anonymous
  • anonymous
what's the trigonometric subtraction formula for sine ?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
It's not (sin(a)(cos(b)) - (cos(a)sin(b)) ?
UnkleRhaukus
  • UnkleRhaukus
Do you mean this one?\[\sin(\theta-\phi)=\sin\theta\cos\phi-\cos\theta\sin\phi\]
anonymous
  • anonymous
yeah , my notes give it with a and b variables . I just wanted to make sure it was that one though

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anonymous
  • anonymous
thank you
anonymous
  • anonymous
how would you use that formula to verify \[\sin(\frac{ \pi }{ 2 }-x)=cosx\]
UnkleRhaukus
  • UnkleRhaukus
\[\sin(\tfrac\pi2-x)\\ =\sin\tfrac\pi2\cos x-\cos\tfrac\pi2\sin x\\[3ex] \qquad\qquad(\sin\tfrac\pi2=\sin90°=1,\qquad \cos\tfrac\pi2=\cos 90°=0\ )\\[3ex] =\]
anonymous
  • anonymous
oooooh okay , it's somewhat making sense
UnkleRhaukus
  • UnkleRhaukus
\[\sin(\tfrac\pi2-x)\\ =\sin\tfrac\pi2\cos x-\cos\tfrac\pi2\sin x\\[3ex] \qquad\qquad(\sin\tfrac\pi2=\sin90°=1,\qquad \cos\tfrac\pi2=\cos 90°=0\ )\\[3ex] =(1)\times\cos x-(0)\times\sin x\\ =\]
anonymous
  • anonymous
so it'd be 1 , meaning that it is an identity ?
UnkleRhaukus
  • UnkleRhaukus
what would be 1?
anonymous
  • anonymous
im so confused :/
UnkleRhaukus
  • UnkleRhaukus
the does the last line (of the above) simplify to become?
anonymous
  • anonymous
idk
UnkleRhaukus
  • UnkleRhaukus
\[=\vdots\\=(1)\times\cos x-(0)\times\sin x\\ =\] What is \(1\times \cos x\)?
anonymous
  • anonymous
cosx
UnkleRhaukus
  • UnkleRhaukus
good, and what is \(0\times \sin x\)
anonymous
  • anonymous
0
UnkleRhaukus
  • UnkleRhaukus
yes, so what is \[=\quad\vdots\\[3ex]=(1)\times\cos x-(0)\times\sin x\\ =\cos x-0\\ =\]
anonymous
  • anonymous
cosx ?
UnkleRhaukus
  • UnkleRhaukus
yeah. So. We have shown that: \[\sin(\tfrac\pi2-x) = \cos x\]
UnkleRhaukus
  • UnkleRhaukus
All clear?
anonymous
  • anonymous
I think I'm just tired and that's why I'm not comprehending it completely lol
UnkleRhaukus
  • UnkleRhaukus
\[\sin(\tfrac\pi2-x) =\sin\tfrac\pi2\cos x-\cos\tfrac\pi2\sin x\\%[3ex] %\qquad\qquad(\sin\tfrac\pi2=\sin90°=1,\qquad \cos\tfrac\pi2=\cos 90°=0\ )\\[3ex] \qquad\qquad\,\, =(1)\times\cos x-(0)\times\sin x\\ \qquad\qquad\,\, =\cos x\]
anonymous
  • anonymous
and the 1 and 0 are the points correct ?
UnkleRhaukus
  • UnkleRhaukus
the 1 and 0, are the values of the sine and cosine of the the first term in the sine of the difference: (pi/2-x), since the first term in this difference is a constant (pi/2), the sines and cosines of this term are necessarily constants. [and can be determined from points on a unit circle]
anonymous
  • anonymous
yeah , that's what I meant by points . On the unit circle lol
UnkleRhaukus
  • UnkleRhaukus
i was hoping so .
anonymous
  • anonymous
I just looked at the one that have right now to make sure that's what it's from
UnkleRhaukus
  • UnkleRhaukus
i suppose they could also be points on the plots of the function sine and cosine: |dw:1440307933477:dw|
anonymous
  • anonymous
My tutor said they're points so that's what I've stuck to calling them lol
UnkleRhaukus
  • UnkleRhaukus
|dw:1440308112848:dw|
UnkleRhaukus
  • UnkleRhaukus
|dw:1440308149960:dw| actually they are more like the lengths on the unit circle, and points on the functions
UnkleRhaukus
  • UnkleRhaukus
do you see what i mean?, on the unit circle, at π/2 the length of the cosine side of the triangle goes to 0, as the length of the sine side , goes to equal 1. whereas on the function plots, the points at π/2 on cosine and sine 0, and 1 respectively
UnkleRhaukus
  • UnkleRhaukus
the point on the unit circle is at theta = π/2 ie (0,1)
anonymous
  • anonymous
anonymous
  • anonymous
thats what I'm looking at
UnkleRhaukus
  • UnkleRhaukus
where each point is \((x,y)=(\cos\theta,\sin\theta)\)
anonymous
  • anonymous
yeah

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