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Mathematics 13 Online
OpenStudy (anonymous):

Why does sin(pi-x)=sin(x)? Where x is angle measure in radians.

OpenStudy (anonymous):

\[\sin (\pi-\theta)=\sin \theta \]

OpenStudy (tyteen4a03):

If I remember correctly \(sin(\pi - \theta) = -sin\theta\). To verify this identity we can use our beloved Addition Identity: \(sin(A - B) = sinAcosB - cosAsinB\). Substitute 360 degrees and theta in, we have \(sin(360)cos\theta - cos(360)sin(\theta)\) \(= 0 cos\theta - (1) sin\theta\) \(= -sin\theta\)

OpenStudy (anonymous):

radians not degrees right?

OpenStudy (anonymous):

if you convert, which you should not, you would use 180 degrees instead of 360

OpenStudy (tyteen4a03):

@satellite73 If didn't memorize the radians to degrees chart wrong, \(\pi / 2 = 180\), which would make \(\pi = 360\).

OpenStudy (anonymous):

a complete rotation is 360 degrees, which in radians is \(2\pi\) a straight angle is \(\pi\) or in degrees 180 a right angle is \(\frac{\pi}{2}\) or in degrees 90

OpenStudy (tyteen4a03):

@satellite73 I see (never used radians in my maths curriculum so I like conversion better) So back to the drawing board: \(sin(A - B) = sinAcosB - cosAsinB\) Substitute 180 - theta into this equation, we have \(sin180cos\theta - cos180sin\theta\) \(= 0 cos\theta - (-1) sin\theta\) \(= sin(\theta)\)

OpenStudy (anonymous):

the answer is sin x according to my textbook and teacher

OpenStudy (anonymous):

Can you describe why in terms of a unit circle. I don't really understand your proof

OpenStudy (anonymous):

sin(A−B)=sinAcosB−cosAsinB that part

OpenStudy (tyteen4a03):

|dw:1358400364879:dw| I don't have my textbook on hand so I'll try my best at explaining why sin(180 - x) = sin(x). For the addition formula, the complete proof is pretty hard to understand (but if you want to try your luck, it's here: http://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities ), so your best bet is to memorize it for now. (Proof below - need access to my graph)

OpenStudy (tyteen4a03):

|dw:1358400942158:dw| |dw:1358401009614:dw| As you can see, the formation of theta in quadrant I is simply mirrored over to form 180 - theta over at quadrant II. (Somebody probably can explain better than me)

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