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Question: Solve the trigonometric expression \({\sin ^6}A + {\cos ^6}A + 3{\sin ^2}A{\cos ^2}A - 1 = ? \)...

Solve the trigonometric expression sin⁡6A+cos⁡6A+3sin⁡2Acos⁡2A−1=?{\sin ^6}A + {\cos ^6}A + 3{\sin ^2}A{\cos ^2}A - 1 = ?

Explanation

Solution

Here in the given question we need to simplify the given expression by using the trigonometric formulae which is:

\sin 2A = 2\sin A\cos A \\\ \sin (A + B) = \sin A\cos B + \cos A\sin B $$ Using the above formula we can simplify the terms and get the final answer for the given expression. **Formulae:** Here we have to use the formula which are:

\left( {{{\sin }^2}A + {{\cos }^2}A} \right) = 1 \\
{a^3} + {b^3} = (a + b)({a^2} - 2ab + {b^2}) \\

**Complete step-by-step solution:** Here to simplify the given expression, we need to solve by using the formula mentioned above and then simplify step by step, on solving we get: $$ {\sin ^6}A + {\cos ^6}A + 3{\sin ^2}A{\cos ^2}A - 1 \\\ = {\left( {{{\sin }^2}A} \right)^3} + {\left( {{{\cos }^2}A} \right)^3} + 3{\sin ^2}A{\cos ^2}A - 1 $$ Here first we will use the formulae of : $$ \Rightarrow {a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})$$ On applying the formulae to our equation we get: $${\sin ^6}A + {\cos ^6}A + 3{\sin ^2}A{\cos ^2}A - 1 \\\ = \left( {{{\sin }^2}A + {{\cos }^2}A} \right)\left( {{{\left( {{{\sin }^2}A} \right)}^2} - {{\sin }^2}A{{\cos }^2}A + {{\left( {{{\cos }^2}A} \right)}^2}} \right) + 3{\sin ^2}A{\cos ^2}A - 1 \\\ = \left( 1 \right)\left( {\left( {{{\sin }^4}A} \right) - {{\sin }^2}A{{\cos }^2}A + \left( {{{\cos }^4}A} \right)} \right) + 3{\sin ^2}A{\cos ^2}A - 1 \\\ = {\sin ^4}A + {\cos ^4}A - {\sin ^2}A{\cos ^2}A + 3{\sin ^2}A{\cos ^2}A - 1 \\\ = {\sin ^4}A + {\cos ^4}A + 2{\sin ^2}A{\cos ^2}A - 1 \\\ = {\left( {{{\sin }^2}A} \right)^2} + {\left( {{{\cos }^2}A} \right)^2} + 2{\sin ^2}A{\cos ^2}A - 1 \\\ = {\left( {{\sin }^2}A + {{\cos }^2}A \right)^2} - 1 \\\ = {1^2} - 1 = 1 - 1 = 0 $$ **Hence we got the answer as $$0$$.** **Note:** In mathematics, a trigonometric sign of an angle is sine. Of a right angled triangle, sine is defined as an acute angle for a specified angle. To model the periodic phenomena sine function is commonly used. It must be noted that, The sine function can also be represented as a generalised continued fraction.