Prove that: \[\text{co}{{\text{s}}^{6}}A+{{\sin }^{6}}A=1-3{{\sin }^{2}}A{{\cos }^{2}}A\]

Hint: We will begin with the left hand side of the equation (1) and factorize the terms in the left hand side using \[{{a}^{3}}+{{b}^{3}}={{(a+b)}^{3}}-3ab(a+b)\]. But here the two terms are to the power 6 which we can write as to the power cube making a equal to \[{{\cos }^{2}}A\] and b equal to \[{{\sin }^{2}}A\] in the above mentioned formula. And then we will use the formula \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\] a few number of times which will help us in proving the given expression.

Complete step-by-step answer:

It is mentioned in the question that \[\text{co}{{\text{s}}^{6}}A+{{\sin }^{6}}A=1-3{{\sin }^{2}}A{{\cos }^{2}}A.......(1)\]

Now beginning with the left hand side of the equation (1) we get,

\[\Rightarrow \text{co}{{\text{s}}^{6}}A+{{\sin }^{6}}A.......(2)\]

Writing \[{{\cos }^{6}}A\] in terms of \[{{({{\cos }^{2}}A)}^{3}}\] and \[{{\sin }^{6}}A\] in terms of \[{{({{\sin }^{2}}A)}^{3}}\] in equation (2) we get,

\[\Rightarrow {{(\text{co}{{\text{s}}^{2}}A)}^{3}}+{{({{\sin }^{2}}A)}^{3}}.......(3)\]

Now we know \[{{a}^{3}}+{{b}^{3}}={{(a+b)}^{3}}-3ab(a+b)\]. So applying this formula in equation (3) we get,

\[\Rightarrow {{(\text{co}{{\text{s}}^{2}}A+{{\sin }^{2}}A)}^{3}}-3({{\cos }^{2}}A{{\sin }^{2}}A)({{\sin }^{2}}A+{{\cos }^{2}}A)....(4)\]

Now we know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]. So applying this in equation (4) we get,
\[\Rightarrow 1-3{{\cos }^{2}}A{{\sin }^{2}}A....(5)\]

Hence from equation (5) we can say that the left hand side is equal to the right hand side in equation (1). Hence we have proved the given expression.

Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We may get confused about how to proceed further after equation (3) but here the key is to factorize the two terms using \[{{a}^{3}}+{{b}^{3}}\]. Also to make things clearer we have done grouping which will directly help us to solve using the formula \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\].