If x + 1 x = 2 c o s θ , then x 3 + 1 x 3 = Options: A . &nbsp c o s 3 θ B . &nbsp 2 c o s 3 θ C . &nbsp 1 2 c o s 3 θ D . &nbsp 1 3 c o s 3 θ

Answer: Option B : B We have x + 1 x = 2 c o s θ , Now x 3 + 1 x 3 = ( x + 1 x ) 3 - 3 x 1 x ( x + 1 x ) = ( 2 c o s θ ) 3 − 3 ( 2 c o s θ ) = 8 c o s 3 θ − 6 c o s θ = 2 ( 4 c o s 3 θ − 3 c o s θ ) = 2 c o s 3 θ . Trick: Put x = 1 ⇒ θ = 0 ∘ . Then x 3 + 1 x 3 = 2 = 2 c o s 3 θ .

If x + 1x = 2cosθ, then x3 + 1x3 =

Discussion Forum : Trigonometric Functions

Question -

If x + $\frac{1}{x}$ = $2 c o s θ$ , then $x^{3}$ + $\frac{1}{x^{3}}$ =

Options:

A .

c o s 3 θ

B .

2 c o s 3 θ

C .

\frac{1}{2}

c o s 3 θ

D .

\frac{1}{3}

c o s 3 θ

Answer: Option B
:
B

We have x + $\frac{1}{x}$ = $2 c o s θ$ ,

Now $x^{3}$ + $\frac{1}{x^{3}}$ = ${(x + \frac{1}{x})}^{3}$ - $3 x \frac{1}{x} (x + \frac{1}{x})$

= $(2 c o s θ)^{3} - 3 (2 c o s θ) = 8 c o s^{3} θ - 6 c o s θ$

= $2 (4 c o s^{3} θ - 3 c o s θ) = 2 c o s 3 θ$ .

Trick: Put x = 1 $\Rightarrow$ $θ = 0^{\circ}$ .

Then $x^{3}$ + $\frac{1}{x^{3}}$ = 2 = $2 c o s 3 θ$ .

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Discussion Forum : Trigonometric Functions

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