# 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 .

tiny-tim
Homework Helper
Welcome to PF!

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 ???

cristo
Staff Emeritus
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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tiny-tim
Homework Helper
Hi 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 <AOC be AD/OC ... if I know better , its probably a typo .

and in triangle-DOC (which lies squarely on the second quadrant of my graph) how is it that the sine of <AOC is equivalent to the sine of <DOC ... This is the question in the first place .

Many thanks .

tiny-tim
Homework Helper
Im confused ...

how can the cosine of <AOC be AD/OC ... if I know better , its probably a typo .

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 <AOC is equivalent to the sine of <DOC ... This is the question in the first place .

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 <DOC ... ? Im not trying to sound dumb but what youre trying to say is exactly what the question is .

Why should sineAOC =CD/OC , when its perfectly legible that sineDOC = CD/OC .

In essence what youre trying to say is for any angle greater than 90 degrees , the ratio of that angle is equal to the ratio of the acute angle formed with the nearest x axis ... ?? but why? is there a geometric proof of some kind ?? instead of just guessing that the ratios are equal ?