for any triangle abc,using sine formula determine the cosine formula

Question

how can we derive cosine formula by using sine formula?

Answer 1

In triangle ABC draw the perpendicular BX from B to AC. The length of this is csinA or asinC. So a/sinA=c/sinC (part of sine rule). Draw a perpendicular from CY to AB, length=bsinA=asinB, so a/sinA=bsinB (completing sine rule): c/sinC=a/sinA=b/sinB.

Let AX=x, then XC=b-x, because AC=b, and x=ccosA. In right-angled triangle BXC, BX^2=a^2-(b-x)^2 (Pythagoras); but in right-angled triangle BXA, BX^2=c^2-x^2, so a^2-(b-x)^2=c^2-x^2; a^2-b^2+2bx-x^2=c^2-x^2; a^2=b^2+c^2-2bx=b^2+c^2-2bccosA, substituting for x. This is the cosine rule.

In each case, sine or cosine rule, the perpendicular BX has been eliminated in the course of proving the rule.