Question

prove it cos^4-sin^4=2cos^2A-1
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Answer 1

Answer:

refer to the attachment for the solution.

if it helps you then mark as brainliest.

Explanation:

formula used:

sin²∅ + cos²∅ = 1.

and

a² - b² = ( a + b)( a - b).

Answer 2

To Prove :

• ${ \cos }^{4} A \: - { \sin }^{4} A = 2 { \cos}^{2} A - 1$

LHS :

• ${ \cos}^{4} A - { \sin }^{4} A$

RHS :

• $2 {cos}^{2} A - 1$

Proving LHS :

${cos}^{4} A - {sin}^{4} A = { ({cos}^{2}A )}^{2} - { {(sin}^{2}A )}^{2}$

$\implies {a}^{2} - {b}^{2} = (a + b)(a - b)$

Now, by using this formula , we get :

${ ({cos}^{2}A )}^{2} - { {(sin}^{2}A)}^{2} = ( {cos}^{2}A + {sin}^{2} A)( {cos}^{2}A - {sin}^{2}A )$

${ ({cos}^{2}A )}^{2} - { {(sin}^{2}A )}^{2} = (1)( {cos}^{2}A - {sin}^{2}A$

$\implies \: {sin}^{2}A = 1 - {cos}^{2} A$

Now, by applying this formula we get :

${cos}^{4} A - {sin}^{4} A = {cos}^{2} A - (1 - {cos}^{2} A)$

$\implies {cos}^{2} A - 1 + {cos}^{2} A$

$\implies2 {cos}^{2} A - 1.$

Therefore, LHS = RHS.

Hence proved that,

${ \cos }^{4} A\: - { \sin }^{4} A = 2 { \cos}^{2} A - 1$