Question

If sin x + cos x = a, then |sin x – cos x| is equal to
\[\left( a \right)\sqrt{2-{{a}^{2}}}\]
\[\left( b \right)\sqrt{2+{{a}^{2}}}\]
\[\left( c \right)\sqrt{{{a}^{2}}-2}\]
\[\left( d \right)\sqrt{{{a}^{2}}-4}\]

Answer 1

Hint: We are given that sin x + cos x = a and we are asked to find the value of |sin x – cos x|. We will use modular of any function that is given as \[\left| f\left( x \right) \right|=\sqrt{{{\left( f\left( x \right) \right)}^{2}}}.\] We will then put f(x) as sin x – cos x. Then to simplify, we will need \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] and \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab.\] Then using the identity \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] we will get the accurate answer.

Complete step by step answer:
We are given that sin x + cos x = a and we are asked to find the value of |sin x – cos x|. We are asked to find the modular of sin x – cos x. We know that for any function the modulus is given as
\[\left| f\left( x \right) \right|=\sqrt{{{\left( f\left( x \right) \right)}^{2}}}\]
Now, our function is sin x – cos x. So,
\[\left| \sin x-\cos x \right|=\sqrt{{{\left( \sin x-\cos x \right)}^{2}}}\]
Now as we know, \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] so using this, we get,
\[\Rightarrow \left| \sin x-\cos x \right|=\sqrt{{{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x}\]
As we also know that \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] so we get,
\[\Rightarrow \left| \sin x-\cos x \right|=\sqrt{1-2\sin x\cos x}........\left( i \right)\]
Now as we are given that sin x + cos x = a, so we will use this to solve further. Now,
\[\left( \sin x+\cos x \right)=a\]
Squaring both the sides,
\[\Rightarrow {{\left( \sin x+\cos x \right)}^{2}}={{a}^{2}}\]
As, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab,\] so we get,
\[\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x={{a}^{2}}\]
Again using \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] we get,
\[\Rightarrow 1+2\sin x\cos x={{a}^{2}}\]
So, we get the value of 2 sin x cos x as
\[\Rightarrow 2\sin x\cos x={{a}^{2}}-1......\left( ii \right)\]
Now, using (i) and (ii), we get,
\[\Rightarrow \left| \sin x-\cos x \right|=\sqrt{1-2\sin x\cos x}\]
\[\Rightarrow \left| \sin x-\cos x \right|=\sqrt{1-\left( {{a}^{2}}-1 \right)}\left[ \text{As }2\sin x\cos x={{a}^{2}}-1 \right]\]
Simplifying further, we get,
\[\Rightarrow \left| \sin x-\cos x \right|=\sqrt{1-{{a}^{2}}+1}\]
\[\Rightarrow \left| \sin x-\cos x \right|=\sqrt{2-{{a}^{2}}}\]
So, we get our final value as
\[\Rightarrow \left| \sin x-\cos x \right|=\sqrt{2-{{a}^{2}}}\]
Hence, option (a) is the right answer.

Note:
 Always remember that the square of two terms is given as \[{{a}^{2}}+{{b}^{2}}+2ab,\] i.e. \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab.\] Do not solve it as \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}.\] Similarly for the difference that is \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}},\] do not incorrectly solve as \[{{\left( a-b \right)}^{2}}={{a}^{2}}-{{b}^{2}}.\] This will lead us to incorrect solutions. Also, remember \[-\left( {{a}^{2}}-1 \right)=-{{a}^{2}}-\left( -1 \right)=-{{a}^{2}}+1.\] ‘–‘ sign will be multiplied with both the terms.