Question

Anyone help me here please!!!!​

Answer 1

Answer:

here is a formulae

cos2A= 2cos square a -1

Step-by-step explanation:

as given cos A= 1/2 (a+1/a)

by simplifying

cosA =a²+1/2a

now by following the formulae

cos2A = 2cos²A-1

=2(a²+1/2a)² -1

=2(a⁴+2a²+1/4a²)-1

=a⁴+2a²+1/2a² -1

=a⁴+2a²+1-2a²/2a²

=a⁴+1/2a²

=1/2 (a⁴+1/a²)

=1/2(a⁴/a²+1/a²)

=1/2(a²+1/a²)

{proved}

please mark me as a brainilist...plzzz

thank uuu

Answer 2

Answer:  see proof below

Step-by-step explanation:

Use the Double Angle Identity: cos 2A = 2cos²A - 1

Proof  LHS → RHS

Given:                                $\cos A=\dfrac{1}{2}\bigg(a+\dfrac{1}{a}\bigg)$

LHS:                                   cos 2A

Double Angle Identity:      2(cos²A) - 1

                                     $=2\bigg(\dfrac{1}{2}\bigg(a+\dfrac{1}{a}\bigg)^2\bigg)-1$

Simplify:                           $2\bigg(\dfrac{1}{4}\bigg(a^2+2+\dfrac{1}{a^2}\bigg)\bigg)-1$

                                     $=\dfrac{1}{2}\bigg(a^2+2+\dfrac{1}{a^2}\bigg)-1$

                                    $=\dfrac{1}{2}a^2+1+\dfrac{1}{2a^2}-1$

                                    $=\dfrac{1}{2}a^2+\dfrac{1}{2a^2}$

Factor:                         $=\dfrac{1}{2}\bigg(a^2+\dfrac{1}{a^2}\bigg)$

$\dfrac{1}{2}\bigg(a^2+\dfrac{1}{a^2}\bigg)=\dfrac{1}{2}\bigg(a^2+\dfrac{1}{a^2}\bigg)\quad \checkmark$