To complete the proof that Bobby was working on, let's go through the steps in detail.
### Starting Equation
Bobby has reached the equation:
[tex]\[ b^2 = c^2 \sin^2(B) + c^2 \cos^2(B) + a^2 - 2ac \cos(B) \][/tex]
### Step 1: Factor out [tex]\( c^2 \)[/tex]
First, notice that in the terms [tex]\( c^2 \sin^2(B) \)[/tex] and [tex]\( c^2 \cos^2(B) \)[/tex], we can factor out [tex]\( c^2 \)[/tex]:
[tex]\[ b^2 = c^2 [ \sin^2(B) + \cos^2(B) ] + a^2 - 2ac \cos(B) \][/tex]
### Step 2: Use the Trigonometric Identity
Next, apply the trigonometric identity [tex]\( \sin^2(B) + \cos^2(B) = 1 \)[/tex] to simplify the expression inside the brackets:
[tex]\[ b^2 = c^2 \cdot 1 + a^2 - 2ac \cos(B) \][/tex]
[tex]\[ b^2 = c^2 + a^2 - 2ac \cos(B) \][/tex]
### Conclusion
Thus, the proof steps required are:
- Factor [tex]\( c^2 \)[/tex] for [tex]\( b^2 = c^2 [ \sin^2(B) + \cos^2(B) ] + a^2 - 2ac \cos(B) \)[/tex]
- Use the trigonometric identity [tex]\( \sin^2(B) + \cos^2(B)=1 \)[/tex] for [tex]\( b^2 = c^2 + a^2 - 2ac \cos(B) \)[/tex]
The other given options (use the Pythagorean theorem and FOIL) are not relevant to this proof. Only the steps involving factoring [tex]\( c^2 \)[/tex] and then using the trigonometric identity are necessary to complete Bobby's proof of the law of cosines.
