Sarah5678
Sarah5678 Sarah5678
  • 21-07-2019
  • Mathematics
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Need help with this question please!!

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Tucon
Tucon Tucon
  • 21-07-2019

 

[tex]\displaystyle\\\tan x=1\\ 0^o<x<90^o\\\implies~x=\arctan(1)=45^o\\\\\sin\left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos x}{2}}\\\\\sin\left(\frac{45^o}{2}\right)=\sqrt{\frac{1-\cos 45^o}{2}}\\\\\sin\left(\frac{45^o}{2}\right)=\sqrt{\frac{1-\dfrac{\sqrt{2}}{2}}{2}}\\\\\sin\left(\frac{45^o}{2}\right)=\sqrt{\frac{\dfrac{2-\sqrt{2}}{2}}{2}}\\\\\sin\left(\frac{45^o}{2}\right)=\sqrt{\dfrac{2-\sqrt{2}}{4}}\\\\\sin\left(\frac{45^o}{2}\right)=\frac{1}{2} \sqrt{2-\sqrt{2}}[/tex]

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