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Trigonometric exercise asked in mathematics teacher group LeveL 3


Exercise:

Show that, if cos(BC)=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(πA)=sinA

so cos(BC)sin(B+C)=2sinBsinCsinA

Apply the trigonometric identities from product to sum we get:

cos(BC)sin(B+C)=12(sin(BC+B+C)sin(BCBC))

                              =12(sin2Bsin(2C))=12(sin2B+sin2C)

Hence 12(sin2B+sin2C)=2sinBsinCsinAsin2B+sin2C2sinBsinC=2sinA

So 2sinBcosB2sinBsinC+2sinCcosC2sinBsinC=2SinAcosBsinC+cosCsinB=2sinA

cosBsinAsinC+cosCsinAsinB=2

Take AB=a,BC=b&AC=c then by Using cosine law :

c2=a2+b22abcosBcosB=a2+b2c22ab

a2=c2+b22cbcosCcosC=b2+c2a22bc

Also sinAsinC=hahb=ba and sinAsinB=hchb=bc

So a2+b2c22ab×ba+b2+c2a22bc×bc=2 a2+b2c22a2+b2+c2a22c2=2

c2(a2+b2c2)+a2(b2+c2a2)2a2c2=2 c2a2+c2b2c4+a2b2+a2c2a4=4a2c2

c2a2+c2b2+a2b2+a2c2=a4+c4+2a2c2+2a2c2=(a2+c2)2+2a2c2

b2(a2+c2)+2a2c2=(a2+c2)2+2a2c2b2(a2+c2)=(a2+c2)2

b2=a2+c2 hence the triangle is right at A and satisfied the Pythagoras theorem

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