Mutual transformation between sin and COS Transformation of COS a into sin acos A＝bcos B a/cosA=b/cosB=c/cosC According to the given conditions, the shape of △ ABC is judged Write the process down And explain

Mutual transformation between sin and COS Transformation of COS a into sin acos A＝bcos B a/cosA=b/cosB=c/cosC According to the given conditions, the shape of △ ABC is judged Write the process down And explain

How can I feel that I can't use acosa = bcosb = = known from the sine theorem: A / Sina = B / SINB = C / sinc from the question: A / cosa = B / CoSb = C / COSC  A / b = Sina / SINB = cosa / CoSb diagonal multiplication: sinacosb = cosasinb, that is sin (a-b) = 0 ∧ a = ∧ B, similarly: ∧ B = ∧ C, so triangle AB

How does sin cos transform into each other?
For example, how is sin5 / 2 π equal to Cos4 π

sin(5π/2)=sin(2π+π/2)=sinπ/2=1=cos0=cos(2π）=cos（2π+2π）=cos4π

How to transform sin and COS

sin(x)=cos((π/2)-x)