1- n C 1cos- n C 2cos 2 - n C n cos n - equalsA- 2 cos 2 n -2B- 2 cos-2n-cosn -2C- 2 cos 2 n -2D- 2 cos 2-2n

1+ ^nC_1cosθ+^nC_2cos2θ+.........+^nC_ncos nθ can be written as nbsp; Re( nbsp;1+^nC_1e^iθ+^nC_2e^i2θ+…)=Re(1+e^iθ)^n =Re(cosθ+1+isinθ)^n =(2cosθ2)^ncosnθ2 ...

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Standard XII

Mathematics

Integration by Substitution

1+ n C 1cosθ+...

Question

$1 +^{n} C_{1} cos θ +^{n} C_{2} cos 2 θ + . . . . . . . . . +^{n} C_{n} cos n θ$ equals

{(2 cos \frac{θ}{2})}^{n} \cdot cos \frac{n θ}{2}

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2 {cos}^{2} \frac{n θ}{2}

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2 {cos}^{2 n} \frac{θ}{2}

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{(2 {cos}^{2} \frac{θ}{2})}^{n}

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Solution

The correct option is A ${(2 cos \frac{θ}{2})}^{n} \cdot cos \frac{n θ}{2}$
$1 +^{n} C_{1} cos θ +^{n} C_{2} cos 2 θ + . . . . . . . . . +^{n} C_{n} cos n θ$ can be written as
$R e (1 +^{n} C_{1} e^{i θ} +^{n} C_{2} e^{i 2 θ} + \dots) = R e (1 + e^{i θ})^{n} = R e (cos θ + 1 + i sin θ)^{n} = {(2 cos \frac{θ}{2})}^{n} cos \frac{n θ}{2}$

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Similar questions

1 +^{n} C_{1} cos θ +^{n} C_{2} cos 2 θ + . . . . . . . . +^{n} C_{n} cos n θ

equals

(2 c o s θ - 1) (2 c o s 2 θ - 1) (2 c o s 2^{2} θ - 1) . . . . . . (2 c o s 2^{n - 1} θ - 1) = \frac{2 c o s 2^{n} θ + 1}{2 c o s θ + 1}

Q. Show that

\frac{2 cos 2^{n} θ + 1}{2 cos θ + 1} = (2 cos θ - 1) (2 cos 2 θ - 1) (2 cos 2^{2} θ - 1) \dots (2 cos 2^{n - 1} θ - 1)

Q. The value of

(1 + cos θ + i sin θ)^{n} + (1 + cos θ - i sin θ)^{n}

Q. If

(l_{1}, m_{1}, n_{1})

and

(l_{2}, m_{2}, n_{2},)

are d.c.'s of

¯ ¯¯¯¯¯¯¯¯ ¯ O A,

¯ ¯¯¯¯¯¯ ¯ O B

such that

∠ A O B = θ

where ‘O’ is the origin, then the d.c.’s of the internal bisector of the angle

∠ A O B

are

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1+nC1cosθ+nC2cos2θ+.........+nCncosnθ equals

The correct option is A (2cosθ2)n⋅cosnθ21+nC1cosθ+nC2cos2θ+.........+nCncosnθ can be written as Re(1+nC1eiθ+nC2ei2θ+…)=Re(1+eiθ)n=Re(cosθ+1+isinθ)n=(2cosθ2)ncosnθ2

$1 +^{n} C_{1} cos θ +^{n} C_{2} cos 2 θ + . . . . . . . . . +^{n} C_{n} cos n θ$ equals