anonymous
  • anonymous
can someone explain this to me???
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
simplify
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kropot72
  • kropot72
The first step in simplifying is to multiply out the numerator: \[\large (1-\cos \theta)(1+\cos \theta)=?\] Can you do that and post the result?
anonymous
  • anonymous
\[1+\cos \theta -\cos \theta + \cos^2 ?\]

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DecentNabeel
  • DecentNabeel
(1−cosθ)(1+cosθ)=1-cos^2x \[\sqrt{\frac{1-\cos^2x}{\cos^2x}}\] \[\mathrm{Use\:the\:following\:identity}:\quad \:1-\cos ^2\left(x\right)=\sin ^2\left(x\right)\] \[=\sqrt{\frac{\sin ^2\left(x\right)}{\cos ^2\left(x\right)}}\]
anonymous
  • anonymous
ok. so im left with sin(x) over cos(x)?
DecentNabeel
  • DecentNabeel
yes
kropot72
  • kropot72
Nearly correct! However the sign in front of the cos^2 (theta) term should be negative. Therefore we get: \[\large \sqrt{\frac{(1-\cos^{2} \theta)}{\cos^{2} \theta}}\] Does that make sense?
anonymous
  • anonymous
yes. but cant i break that down further?
DecentNabeel
  • DecentNabeel
are you understand @kapukawai
kropot72
  • kropot72
You need to use the relationship: \[\large \sin^{2}\theta+\cos^{2}\theta=1\] From which you get: \[\large 1-\cos^{2}\theta=\sin^{2}\theta\] and finally by substitution we can write: \[\large \sqrt{\frac{\sin^{2}\theta}{\cos^{2}\theta}}\]
kropot72
  • kropot72
Next you can use the relationship \[\large \frac{\sin\theta}{\cos\theta}=\tan\theta\]
anonymous
  • anonymous
so tan theta is the simplified form of ...... ok that makes sence. i just have to memorise the relation ships?
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kropot72
  • kropot72
Yes, those relationships are some of the basic ones in trigonometry.
anonymous
  • anonymous
thank you!
kropot72
  • kropot72
You're welcome :)

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