# Prove the following trigonometric equation

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

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Hint: We are going to relate the given trigonometric equation to the algebraic equation to solve it easily.

Complete step-by-step answer:

Given trigonometric equation is ${\sin ^4}A + {\cos ^4}A = 1 - 2{\sin ^2}A \cdot {\cos ^2}A$

Taking the LHS of the above equation i.e., ${\sin ^4}A + {\cos ^4}A$

From algebra, we have ${\left( {{a^2} + {b^2}} \right)^2} = {a^4} + {b^4} + 2{a^2}{b^2}$

We can write

${a^4} + {b^4} = {\left( {{a^2} + {b^2}} \right)^2} - 2{a^2}{b^2}$

Take $a = \sin A$ and $b = \cos A$ we get

$${\sin ^4}A + {\cos ^4}A = {\left( {{{\sin }^2}A + {{\cos }^2}A} \right)^2} - 2{\sin ^2}A{\cos ^2}A$$

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

= RHS

Hence proved.

Note: We take ‘a’ as sin A and ‘b’ as sin B. Then we use these values in the ${a^4} + {b^4}$ formula to solve the given problem. We used one of the basic trigonometric identity $${\sin ^2}A + {\cos ^2}A = 1.$$

Complete step-by-step answer:

Given trigonometric equation is ${\sin ^4}A + {\cos ^4}A = 1 - 2{\sin ^2}A \cdot {\cos ^2}A$

Taking the LHS of the above equation i.e., ${\sin ^4}A + {\cos ^4}A$

From algebra, we have ${\left( {{a^2} + {b^2}} \right)^2} = {a^4} + {b^4} + 2{a^2}{b^2}$

We can write

${a^4} + {b^4} = {\left( {{a^2} + {b^2}} \right)^2} - 2{a^2}{b^2}$

Take $a = \sin A$ and $b = \cos A$ we get

$${\sin ^4}A + {\cos ^4}A = {\left( {{{\sin }^2}A + {{\cos }^2}A} \right)^2} - 2{\sin ^2}A{\cos ^2}A$$

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

= RHS

Hence proved.

Note: We take ‘a’ as sin A and ‘b’ as sin B. Then we use these values in the ${a^4} + {b^4}$ formula to solve the given problem. We used one of the basic trigonometric identity $${\sin ^2}A + {\cos ^2}A = 1.$$

Last updated date: 18th Sep 2023

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