anonymous
  • anonymous
Help with identities: 1. ( sin2x/sin x)-(cos2x/cos x)=sec 2. (cos x+cos y)^2 +(sin x-sin y)^2 =2+2cos(x+y)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
For question 1. it says the steps should be: apply the double-angle formulas for sine and cosine, then simplify the expression. I got this far. \[(2\sin x \cos x)/\sin x)-(2 \cos^2x-1)/\cos x)\] \[2 \cos x=(2 \cos^2 x-1)/\cos x\]
UnkleRhaukus
  • UnkleRhaukus
i don't think i am reading question 1. properly, can you type out the equation again (or draw it) please?
anonymous
  • anonymous
\[\frac{ \sin (2x) }{ sinx }-\frac{ \cos(2x) }{ \cos x }=\sec x\]

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UnkleRhaukus
  • UnkleRhaukus
ah, i can read it properly now, thakyou
UnkleRhaukus
  • UnkleRhaukus
your working so far it good
UnkleRhaukus
  • UnkleRhaukus
you have \[LHS = 2 \cos x-\frac{2 \cos^2 x-1}{\cos x}\]
UnkleRhaukus
  • UnkleRhaukus
now break up the fraction like this \[\frac{a+b}{c}=\frac ac+\frac bc\]
anonymous
  • anonymous
\[\frac{ 2\cos^2 }{ \cos }=2 \cos\] \[\frac{ -1 }{ \cos}=-\sec\] Is this right?
UnkleRhaukus
  • UnkleRhaukus
right!, (becareful with the extra -sign, outside of the big fraction)
UnkleRhaukus
  • UnkleRhaukus
so you've got \[LHS = 2\cos x-\frac{2 \cos^2 x-1}{\cos x}\\ \qquad=2\cos x-\left(\frac{2 \cos^2 x}{\cos x}+\frac{-1}{\cos x}\right)\\ \qquad=2\cos x-\left(2\cos x-\sec x\right)\\\qquad=\]
anonymous
  • anonymous
2 cos x-2cos x=0 -sec x=sec x ?
UnkleRhaukus
  • UnkleRhaukus
be CAREFUL with the minus signs. \[\qquad=2\cos x-\left(2\cos x-\sec x\right)\\\qquad=2\cos x-2\cos x+\sec x\\\qquad=\]
anonymous
  • anonymous
Ahhhh okay, I got it now!

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