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

1. anonymous

It's not (sin(a)(cos(b)) - (cos(a)sin(b)) ?

2. UnkleRhaukus

Do you mean this one?$\sin(\theta-\phi)=\sin\theta\cos\phi-\cos\theta\sin\phi$

3. anonymous

yeah , my notes give it with a and b variables . I just wanted to make sure it was that one though

4. anonymous

thank you

5. anonymous

how would you use that formula to verify $\sin(\frac{ \pi }{ 2 }-x)=cosx$

6. UnkleRhaukus

$\sin(\tfrac\pi2-x)\\ =\sin\tfrac\pi2\cos x-\cos\tfrac\pi2\sin x\\[3ex] \qquad\qquad(\sin\tfrac\pi2=\sin90°=1,\qquad \cos\tfrac\pi2=\cos 90°=0\ )\\[3ex] =$

7. anonymous

oooooh okay , it's somewhat making sense

8. UnkleRhaukus

$\sin(\tfrac\pi2-x)\\ =\sin\tfrac\pi2\cos x-\cos\tfrac\pi2\sin x\\[3ex] \qquad\qquad(\sin\tfrac\pi2=\sin90°=1,\qquad \cos\tfrac\pi2=\cos 90°=0\ )\\[3ex] =(1)\times\cos x-(0)\times\sin x\\ =$

9. anonymous

so it'd be 1 , meaning that it is an identity ?

10. UnkleRhaukus

what would be 1?

11. anonymous

im so confused :/

12. UnkleRhaukus

the does the last line (of the above) simplify to become?

13. anonymous

idk

14. UnkleRhaukus

$=\vdots\\=(1)\times\cos x-(0)\times\sin x\\ =$ What is $$1\times \cos x$$?

15. anonymous

cosx

16. UnkleRhaukus

good, and what is $$0\times \sin x$$

17. anonymous

0

18. UnkleRhaukus

yes, so what is $=\quad\vdots\\[3ex]=(1)\times\cos x-(0)\times\sin x\\ =\cos x-0\\ =$

19. anonymous

cosx ?

20. UnkleRhaukus

yeah. So. We have shown that: $\sin(\tfrac\pi2-x) = \cos x$

21. UnkleRhaukus

All clear?

22. anonymous

I think I'm just tired and that's why I'm not comprehending it completely lol

23. UnkleRhaukus

$\sin(\tfrac\pi2-x) =\sin\tfrac\pi2\cos x-\cos\tfrac\pi2\sin x\\%[3ex] %\qquad\qquad(\sin\tfrac\pi2=\sin90°=1,\qquad \cos\tfrac\pi2=\cos 90°=0\ )\\[3ex] \qquad\qquad\,\, =(1)\times\cos x-(0)\times\sin x\\ \qquad\qquad\,\, =\cos x$

24. anonymous

and the 1 and 0 are the points correct ?

25. 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]

26. anonymous

yeah , that's what I meant by points . On the unit circle lol

27. UnkleRhaukus

i was hoping so .

28. anonymous

I just looked at the one that have right now to make sure that's what it's from

29. UnkleRhaukus

i suppose they could also be points on the plots of the function sine and cosine: |dw:1440307933477:dw|

30. anonymous

My tutor said they're points so that's what I've stuck to calling them lol

31. UnkleRhaukus

|dw:1440308112848:dw|

32. UnkleRhaukus

|dw:1440308149960:dw| actually they are more like the lengths on the unit circle, and points on the functions

33. 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

34. UnkleRhaukus

the point on the unit circle is at theta = π/2 ie (0,1)

35. anonymous

36. anonymous

thats what I'm looking at

37. UnkleRhaukus

where each point is $$(x,y)=(\cos\theta,\sin\theta)$$

38. anonymous

yeah