what's the trigonometric subtraction formula for sine ?

- anonymous

what's the trigonometric subtraction formula for sine ?

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

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

- UnkleRhaukus

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

thank you

- anonymous

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

- 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

oooooh okay , it's somewhat making sense

- 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

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

- UnkleRhaukus

what would be 1?

- anonymous

im so confused :/

- UnkleRhaukus

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

- anonymous

idk

- UnkleRhaukus

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

- anonymous

cosx

- UnkleRhaukus

good, and
what is \(0\times \sin x\)

- anonymous

0

- UnkleRhaukus

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

- anonymous

cosx ?

- UnkleRhaukus

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

- UnkleRhaukus

All clear?

- anonymous

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

- 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

and the 1 and 0 are the points correct ?

- 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

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

- UnkleRhaukus

i was hoping so .

- anonymous

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

- UnkleRhaukus

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

- anonymous

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

- UnkleRhaukus

|dw:1440308112848:dw|

- UnkleRhaukus

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

- 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

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

- anonymous

##### 1 Attachment

- anonymous

thats what I'm looking at

- UnkleRhaukus

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

- anonymous

yeah

Looking for something else?

Not the answer you are looking for? Search for more explanations.