Question

what's the trigonometric subtraction formula for sine ?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

thank you

Answer 5

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

Answer 6

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

Answer 7

oooooh okay , it's somewhat making sense

Answer 8

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

Answer 9

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

Answer 10

what would be 1?

Answer 11

im so confused :/

Answer 12

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

Answer 13

idk

Answer 14

\[=\vdots\\=(1)\times\cos x-(0)\times\sin x\\
=\]


What is \(1\times \cos x\)?

Answer 15

cosx

Answer 16

good, and

what is \(0\times \sin x\)

Answer 17

0

Answer 18

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

Answer 19

cosx ?

Answer 20

yeah.


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

Answer 21

All clear?

Answer 22

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

Answer 23

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

Answer 24

and the 1 and 0 are the points correct ?

Answer 25

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]

Answer 26

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

Answer 27

i was hoping so .

Answer 28

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

Answer 29

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