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Question: If \(\sin 2x = \dfrac{{2024}}{{2025}},{\text{where}}\dfrac{{5\pi }}{4} < x < \dfrac{{9\pi }}{4}\) th...

If sin⁡2x=20242025,where5π4<x<9π4\sin 2x = \dfrac{{2024}}{{2025}},{\text{where}}\dfrac{{5\pi }}{4} < x < \dfrac{{9\pi }}{4} then the value of the sin⁡x−cos⁡x\sin x - \cos x is equal to −145\dfrac{{ - 1}}{{45}}.

Explanation

Solution

For solving this problem, first we need to consider an expression and then square that to get the value of the expression in the range.

Formula used:
(a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab
sin⁡2x−cos⁡2x=1{\sin ^2}x - {\cos ^2}x = 1

Complete step-by-step answer:
Consider the expression (sin⁡x−cos⁡x)2{\left( {\sin x - \cos x} \right)^2}
Here we have to use the formula we get,
=sin⁡2x−cos⁡2x−2sin⁡xcos⁡x= {\sin ^2}x - {\cos ^2}x - 2\sin x\cos x
Again applying the formula, we get
=1−sin⁡2x= 1 - \sin 2x
Putting the value of sin⁡2x\sin 2x which is given above in the question
=1−20242025= 1 - \dfrac{{2024}}{{2025}}
Taking LCM we get,
=2025−20242025= \dfrac{{2025 - 2024}}{{2025}}
On subtracting we get,
== 12025\dfrac{1}{{2025}}
Taking square root of the above term we get,
=12025= \sqrt {\dfrac{1}{{2025}}}
After square root of the expression, we get
+145,−145\dfrac{{ + 1}}{{45}},\dfrac{{ - 1}}{{45}}
Now we need to multiply and divide by 2\sqrt 2
=22(sin⁡x−cos⁡x)= \dfrac{{\sqrt 2 }}{{\sqrt 2 }}\left( {\sin x - \cos x} \right)
=2(12sin⁡x−12cos⁡x)= \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }}\sin x - \dfrac{1}{{\sqrt 2 }}\cos x} \right)
It is a property of sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b\sin (a - b) = \sin a\cos b - \cos a\sin b
Hence we can write the above expression as below
=2sin⁡(x−π4)= \sqrt 2 \sin (x - \dfrac{\pi }{4})
We know that, the range is given in the question
5π4⩽x⩽9π4\dfrac{{5\pi }}{4} \leqslant x \leqslant \dfrac{{9\pi }}{4}
Hence, by subtracting π4\dfrac{\pi }{4} on both sides,
5π4−π4⩽x−π4⩽9π4−π4\dfrac{{5\pi }}{4} - \dfrac{\pi }{4} \leqslant x - \dfrac{\pi }{4} \leqslant \dfrac{{9\pi }}{4} - \dfrac{\pi }{4}
On subtracting the terms we get,
π⩽x−π4⩽2π\pi \leqslant x - \dfrac{\pi }{4} \leqslant 2\pi
We know the graph of sin⁡x\sin x lies between the given ranges is positive
Now we will have a negative value that is −145\dfrac{{ - 1}}{{45}} in the given range but not145\dfrac{1}{{45}}.
Therefore, (sin⁡x−cos⁡x)2=−145{\left( {\sin x - \cos x} \right)^2} = \dfrac{{ - 1}}{{45}}
Thus we attained the required result.
Hence proved.

Note: In such types of questions the key concept is that always remember the trigonometric identities.
In this type of question we need to read the range very carefully and also make sure to note down the same range given in the question.
Here the some of the formula for trigonometry are
sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x
cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x
tan⁡2x=2tan⁡x1−tan⁡2x\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}