Question

The value of $({{\sin }^{6}}x+{{\cos }^{6}}x)-3({{\sin }^{4}}x+{{\cos }^{4}}x)+1$ is equal to –
(a) 0
(b) 1
(c) -2
(d) None of these

Answer 1

Hint: To solve this problem, we will start by using the below algebraic identities –
${{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab$
${{a}^{3}}+{{b}^{3}}={{(a+b)}^{3}}-3ab(a+b)$
The trigonometric identity in concern to be used in this problem is ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. We will use these properties in combination to solve the problem in hand.

Complete step-by-step answer:
To solve this problem, we will first consider the first term of the expression$({{\sin }^{6}}x+{{\cos }^{6}}x)-3({{\sin }^{4}}x+{{\cos }^{4}}x)+1$, thus, we have,
= $({{\sin }^{6}}x+{{\cos }^{6}}x)$
Using the algebraic identity ${{a}^{3}}+{{b}^{3}}={{(a+b)}^{3}}-3ab(a+b)$, we have,
= ${{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{3}}-3{{\sin }^{2}}x\,{{\cos }^{2}}x ({{\sin }^{2}}x+{{\cos }^{2}}x)$
Since, ${{({{\sin }^{6}}x)}^{\dfrac{1}{3}}}={{\sin }^{2}}x$.
Now, we use the fact that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, thus, we have,
= 1 - $3{{\sin }^{2}}x\,{{\cos }^{2}}x$ -- (1)
Now, we consider the second term of the expression$({{\sin }^{6}}x+{{\cos }^{6}}x)-3({{\sin }^{4}}x+{{\cos }^{4}}x)+1$, thus, we have,
= $3({{\sin }^{4}}x+{{\cos }^{4}}x)$
Using the algebraic identity ${{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab$, we have,
= $3[{{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x\,{{\cos }^{2}}x]$
Since, ${{({{\sin }^{4}}x)}^{\dfrac{1}{2}}}={{\sin }^{2}}x$.
Now, we use the fact that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, thus, we have,
= 3[1 - $2{{\sin }^{2}}x\,{{\cos }^{2}}x$] -- (2)
Now, we combine the values obtained from (1) and (2) in the final expression. Thus, we have,
= $({{\sin }^{6}}x+{{\cos }^{6}}x)-3({{\sin }^{4}}x+{{\cos }^{4}}x)+1$
= 1 - $3{{\sin }^{2}}x\, {{\cos }^{2}}x$ - 3[1 - $2{{\sin }^{2}}x\, {{\cos }^{2}}x$] + 1
= 1 - $3{{\sin }^{2}}x\, {{\cos }^{2}}x$ - 3 + $6{{\sin }^{2}}x\, {{\cos }^{2}}x$ + 1
= -1 + $3{{\sin }^{2}}x\, {{\cos }^{2}}x$ -- (A)
Hence, the expression can have different values depending on the value of x. To explain if x = 90 degrees, the answer would be -1, whereas if x = 45 degrees, the answer would be -0.25. Hence, in short, the answer would be represented by the trigonometric expression given in the final result (A). Thus, the correct answer is (d) None of these.

Note: While solving problems related to trigonometric expressions involving multi correct options, we can actually put different values of x in the expression and find the answer. To explain, let’s first put x = 0 degrees in $({{\sin }^{6}}x+{{\cos }^{6}}x)-3({{\sin }^{4}}x+{{\cos }^{4}}x)+1$, we get answer as 1 – 3 + 1 = -1. Now, we put, x = 45 degrees in $({{\sin }^{6}}x+{{\cos }^{6}}x)-3({{\sin }^{4}}x+{{\cos }^{4}}x)+1$, we get $\left( {{\left( \dfrac{1}{\sqrt{2}} \right)}^{6}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{6}} \right)-3\left( {{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}} \right)+1$ = $\dfrac{1}{4}-3\left( \dfrac{1}{2} \right)+1$ = $-\dfrac{1}{4}$. Thus, we get two different answers for different values of x, thus the trigonometric expression is dependent on x, thus, since all the options are constant, the answer has to be none of these.