Derive the cosine formula a^2=b^2+c^2-2bc cosA and the sine formula (sin A/a) = (sinB/b) = (sinC/c) using dot product and cross product, respectively.

Question

Derive the cosine formula

[latex]a^{2}=b^{2}+c^{2}-2bc\cos A[\latex]

and the sine formula

[latex]\frac{\sin A}{2}=\frac{\sin B}{b}=\frac{\sin C}{c}[\latex]

using dot product and cross product, respectively.

Accepted Answer

Consider a triangle as shown in Figure. From the figure, we notice that

a + b + c = 0

that is,

b + c = -a

Hence,

[latex]a^{2}= a\cdot a=\left(b + c ight) \cdot \left(b+c ight)[/latex]

[latex]=b\cdot b+c\cdot c+2b\cdot c[/latex]

[latex]a^{2}=b^{2}+c^{2}-2bc\cos A[/latex]

where (π - A) is the angle between b and c. The area of a triangle is half of the product of its height and base. Hence,

[latex]\left|\frac{1}{2}a imes b ight| =\left|\frac{1}{2} b imes c ight| =\left|\frac{1}{2} c imes a ight|[/latex]

ab sinC = bc sinA = ca sinB

Dividing through by abc gives

[latex]\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}[/latex]

Question 1.6: Derive the cosine formula a^2=b^2+c^2-2bc cosA and the sine ...