## Elementary Trig Question

Say I have an acute angle A in standard position in the first quadrant with a rotating arm of length r , terminating at coordinate P(a,b) . Now say I rotate it 90 degrees further from that position to the second quadrant , this ends up at coordinate Q(-a,b) . I draw perpendiculars from those two positions to the x axis .

for the first triangle , sin A = b/r and cos A = a/r

as for the second triangle , why is sin (90+A) = a/r = cos A
and why is cos(90+A) = -b/r = -sin A

Im confused ... how can you compute the trig ratio for an angle greater than 90 degress ? Why not take the ratio for the acute angle (180 - (90+a))

Pls Help .
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 Quote by pokemonDoom Say I have an acute angle A in standard position in the first quadrant with a rotating arm of length r , terminating at coordinate P(a,b) . Now say I rotate it 90 degrees further from that position to the second quadrant , this ends up at coordinate Q(-a,b) .
Hi ! Welcome to PF!

No … (-a,b) is the reflection of (a,b) in the y-axis.

You want Q(b,-a).

Then everythihng will make sense!
 hmm ... but that doesn't solve my problem ... I need to know why the so called "reference angle" or the "corresponding acute angle" actually works for angles greater than 90 degrees , when taking sines and cosines or tangents . Is there a geometric argument for it ???

Mentor

## Elementary Trig Question

Im not entirely sure what your question is; but if it boils down to "why is cos(x)=sin(x+90) ?", then you can answer this by looking at the graphs of the sine and cosine functions.
 Heres my problem ... For the triangle in the second quadrant , why does my book state sin(90+theta) = a/r and cos(90+theta) = -b/r when clearly (90+theta) isn't even one of the angles in the triangle ... ??? This was what I meant when I said why not calculate (180-90-theta) .... I dont get this part .... !!! Ive looked up 2/3 trig books by native authors (I live in Nepal) and they don't seem to explain why this is so ... and I have yet to find a good trig book on the internet .... Thanks a mil .

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Hi pokemonDoom!
 Quote by pokemonDoom For the triangle in the second quadrant …
No … the triangle is not in the second quadrant … its base is still in the first quadrant, and its tip is in the second quadrant.

For an easy example, consider the two triangles OAB and OAC, where O = (0,0), A = (1,0), B = (1,1), and C = (-1,1). So OA = 1, and OB = OC = √2. Put D = (-1,0).

Then angle AOB = 45º, and angle AOC = 135º.

And cosAOB = OA/OB = 1/√2, cosAOC = AD/OC = -1/√2.

But sinAOB = AB/OB = 1/√2, sinAOC = DC/OC = 1/√2.
 Hello Tim , Im confused ... how can the cosine of

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 Quote by pokemonDoom Im confused ... how can the cosine of
Hi pokemonDoom!

Yes … sorry … it should be cosAOC = OD/OC = -1/√2.
 and in triangle-DOC (which lies squarely on the second quadrant of my graph) how is it that the sine of
Because sin = opposite/hypotenuse.

The side opposite angle AOB is AB, and the side opposite angle AOC is DC.

And AB = DC, so sinAOB = sinAOC.

(and of course, the diagram works for any angle, not just 45º)
 ... I get it !!! But isnt the angle opposite side DC

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