Proof of sin(x+pi/2)=cosx and -cos(x+pi/2)=sinx

[Ozma]

Hi, I would like to ask to you a confirm if my proof is correct (since my textbook does this in a different way); I will upload a picture in the post.
To prove that [MATH]\sin \left(x+\frac{\pi}{2} \right) = \cos x[/MATH] and [MATH]-\cos \left(x+\frac{\pi}{2} \right)=\sin x[/MATH] I've used the following reasoning: the triangle [MATH]OAB[/MATH] and [MATH]OCD[/MATH] are congruent since [MATH]OB \cong OD[/MATH] (because they are on the goniometric circumference who has radius [MATH]1[/MATH], so they both equal [MATH]1[/MATH]) and they have two angles (the angle [MATH]x[/MATH] and the right angle) who are congruent (because the angle in the second quadrant is [MATH]\frac{\pi}{2}+x-\frac{\pi}{2}=x[/MATH]). So by one of the congruence theorems of the triangles (the one who says: "If two triangles have one congruent side, the angle adiacent to that side congruent and the opposite angle to the latter congruent then they are congruent") they are congruent and so we have that [MATH]OA \cong OC[/MATH], that is [MATH]\cos x = \sin \left(x+\frac{\pi}{2} \right)[/MATH], and we have that [MATH]CD \cong AB[/MATH], that is [MATH]-\cos \left(x+\frac{\pi}{2} \right)=\sin x[/MATH].
Is this correct?
Another question: in my book there is written that those sides and angles must be ordered, does this mean that if, for example in the triangle [MATH]ABC[/MATH], [MATH]x[/MATH] was the angle [MATH]O\hat{B}C[/MATH] the congruence wasn't true?
Thanks and sorry if my English is bad (it's not my first language).

 

[pka]

\(\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)~\&~\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)\)

\(\sin\left(\frac{\pi}{2}\right)=1~\&~\cos\left(\frac{\pi}{2}\right)=0\)

 

[Ozma]

@pka: Thanks for your answer, actually I still haven't seen those identities. I asked for a confirmation of my work, can you see if it is correct or no, please?

 

[JeffM]

Your proof looks sound to me but maybe a bit too complex. The two triangles are similar because, by construction, two of the angles are the same in each triangle. And one side of each triangle (the hypotenuse that equals the circle's radius) is necessarily equal to a side of the other. Therefore the triangles are congruent.

However, once you are dealing angles that exceed pi/2, it's probably easier to deal with the trigonometric functions without reference to triangles.

 

[Ozma]

@JeffM: Thanks for your confirmation. Actually I am learning now the trigonometric functions so the proof must be done without them for having a logical order.