Trigonometry

Geometry Level 3

1 + sin ⁡ α − 1 − sin ⁡ α \large \sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha}

If α ∈ [ π 2 , π ] \alpha \in \left[ \dfrac{\pi}{2},\pi\right] , then find the value of the above expression.

2 cos ⁡ α 2 2 \cos \frac{\alpha}{2} 2 sin ⁡ α 2 2 \sin \frac{\alpha}{2} None of these choices

Akshat Sharda by Akshat Sharda , 21, India

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1 solution

Chew-Seong Cheong
Jun 14, 2016

Solution suggested by @Vighnesh Shenoy

1 + sin ⁡ α − 1 − sin ⁡ α = 1 + 2 sin ⁡ α 2 cos ⁡ α 2 − 1 − 2 sin ⁡ α 2 cos ⁡ α 2 = sin ⁡ 2 α 2 + cos ⁡ 2 α 2 + 2 sin ⁡ α 2 cos ⁡ α 2 − sin ⁡ 2 α 2 + cos ⁡ 2 α 2 − 2 sin ⁡ α 2 cos ⁡ α 2 = ( sin ⁡ α 2 + cos ⁡ α 2 ) 2 − ( sin ⁡ α 2 − cos ⁡ α 2 ) 2 = sin ⁡ α 2 + cos ⁡ α 2 − sin ⁡ α 2 + cos ⁡ α 2 = 2 cos ⁡ α 2 \begin{aligned} \sqrt{1+\sin \alpha} - \sqrt{1- \sin \alpha} & = \sqrt{1+2\sin \frac \alpha 2 \cos \frac \alpha 2} - \sqrt{1 - 2 \sin \frac \alpha 2 \cos \frac \alpha 2} \\ & = \sqrt{\sin^2 \frac \alpha 2 + \cos^2 \frac \alpha 2+2 \sin \frac \alpha 2 \cos \frac \alpha 2} - \sqrt{\sin^2 \frac \alpha 2 + \cos^2 \frac \alpha 2 -2 \sin \frac \alpha 2 \cos \frac \alpha 2} \\ & = \sqrt{\left(\sin \frac \alpha 2 + \cos \frac \alpha 2\right)^2} - \sqrt{\left(\sin \frac \alpha 2 - \cos \frac \alpha 2\right)^2} \\ & = \sin \frac \alpha 2 + \cos \frac \alpha 2 - \sin \frac \alpha 2 + \cos \frac \alpha 2 \\ & = \boxed{2 \cos \dfrac \alpha 2} \end{aligned}

My solution

1 + sin ⁡ α − 1 − sin ⁡ α = 1 + 2 tan ⁡ α 2 1 + tan ⁡ 2 α 2 − 1 − 2 tan ⁡ α 2 1 + tan ⁡ 2 α 2 = tan ⁡ 2 α 2 + 2 tan ⁡ α 2 + 1 1 + tan ⁡ 2 α 2 − tan ⁡ 2 α 2 − 2 tan ⁡ α 2 + 1 1 + tan ⁡ 2 α 2 = tan ⁡ α 2 + 1 1 + tan ⁡ 2 α 2 − tan ⁡ α 2 − 1 1 + tan ⁡ 2 α 2 = 2 1 + tan ⁡ 2 α 2 = 2 sec ⁡ α 2 = 2 cos ⁡ α 2 \begin{aligned} \sqrt{1+\sin \alpha} - \sqrt{1- \sin \alpha} & = \sqrt{1+\frac{2 \tan \frac \alpha 2}{1 + \tan^2 \frac \alpha 2}} - \sqrt{1-\frac{2 \tan \frac \alpha 2}{1 + \tan^2 \frac \alpha 2}} \\ & = \sqrt{\frac{\tan ^2 \frac \alpha 2 + 2 \tan \frac \alpha 2 + 1}{1 + \tan^2 \frac \alpha 2}} - \sqrt{\frac{\tan ^2 \frac \alpha 2 - 2 \tan \frac \alpha 2 + 1}{1 + \tan^2 \frac \alpha 2}} \\ & = \frac{\tan \frac \alpha 2 + 1}{\sqrt{1 + \tan^2 \frac \alpha 2}} - \frac{\tan \frac \alpha 2 - 1}{\sqrt{1 + \tan^2 \frac \alpha 2}} \\ & = \frac{2}{\sqrt{1 + \tan^2 \frac \alpha 2}} \\ & = \frac{2}{ \sec \frac \alpha 2} \\ & = \boxed{2 \cos \dfrac \alpha 2} \end{aligned}

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Instead if converting to tan one could directly use,
1 ± sin ⁡ ( α ) = ∣ sin ⁡ ( α 2 ) ± cos ⁡ ( α 2 ) ∣ \sqrt{1 \pm \sin(\alpha)} = \left|\sin\left(\dfrac{\alpha}{2}\right) \pm\cos\left(\dfrac{\alpha}{2}\right)\right|

A Former Brilliant Member - 5 years ago

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Good solution. I will present it for you.

Chew-Seong Cheong - 5 years ago

Did the same way +1!

Rishabh Tiwari - 5 years ago

Nice solution sir,+1!

Rishabh Tiwari - 5 years ago

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