Question

If we have a trigonometric expression \[\sin x-\cos x=0\], then what is the value of \[{{\sin }^{4}}x+{{\cos }^{4}}x\]?
A). 1
B). \[\dfrac{3}{4}\]
C). \[\dfrac{1}{2}\]
D). \[\dfrac{1}{4}\]

Answer 1

Hint: Take the relation \[\sin x-\cos x=0\] as \[\sin x=\cos x\] then put this relation in the identity \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] then from that get the value of \[{{\sin }^{2}}x\] and \[{{\cos }^{2}}x\] respectively and then further put it back in the expression \[{{\sin }^{4}}x+{{\cos }^{4}}x\] to get the answer.

Complete step-by-step solution:
In the question we are given an equation of \[\sin x\] and \[\cos x\] which is \[\sin x-\cos x=0\] and we have to find the value of \[{{\sin }^{4}}x+{{\cos }^{4}}x\].
So we are given that \[\sin x-\cos x=0\] or \[\sin x=\cos x\]. So we have to find the value such that \[\sin x\] is equal to \[\cos x\].
We know a universal satisfying identity for all ‘x’, \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] which is applicable for all values of ‘x’. So, we are given, \[\sin x=\cos x\]. Now we will take \[\sin x\] as \[\cos x\] and replace it in the identity \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] to proceed. So we are given,
\[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]
On replacing \[\sin x\] by \[\cos x\] we get,
\[{{\cos }^{2}}x+{{\cos }^{2}}x=1\]
Or, \[2{{\cos }^{2}}x=1\]
So, the value of \[{{\cos }^{2}}x=\dfrac{1}{2}\].
If \[\sin x=\cos x\], then \[{{\sin }^{2}}x={{\cos }^{2}}x\]. So, we found out \[{{\cos }^{2}}x=\dfrac{1}{2}\]. So, \[{{\sin }^{2}}x=\dfrac{1}{2}\].
Hence, we were asked to find \[{{\sin }^{4}}x+{{\cos }^{4}}x\] which we can write as, \[{{\left( {{\sin }^{2}}x \right)}^{2}}+{{\left( {{\cos }^{2}}x \right)}^{2}}\].
Now, as we know, \[{{\sin }^{2}}x={{\cos }^{2}}x=\dfrac{1}{2}\].
So, we get,
\[{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{1}{2} \right)}^{2}}\] or \[\dfrac{1}{4}+\dfrac{1}{4}\] or \[\dfrac{1}{2}\].
Hence for given condition \[\sin x-\cos x=0\] then \[{{\sin }^{4}}x+{{\cos }^{4}}x\] will be \[\dfrac{1}{2}\].
So, the correct option is (c).

Note: There is an another method to solve this problem by writing given \[\sin x-\cos x=0\] then\[\sin x=\cos x\] as $\dfrac{\sin x}{\cos x}=1$ and then \[\tan x=1\] and hence we will finding the value of x. Then put the value of x in the expression \[{{\sin }^{4}}x+{{\cos }^{4}}x\] then after solving this equation we will get the final answer which is $\dfrac{1}{2}$.