Answer
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Hint: In this question we will be using very basic formulas. We will step by step simplify this equation by using formulas given below. This type of question you should understand because if you understand one question you can solve another question using this method and formulas.
Formula used:
$
\Rightarrow {a^2} - {b^2} = (a + b)(a - b) \\
\Rightarrow {a^4} - {b^4} = {({a^2})^2} - {({b^2})^2} = ({a^2} - {b^2})({a^2} + {b^2}) \\
\Rightarrow {\sin ^2}x + {\cos ^2}x = 1 \\
\Rightarrow 2{\sin ^2}x - 1 = \cos 2x \\
\Rightarrow 1 - 2{\cos ^2}x = \cos 2x \\
\Rightarrow {\cos ^2}x - {\sin ^2}x = \cos 2x \\
$
Complete step-by-step answer:
For solving this question we are using many formulas, which are already studied and also given in the formula used.
So first we will use this formula
${a^4} - {b^4} = {({a^2})^2} - {({b^2})^2} = ({a^2} - {b^2})({a^2} + {b^2})$
Here, the value of a is $\sin x$ and the value of b is $\cos x$.
So after substituting the value of a and b in formula we get,
${\sin ^4}x - {\cos ^4}x = {({\sin ^2}x)^2} - {({\cos ^2}x)^2} = ({\sin ^2}x - {\cos ^2}x)({\sin ^2}x + {\cos ^2}x)$
Now in the above equation we will use ${\sin ^2}x + {\cos ^2}x = 1$.
So after using ${\sin ^2}x + {\cos ^2}x = 1$ we will get,
$ \Rightarrow ({\sin ^2}x - {\cos ^2}x)({\sin ^2}x + {\cos ^2}x) = ({\sin ^2}x - {\cos ^2}x)(1)$
Now, we know that ${\cos ^2}x - {\sin ^2}x = \cos 2x$.
So after substituting the value of ${\cos ^2}x - {\sin ^2}x = \cos 2x$in equation we get,
$ \Rightarrow {\sin ^2}x - {\cos ^2}x = - ({\cos ^2}x - {\sin ^2}x) = - \cos 2x$
So, here the answer of this question is ${\sin ^4}x - {\cos ^4}x = - \cos 2x$
Note: So, we have seen to get an answer to such a type of question you have to follow steps. Because if you make a mistake in one step then you will not get the right answer. And in this type of question finding mistakes is very difficult. So always be focused when we are solving such a type of question.
Formula used:
$
\Rightarrow {a^2} - {b^2} = (a + b)(a - b) \\
\Rightarrow {a^4} - {b^4} = {({a^2})^2} - {({b^2})^2} = ({a^2} - {b^2})({a^2} + {b^2}) \\
\Rightarrow {\sin ^2}x + {\cos ^2}x = 1 \\
\Rightarrow 2{\sin ^2}x - 1 = \cos 2x \\
\Rightarrow 1 - 2{\cos ^2}x = \cos 2x \\
\Rightarrow {\cos ^2}x - {\sin ^2}x = \cos 2x \\
$
Complete step-by-step answer:
For solving this question we are using many formulas, which are already studied and also given in the formula used.
So first we will use this formula
${a^4} - {b^4} = {({a^2})^2} - {({b^2})^2} = ({a^2} - {b^2})({a^2} + {b^2})$
Here, the value of a is $\sin x$ and the value of b is $\cos x$.
So after substituting the value of a and b in formula we get,
${\sin ^4}x - {\cos ^4}x = {({\sin ^2}x)^2} - {({\cos ^2}x)^2} = ({\sin ^2}x - {\cos ^2}x)({\sin ^2}x + {\cos ^2}x)$
Now in the above equation we will use ${\sin ^2}x + {\cos ^2}x = 1$.
So after using ${\sin ^2}x + {\cos ^2}x = 1$ we will get,
$ \Rightarrow ({\sin ^2}x - {\cos ^2}x)({\sin ^2}x + {\cos ^2}x) = ({\sin ^2}x - {\cos ^2}x)(1)$
Now, we know that ${\cos ^2}x - {\sin ^2}x = \cos 2x$.
So after substituting the value of ${\cos ^2}x - {\sin ^2}x = \cos 2x$in equation we get,
$ \Rightarrow {\sin ^2}x - {\cos ^2}x = - ({\cos ^2}x - {\sin ^2}x) = - \cos 2x$
So, here the answer of this question is ${\sin ^4}x - {\cos ^4}x = - \cos 2x$
Note: So, we have seen to get an answer to such a type of question you have to follow steps. Because if you make a mistake in one step then you will not get the right answer. And in this type of question finding mistakes is very difficult. So always be focused when we are solving such a type of question.
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