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Question: Prove that \[{\sin ^6}A + {\cos ^6}A = 1 - 3{\sin ^2}A{\cos ^2}A\]....

Prove that sin⁡6A+cos⁡6A=1−3sin⁡2Acos⁡2A{\sin ^6}A + {\cos ^6}A = 1 - 3{\sin ^2}A{\cos ^2}A.

Explanation

Solution

We will prove this identity using the trigonometric identity i.e. sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1. The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS). We use the formula (a+b)3=a3+b3+3ab(a+b){(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)anda6=(a2)3{a^6} = {({a^2})^3}to solve this expression.

Complete step by step answer:
Given the following equation as below,
sin⁡6A+cos⁡6A=1−3sin⁡2Acos⁡2A{\sin ^6}A + {\cos ^6}A = 1 - 3{\sin ^2}A{\cos ^2}A
Here, we will solve this in two parts. We will solve the first LHS part and then the RHS part (if needed). And will see if both are equal or not. So, let's start this by solving the LHS part first.
LHS=sin⁡6A+cos⁡6ALHS = {\sin ^6}A + {\cos ^6}A
Using the rule, a6=(a2)3{a^6} = {({a^2})^3} in the above expression we will reduce it as below,
sin⁡6A=(sin⁡2A)3{\sin ^6}A = {({\sin ^2}A)^3} and cos⁡6A=(cos⁡2A)3{\cos ^6}A = {({\cos ^2}A)^3}

Using this, we will get,
LHS=(sin⁡2A)3+(cos⁡2A)3LHS= {({\sin ^2}A)^3} + {({\cos ^2}A)^3}
Using this formula (a+b)3=a3+b3+3ab(a+b){(a + b)^3} = {a^3} + {b^3} + 3ab(a + b) in the above expression.
The new expression is expanded as below,
LHS=(sin⁡2A+cos⁡2A)3−3sin⁡2Acos⁡2A(sin⁡2A+cos⁡2A)LHS = {({\sin ^2}A + {\cos ^2}A)^3} - 3{\sin ^2}A{\cos ^2}A({\sin ^2}A + {\cos ^2}A)
Using this formulasin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1in the above expression, we get,
LHS=(1)3−3sin⁡2Acos⁡2A(1)LHS = {(1)^3} - 3{\sin ^2}A{\cos ^2}A(1)
Since, we know that(1)3=1{(1)^3} = 1, the expression is reduced as below,
LHS=1−3sin⁡2Acos⁡2ALHS = 1 - 3{\sin ^2}A{\cos ^2}A
Next, we will solve the RHS part.
RHS=1−3sin⁡2Acos⁡2ARHS = 1 - 3{\sin ^2}A{\cos ^2}A
Here, LHS = RHS. Thus, the given expression is solved.
Hence, prove that sin⁡6A+cos⁡6A=1−3sin⁡2Acos⁡2A{\sin ^6}A + {\cos ^6}A = 1 - 3{\sin ^2}A{\cos ^2}A

We can solve this using other methods too. We will use the trigonometry identity to solve this.
sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1
Cubing on both the sides, we get,
(sin⁡2A)3+(cos⁡2A)3=13{({\sin ^2}A)^3} + {({\cos ^2}A)^3} = {1^3}
Using this formula (a+b)3=a3+b3+3ab(a+b){(a + b)^3} = {a^3} + {b^3} + 3ab(a + b) and(1)3=1{(1)^3} = 1 in the above expression. The new expression is expanded as below,
(sin⁡2A+cos⁡2A)3−3sin⁡2Acos⁡2A(sin⁡2A+cos⁡2A)=1{({\sin ^2}A + {\cos ^2}A)^3} - 3{\sin ^2}A{\cos ^2}A({\sin ^2}A + {\cos ^2}A) = 1
Using this formulasin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1in the above expression.
Also, we will use the rule a6=(a2)3{a^6} = {({a^2})^3}too, we will get,
⇒sin⁡6A+cos⁡6A−3sin⁡2Acos⁡2A(1)=1\Rightarrow {\sin ^6}A + {\cos ^6}A - 3{\sin ^2}A{\cos ^2}A(1) = 1
Using the transposition in the above expression, we will get,
∴sin⁡6A+cos⁡6A=1−3sin⁡2Acos⁡2A\therefore {\sin ^6}A + {\cos ^6}A = 1 - 3{\sin ^2}A{\cos ^2}A
Hence proved.

Note: Any term of an equation may be taken from one side to another with the change in its sign, this does not affect the equality of the statement and this process is called transposition. We can make use of cubic formulas to expand the expression into a simplified form. We can solve using the trigonometry identity sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1, to solve this expression too.