Question

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

Answer 1

Hint: 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.