Exercise:
Show that, if cos(B−C)=2sinBsinCsin2A then △ABC is right
Solution: we know that the sum of angle in any triangle is 180
hence A+B+C=π so sin(B+C)=sin(π−A)=sinA
so cos(B−C)sin(B+C)=2sinBsinCsinA
Apply the trigonometric identities from product to sum we get:
cos(B−C)sin(B+C)=12(sin(B−C+B+C)−sin(B−C−B−C))
=12(sin2B−sin(−2C))=12(sin2B+sin2C)
Hence 12(sin2B+sin2C)=2sinBsinCsinA⇔sin2B+sin2C2sinBsinC=2sinA
So 2sinBcosB2sinBsinC+2sinCcosC2sinBsinC=2SinA⇔cosBsinC+cosCsinB=2sinA
⇔cosBsinAsinC+cosCsinAsinB=2

c2=a2+b2−2abcosB⇔cosB=a2+b2−c22ab
a2=c2+b2−2cbcosC⇔cosC=b2+c2−a22bc
Also sinAsinC=hahb=ba and sinAsinB=h∗ch∗b=bc
So a2+b2−c22ab×ba+b2+c2−a22bc×bc=2 ⇔a2+b2−c22a2+b2+c2−a22c2=2
⇔c2(a2+b2−c2)+a2(b2+c2−a2)2a2c2=2 ⇔c2a2+c2b2−c4+a2b2+a2c2−a4=4a2c2
⇔c2a2+c2b2+a2b2+a2c2=a4+c4+2a2c2+2a2c2=(a2+c2)2+2a2c2
⇔b2(a2+c2)+2a2c2=(a2+c2)2+2a2c2⇔b2(a2+c2)=(a2+c2)2
⇔b2=a2+c2 hence the triangle is right at A and satisfied the Pythagoras theorem
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