Prove the following trigonometric equation
${\sin ^4}A + {\cos ^4}A = 1 - 2{\sin ^2}A \cdot {\cos ^2}A$
Answer
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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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