Question

If \[\sin 2x = \dfrac{{2024}}{{2025}}\], where \[\dfrac{{5\pi }}{4} < x < \dfrac{{9\pi }}{4}\]. The value of \[\sin x - \cos x\] is equal to?

Answer 1

Hint: Here we use formula \[{(a - b)^2} = {a^2} + {b^2} - 2ab\] to open up the square of \[\sin x - \cos x\]. Put the value of \[\sin 2x\] in the expansion from the statement of the question and solve for the value using the trigonometric formulas. We check for the sign of the value obtained by converting \[\sin x - \cos x\] into a sine function using the formula for \[\sin (A - B)\] and checking if it is positive or negative with the help of quadrant diagram.
* \[{\sin ^2}x + {\cos ^2}x = 1\]
* \[2\sin x\cos x = \sin 2x\]
*\[\sin (A - B) = \sin A\cos B - \sin B\cos A\]

Complete step-by-step answer:
We are given \[\sin 2x = \dfrac{{2024}}{{2025}}\].
We have to find the value of \[\sin x - \cos x\].
We know \[{(a - b)^2} = {a^2} + {b^2} - 2ab\].
Substitute the value of \[a = \sin x,b = \cos x\]
\[ \Rightarrow {(\sin x - \cos x)^2} = {\sin ^2}x + {\cos ^2}x - 2\sin x\cos x\]
Use the trigonometric identity \[{\sin ^2}x + {\cos ^2}x = 1\] in the RHS of the equation.
\[ \Rightarrow {(\sin x - \cos x)^2} = 1 - 2\sin x\cos x\]
Now substitute the value of \[2\sin x\cos x = \sin 2x\] in the RHS of the equation.
\[ \Rightarrow {(\sin x - \cos x)^2} = 1 - \sin 2x\]
Substitute the value of \[\sin 2x = \dfrac{{2024}}{{2025}}\] in the RHS of the equation.
\[ \Rightarrow {(\sin x - \cos x)^2} = 1 - \dfrac{{2024}}{{2025}}\]
Take LCM on the RHS of the equation.
\[ \Rightarrow {(\sin x - \cos x)^2} = \dfrac{{2025 - 2024}}{{2025}}\]
\[ \Rightarrow {(\sin x - \cos x)^2} = \dfrac{1}{{2025}}\]
We can write the value in the denominator as \[2025 = {5^2} \times {9^2}\]
On calculation we get \[2025 = {(9 \times 5)^2} = {(45)^2}\]
\[ \Rightarrow {(\sin x - \cos x)^2} = \dfrac{1}{{{{(45)}^2}}}\]
Take square root on both sides of the equation
\[ \Rightarrow \sqrt {{{(\sin x - \cos x)}^2}} = \sqrt {\dfrac{1}{{{{(45)}^2}}}} \]
Cancel square root by square power on both sides of the equation.
\[ \Rightarrow \sin x - \cos x = \pm \dfrac{1}{{45}}\]
We have to find the sign of \[\sin x - \cos x\]
Let us multiply and divide the value \[\sin x - \cos x\]by \[\sqrt 2 \].
\[ \Rightarrow \dfrac{{\sqrt 2 }}{{\sqrt 2 }}(\sin x - \cos x) = \sqrt 2 \left( {\sin x \times \dfrac{1}{{\sqrt 2 }} - \cos x \times \dfrac{1}{{\sqrt 2 }}} \right)\]
We substitute \[\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\] and \[\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
\[ \Rightarrow \dfrac{{\sqrt 2 }}{{\sqrt 2 }}(\sin x - \cos x) = \sqrt 2 \left( {\sin x \times \cos \dfrac{\pi }{4} - \cos x \times \sin \dfrac{\pi }{4}} \right)\]
Use the formula \[\sin (A - B) = \sin A\cos B - \sin B\cos A\]
\[ \Rightarrow \dfrac{{\sqrt 2 }}{{\sqrt 2 }}(\sin x - \cos x) = \sqrt 2 \left( {\sin (x - \dfrac{\pi }{4})} \right)\]
Now we are given the condition \[\dfrac{{5\pi }}{4} < x < \dfrac{{9\pi }}{4}\]
Therefore subtract \[\dfrac{\pi }{4}\] from all values in the inequality \[\dfrac{{5\pi }}{4} < x < \dfrac{{9\pi }}{4}\]
\[ \Rightarrow \dfrac{{5\pi }}{4} - \dfrac{\pi }{4} < x - \dfrac{\pi }{4} < \dfrac{{9\pi }}{4} - \dfrac{\pi }{4}\]
\[ \Rightarrow \dfrac{{4\pi }}{4} < x - \dfrac{\pi }{4} < \dfrac{{8\pi }}{4}\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow \pi < x - \dfrac{\pi }{4} < 2\pi \]
Now apply sine function to values in the inequality
\[ \Rightarrow \sin \pi < \sin (x - \dfrac{\pi }{4}) < \sin 2\pi \]
We see the quadrant diagram to check if sine is positive or negative in between \[\pi \] and \[2\pi \]

Since, sine function is negative between \[\pi ({180^ \circ })\]and \[2\pi ({360^ \circ })\]
\[ \Rightarrow \dfrac{{\sqrt 2 }}{{\sqrt 2 }}(\sin x - \cos x) = \sqrt 2 \left( {\sin (x - \dfrac{\pi }{4})} \right)\] is a negative value.

Therefore, value of \[\sin x - \cos x = - \dfrac{1}{{45}}\]

Note: Students make mistake of writing the answer after removing the square root as \[\dfrac{1}{{45}}\], which is wrong because we have to take both the possibilities of negative and positive terms after removing square root from any value.
Also, keep in mind while checking the values from the quadrant diagram we move in an anticlockwise direction.