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Question: If we have the value of x as \(\pi < x< 2\pi \) then find the value of \(\dfrac{\sqrt{1+\cos x}+\sqr...

If we have the value of x as π<x<2π\pi < x< 2\pi then find the value of 1+cos⁡x+1−cos⁡x1+cos⁡x−1−cos⁡x\dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}.

Explanation

Solution

We first try to use the formula of submultiple angle. Then for the root value we take modulus and based on the value of π<x<2π\pi < x <2\pi , we take their respective sign. Then using the formula of tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡A.tan⁡B\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B} we convert the equation into a single trigonometric function.

Complete step-by-step solution
We have the trigonometric submultiple angle formula of 1+cos⁡x=2cos⁡2x21+\cos x=2{{\cos }^{2}}\dfrac{x}{2} and also 1−cos⁡x=2sin⁡2x21-\cos x=2{{\sin }^{2}}\dfrac{x}{2}.
We put the values in the given equation of 1+cos⁡x+1−cos⁡x1+cos⁡x−1−cos⁡x\dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}} to get
1+cos⁡x+1−cos⁡x1+cos⁡x−1−cos⁡x=2cos⁡2x2+2sin⁡2x22cos⁡2x2−2sin⁡2x2\dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}=\dfrac{\sqrt{2{{\cos }^{2}}\dfrac{x}{2}}+\sqrt{2{{\sin }^{2}}\dfrac{x}{2}}}{\sqrt{2{{\cos }^{2}}\dfrac{x}{2}}-\sqrt{2{{\sin }^{2}}\dfrac{x}{2}}}
Now for the square root values we take their modulus values.
2cos⁡2x2+2sin⁡2x22cos⁡2x2−2sin⁡2x2=2∣cos⁡x2∣+2∣sin⁡x2∣2∣cos⁡x2∣−2∣sin⁡x2∣=∣cos⁡x2∣+∣sin⁡x2∣∣cos⁡x2∣−∣sin⁡x2∣\dfrac{\sqrt{2{{\cos }^{2}}\dfrac{x}{2}}+\sqrt{2{{\sin }^{2}}\dfrac{x}{2}}}{\sqrt{2{{\cos }^{2}}\dfrac{x}{2}}-\sqrt{2{{\sin }^{2}}\dfrac{x}{2}}}=\dfrac{\sqrt{2}\left| \cos \dfrac{x}{2} \right|+\sqrt{2}\left| \sin \dfrac{x}{2} \right|}{\sqrt{2}\left| \cos \dfrac{x}{2} \right|-\sqrt{2}\left| \sin \dfrac{x}{2} \right|}=\dfrac{\left| \cos \dfrac{x}{2} \right|+\left| \sin \dfrac{x}{2} \right|}{\left| \cos \dfrac{x}{2} \right|-\left| \sin \dfrac{x}{2} \right|}
As it’s given π<x<2π\pi < x< 2\pi , multiplying 12\dfrac{1}{2} we get π2<x2<π\dfrac{\pi }{2}<\dfrac{x}{2}<\pi . In that range value of cos⁡x2<0\cos \dfrac{x}{2}< 0 and value of sin⁡x2>0\sin \dfrac{x}{2}> 0. The modulus values change to ∣cos⁡x2∣→−cos⁡x2\left| \cos \dfrac{x}{2} \right|\to -\cos \dfrac{x}{2} and ∣sin⁡x2∣→sin⁡x2\left| \sin \dfrac{x}{2} \right|\to \sin \dfrac{x}{2} in the range of π2<x2<π\dfrac{\pi }{2}< \dfrac{x}{2}< \pi .
The final equation becomes ∣cos⁡x2∣+∣sin⁡x2∣∣cos⁡x2∣−∣sin⁡x2∣=−cos⁡x2+sin⁡x2−cos⁡x2−sin⁡x2\dfrac{\left| \cos \dfrac{x}{2} \right|+\left| \sin \dfrac{x}{2} \right|}{\left| \cos \dfrac{x}{2} \right|-\left| \sin \dfrac{x}{2} \right|}=\dfrac{-\cos \dfrac{x}{2}+\sin \dfrac{x}{2}}{-\cos \dfrac{x}{2}-\sin \dfrac{x}{2}}.
Now we take the common of −cos⁡x2-\cos \dfrac{x}{2}.
So, −cos⁡x2+sin⁡x2−cos⁡x2−sin⁡x2=−cos⁡x2(1−tan⁡x2)−cos⁡x2(1+tan⁡x2)\dfrac{-\cos \dfrac{x}{2}+\sin \dfrac{x}{2}}{-\cos \dfrac{x}{2}-\sin \dfrac{x}{2}}=\dfrac{-\cos \dfrac{x}{2}\left( 1-\tan \dfrac{x}{2} \right)}{-\cos \dfrac{x}{2}\left( 1+\tan \dfrac{x}{2} \right)}.
We convert the value 1 into tan⁡π4\tan \dfrac{\pi }{4} and get −cos⁡x2(1−tan⁡x2)−cos⁡x2(1+1.tan⁡x2)=tan⁡π4−tan⁡x21+tan⁡π4.tan⁡x2\dfrac{-\cos \dfrac{x}{2}\left( 1-\tan \dfrac{x}{2} \right)}{-\cos \dfrac{x}{2}\left( 1+1.\tan \dfrac{x}{2} \right)}=\dfrac{\tan \dfrac{\pi }{4}-\tan \dfrac{x}{2}}{1+\tan \dfrac{\pi }{4}.\tan \dfrac{x}{2}}.
Then we use the formula tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡A.tan⁡B\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}
tan⁡π4−tan⁡x21+tan⁡π4.tan⁡x2=tan⁡(π4−x2)\dfrac{\tan \dfrac{\pi }{4}-\tan \dfrac{x}{2}}{1+\tan \dfrac{\pi }{4}.\tan \dfrac{x}{2}}=\tan \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right).
Therefore, the value of 1+cos⁡x+1−cos⁡x1+cos⁡x−1−cos⁡x\dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}} is tan⁡(π4−x2)\tan \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right).

Note: Instead of taking −cos⁡x2-\cos \dfrac{x}{2} as common we could have taken −sin⁡x2-\sin \dfrac{x}{2}. The function would have been of cot⁡α\cot \alpha instead of tan⁡α\tan \alpha . For any case of root value, we always need to use the modulus value if otherwise mentioned.