Name: Introduction to Conditional Identity
Uploaded: 2019-07-19
Duration: 2 min 16 s
Description: Watch this video to learn how to find the sum of sines of double angles under the given specific conditions, using different trigonometric identities.

Question

Prove that the equation

x^{3} - 2 x + 1 = 0

is satisfied by putting for

x

, either of the values

\sqrt{2} \sin 45^{°}

,

2 \sin 18^{°}

and

2 \sin 234^{°}

.

Accepted Answer

$x^{3} - 2 x + 1 = 0$

$\Rightarrow x^{2} (x - 1) + x (x - 1) - (x - 1) = 0$

$\Rightarrow (x - 1) (x^{2} + x - 1) = 0$

Hence, either $x - 1 = 0$ or $x^{2} + x - 1 = 0$

If $x - 1 = 0$ then $x = 1$ .

If $x^{2} + x - 1 = 0$ then $x = \frac{- 1 \pm \sqrt{5}}{2}$

Now $\sqrt{2} \sin 45^{°} = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1 \dots (i)$

$2 \sin 18^{°} = 2 \times \frac{\sqrt{5} - 1}{4} = \frac{\sqrt{5} - 1}{2} \dots (ii)$

$2 \sin 234^{°} = 2 \sin (180^{°} + 54^{°}) = - 2 \sin 54^{°}$

$= - 2 (\frac{\sqrt{5} + 1}{4}) = - (\frac{\sqrt{5} + 1}{2}) \dots (iii)$

Clearly $(i), (ii)$ and $(iii)$ are roots of equation $x^{3} - 2 x + 1 = 0$ are same means $\sqrt{2} \sin 45^{°}$ , $2 \sin 18^{°}$ and $2 \sin 234^{°}$ satisfy the equation.

Prove that the equation $x^{3} - 2 x + 1 = 0$ is satisfied by putting for $x$ , either of the values $\sqrt{2} \sin 45^{°}$ , $2 \sin 18^{°}$ and $2 \sin 234^{°}$ .

Important Questions on Identities and Trigonometrical Equations

Learn and Prepare for any exam you want !

Prove that the equation x3-2x+1=0 is satisfied by putting for x, either of the values 2sin45°, 2sin18° and 2sin234°.

Important Questions on Identities and Trigonometrical Equations

Learn and Prepare for any exam you want !

Prove that the equation $x^{3} - 2 x + 1 = 0$ is satisfied by putting for $x$ , either of the values $\sqrt{2} \sin 45^{°}$ , $2 \sin 18^{°}$ and $2 \sin 234^{°}$ .