Question

if A=30, verify that cos 3A=cosA(4cos²A-3)​

Answer 1

Given :

• The value of A = 30°

To find :

To proof that cos3A = cos(4cos²A - 3).

Solution :

First let us solve the LHS of the equation.

By Substituting the value of A in the equation , we get :

$:\implies \bf{LHS = cos3A}$

We know that , cos3θ = 4cos³θ - 3cosθ , so using it and substituting it in the equation , we get :

$:\implies \bf{Cos3A = 4cos^{3}A - 3cosA }$

$:\implies \bf{Cos3A = 4cos^{3}30^{\circ} - 3cos30^{\circ}}$

We know that cos30° = √3/2, by using it and substituting it in the equation, we get :

$:\implies \bf{Cos3A = 4 \times \bigg(\dfrac{\sqrt{3}}{2}\bigg)^{3} - 3 \times \dfrac{\sqrt{3}}{2}}$

$:\implies \bf{Cos3A = 4 \times \dfrac{3\sqrt{3}}{8} - 3 \times \dfrac{\sqrt{3}}{2}}$

$:\implies \bf{Cos3A = \not{4} \times \dfrac{3\sqrt{3}}{\not{8}} - 3 \times \dfrac{\sqrt{3}}{2}}$

$:\implies \bf{Cos3A = \dfrac{3\sqrt{3}}{2} - 3 \times \dfrac{\sqrt{3}}{2}}$

$:\implies \bf{Cos3A = \dfrac{3\sqrt{3}}{2} - \dfrac{3\sqrt{3}}{2}}$

$:\implies \bf{Cos3A = \dfrac{3\sqrt{3} - 3\sqrt{3}}{2}}$

$:\implies \bf{Cos3A = \dfrac{\not{3\sqrt{3}} - \not{3\sqrt{3}}}{2}}$

$:\implies \bf{Cos3A = \dfrac{0}{2}}$

Since , we know that anything divided by zero , gives zero.

$:\implies \bf{Cos3A = 0}$

$\boxed{\therefore \bf{Cos3A = 0}}$

Hence the LHS of the equation is 0.

Now let's solve the RHS of the Equation :

By Substituting the value of A in the equation , we get :

$:\implies \bf{RHS = cosA(4cos^{2}A - 3)}$

$:\implies \bf{cosA(4cos^{2}A - 3) = cos30^{\circ}(4cos^{2}30^{\circ} - 3)}$

We know that cos30° = √3/2, by using it and substituting it in the equation, we get :

$:\implies \bf{cosA(4cos^{2}A - 3) = \dfrac{\sqrt{3}}{2} \times \bigg[4 \times \bigg(\dfrac{\sqrt{3}}{2}\bigg)^{2} - 3\bigg]}$

$:\implies \bf{cosA(4cos^{2}A - 3) = \dfrac{\sqrt{3}}{2} \times \bigg[4 \times \dfrac{3}{4} - 3\bigg]}$

$:\implies \bf{cosA(4cos^{2}A - 3) = \dfrac{\sqrt{3}}{2} \times \bigg[\not{4} \times \dfrac{3}{\not{4}} - 3\bigg]}$

$:\implies \bf{cosA(4cos^{2}A - 3) = \dfrac{\sqrt{3}}{2} \times (3 - 3)}$

$:\implies \bf{cosA(4cos^{2}A - 3) = \dfrac{\sqrt{3}}{2} \times (\not{3} - \not{3})}$

$:\implies \bf{cosA(4cos^{2}A - 3) = \dfrac{\sqrt{3}}{2} \times 0}$

Since , we know that anything multiplied by zero , gives zero.

$:\implies \bf{cosA(4cos^{2}A - 3) = 0}$

$\boxed{\therefore \bf{cosA(4cos^{2}A - 3) = 0}}$

Hence the RHS of the equation is 0.

Now by putting LHS and RHS together , we get :

$:\implies \bf{LHS = RHS}$

$:\implies \bf{0 = 0}$

Hence it is proved that ,

$\boxed{\therefore \bf{cos3A = cosA(4cos^{2}A - 3)}}$

Proved !!