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anonymous
 one year ago
what's the trigonometric subtraction formula for sine ?
anonymous
 one year ago
what's the trigonometric subtraction formula for sine ?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's not (sin(a)(cos(b))  (cos(a)sin(b)) ?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0Do you mean this one?\[\sin(\theta\phi)=\sin\theta\cos\phi\cos\theta\sin\phi\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah , my notes give it with a and b variables . I just wanted to make sure it was that one though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how would you use that formula to verify \[\sin(\frac{ \pi }{ 2 }x)=cosx\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\sin(\tfrac\pi2x)\\ =\sin\tfrac\pi2\cos x\cos\tfrac\pi2\sin x\\[3ex] \qquad\qquad(\sin\tfrac\pi2=\sin90°=1,\qquad \cos\tfrac\pi2=\cos 90°=0\ )\\[3ex] =\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oooooh okay , it's somewhat making sense

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\sin(\tfrac\pi2x)\\ =\sin\tfrac\pi2\cos x\cos\tfrac\pi2\sin x\\[3ex] \qquad\qquad(\sin\tfrac\pi2=\sin90°=1,\qquad \cos\tfrac\pi2=\cos 90°=0\ )\\[3ex] =(1)\times\cos x(0)\times\sin x\\ =\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it'd be 1 , meaning that it is an identity ?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0the does the last line (of the above) simplify to become?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[=\vdots\\=(1)\times\cos x(0)\times\sin x\\ =\] What is \(1\times \cos x\)?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0good, and what is \(0\times \sin x\)

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0yes, so what is \[=\quad\vdots\\[3ex]=(1)\times\cos x(0)\times\sin x\\ =\cos x0\\ =\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0yeah. So. We have shown that: \[\sin(\tfrac\pi2x) = \cos x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I'm just tired and that's why I'm not comprehending it completely lol

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\sin(\tfrac\pi2x) =\sin\tfrac\pi2\cos x\cos\tfrac\pi2\sin x\\%[3ex] %\qquad\qquad(\sin\tfrac\pi2=\sin90°=1,\qquad \cos\tfrac\pi2=\cos 90°=0\ )\\[3ex] \qquad\qquad\,\, =(1)\times\cos x(0)\times\sin x\\ \qquad\qquad\,\, =\cos x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and the 1 and 0 are the points correct ?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0the 1 and 0, are the values of the sine and cosine of the the first term in the sine of the difference: (pi/2x), 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
 one year ago
Best ResponseYou've already chosen the best response.0yeah , that's what I meant by points . On the unit circle lol

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0i was hoping so .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just looked at the one that have right now to make sure that's what it's from

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0i suppose they could also be points on the plots of the function sine and cosine: dw:1440307933477:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My tutor said they're points so that's what I've stuck to calling them lol

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440308112848:dw

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440308149960:dw actually they are more like the lengths on the unit circle, and points on the functions

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0do 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
 one year ago
Best ResponseYou've already chosen the best response.0the point on the unit circle is at theta = π/2 ie (0,1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats what I'm looking at

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0where each point is \((x,y)=(\cos\theta,\sin\theta)\)
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