii) Proof is similar to (i) .
(iii) Proof is similar to (i) .
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5. Problem
If
sinθ
θ = 863
864, then the value of θ is
1
12 radians (approaximately).
Problem
If
tanθ
θ = 2524
2523, then the value of θ is 1◦
58 radians (approaximately).
Problem
If x is small cos(α + x) = cos α − x sin α−
x2
2 cos α +
x3
6 sin α
Problem
If cos (
π
3 + θ) = 0.49 then the value of θ is 40 minutes (approaximately).
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6. Problem
If x =
2
1! -
4
3! +
6
5! -
8
7! +.... and y = 1 +
2
1! -
23
3! +
25
5! -.... then prove that
x2
= y.
Problem
If θ is small then
(i) θ cot θ = 1 -
θ2
3 -
θ4
45 approximately.
(ii)
1
6 sin3
θ =
θ3
3! - (1 + 32
)
θ5
5! + (1 + 32
+ 34
)
θ7
7! -...
Problem
(i) limx→0
3 sin x−sin 3x
x−sin x = 24
(ii) limx→0
cos2
ax−cos2
bx
1−cos cx =
2(b2
−a2
)
c2 .
(ii) limx→π/2 (
sin x+cos 2x
cos2x
) =
3
2.
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7. Hyperbolic Functions
Definition
The hyperbolic functions are defined by
(i) sinh x =
ex
−e−x
2
(ii) cosh x =
ex
+e−x
2
(iii) tanh x =
sinh x
cosh x
(iii) coth x =
cosh x
sinh x
(iii) cosech x =
1
sinh x
(iii) sech x =
1
cosh x
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8. Theorem
(i) sinh−1
x = loge(x + x2 + 1)
(ii) cosh−1
x = loge(x + x2 − 1)
(iii) tanh−1
x = 1
2 log (1+x
1−x ).
Problem
(i) (cosh x + sinh x)n
= cosh nx + sinh nx
(ii) (1+tanh x
1−tanh x ) = cosh 2x + sinh 2x
(iii) If tan(ai + b) = x + iy then
x
y =
sin 2a
sinh 2b
(iv) If tan(Ai + B) = x + iy then x2
+ y2
+ 2x cot 2A = 1
(v) If sin(Ai + B) = x + iy then
x2
sin2A
-
y2
cos2A
=1
(vi) If cos(x + iy) = cos θ + i sin θ then cos 2x + cosh 2y = 2
(vii) If sin(θ + iψ) = tan α + i sin α then cos 2θ cosh 2ψ = 3
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9. Problem
(viii) If cos(x + iy) = r(cos α + i sin α) then y = 1
2 log [sin(x−α)
sin(x+α)]
(ix) If tan(x
2 ) = tanh(x
2 ) then cos x cosh x = 1
(x) Let u = logetan(π
4 + θ
2 ) iff cosh u = sec θ
(xi) Let cos α + i sin α = logetan(π
4 + θ
2 ) iff cosh u = sec θ
(xii) If cos(θ + iψ) = cos(θ + iψ) then sin2
θ = ±sin α
(xii) If tan(θ + iψ) = (cos α + i sin α) then θ = 1
2 nπ + 1
4 π and
ψ = 1
2 log tann(1
4 π + 1
2 α)
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10. Thank You
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