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Basic Trigonometric Identities

Prove that: ...

Question

Prove that: ${cos}^{4} A - {sin}^{4} A + 1 = 2 {cos}^{2} A .$

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Solution

Given:
$c o s^{4} A - s i n^{4} A + 1 = (c o s^{2} A - s i n^{2} A) (c o s^{2} A + s i n^{2} A) + 1$

$= c o s^{2} A - s i n^{2} A + 1$

$= c o s^{2} A + c o s^{2} A$

$= 2 c o s^{2} A$

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Similar questions

Q. Prove that :

{sin}^{4} A - {cos}^{4} A = {sin}^{2} A - {cos}^{2} A = 2 {sin}^{2} A - 1 = 1 - 2 {cos}^{2} A

Q. Prove the following statements :

{cos}^{4} A - {sin}^{4} A + 1 = 2 {cos}^{2} A

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

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

Q. Prove that :

{sin}^{4} A + {cos}^{4} A = 1 - 2 {sin}^{2} A {cos}^{2} A

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