Answer:
Step-by-step explanation:
Law of sines is:
(sin A) / a = (sin B) / b = (sin C) / c
Law of cosines is:
c² = a² + b² − 2ab cos C
Note that a, b, and c are interchangeable, so long as the angle in the cosine corresponds to the side on the left of the equation (for example, angle C is opposite of side c).
Also, angles of a triangle add up to 180° or π.
(i) sin(B−C) / sin(B+C)
Since A+B+C = π, B+C = π−A:
sin(B−C) / sin(π−A)
Using angle shift property:
sin(B−C) / sin A
Using angle sum/difference identity:
(sin B cos C − cos B sin C) / sin A
Distribute:
(sin B cos C) / sin A − (cos B sin C) / sin A
From law of sines, sin B / sin A = b / a, and sin C / sin A = c / a.
(b/a) cos C − (c/a) cos B
From law of cosines:
c² = a² + b² − 2ab cos C
(c/a)² = 1 + (b/a)² − 2(b/a) cos C
2(b/a) cos C = 1 + (b/a)² − (c/a)²
(b/a) cos C = ½ + ½ (b/a)² − ½ (c/a)²
Similarly:
b² = a² + c² − 2ac cos B
(b/a)² = 1 + (c/a)² − 2(c/a) cos B
2(c/a) cos B = 1 + (c/a)² − (b/a)²
(c/a) cos B = ½ + ½ (c/a)² − ½ (b/a)²
Substituting:
[ ½ + ½ (b/a)² − ½ (c/a)² ] − [ ½ + ½ (c/a)² − ½ (b/a)² ]
½ + ½ (b/a)² − ½ (c/a)² − ½ − ½ (c/a)² + ½ (b/a)²
(b/a)² − (c/a)²
(b² − c²) / a²
(ii) a (cos B + cos C)
a cos B + a cos C
From law of cosines, we know:
b² = a² + c² − 2ac cos B
2ac cos B = a² + c² − b²
a cos B = 1/(2c) (a² + c² − b²)
Similarly:
c² = a² + b² − 2ab cos C
2ab cos C = a² + b² − c²
a cos C = 1/(2b) (a² + b² − c²)
Substituting:
1/(2c) (a² + c² − b²) + 1/(2b) (a² + b² − c²)
Common denominator:
1/(2bc) (a²b + bc² − b³) + 1/(2bc) (a²c + b²c − c³)
1/(2bc) (a²b + bc² − b³ + a²c + b²c − c³)
Rearrange:
1/(2bc) [a²b + a²c + bc² + b²c − (b³ + c³)]
Factor (use sum of cubes):
1/(2bc) [a² (b + c) + bc (b + c) − (b + c)(b² − bc + c²)]
(b + c)/(2bc) [a² + bc − (b² − bc + c²)]
(b + c)/(2bc) (a² + bc − b² + bc − c²)
(b + c)/(2bc) (2bc + a² − b² − c²)
Distribute:
(b + c)/(2bc) (2bc) + (b + c)/(2bc) (a² − b² − c²)
(b + c) + (b + c)/(2bc) (a² − b² − c²)
From law of cosines, we know:
a² = b² + c² − 2bc cos A
2bc cos A = b² + c² − a²
cos A = (b² + c² − a²) / (2bc)
-cos A = (a² − b² − c²) / (2bc)
Substituting:
(b + c) + (b + c)(-cos A)
(b + c)(1 − cos A)
From half angle formula, we can rewrite this as:
2(b + c) sin²(A/2)
(iii) (b + c) cos A + (a + c) cos B + (a + b) cos C
From law of cosines, we know:
cos A = (b² + c² − a²) / (2bc)
cos B = (a² + c² − b²) / (2ac)
cos C = (a² + b² − c²) / (2ab)
Substituting:
(b + c) (b² + c² − a²) / (2bc) + (a + c) (a² + c² − b²) / (2ac) + (a + b) (a² + b² − c²) / (2ab)
Common denominator:
(ab + ac) (b² + c² − a²) / (2abc) + (ab + bc) (a² + c² − b²) / (2abc) + (ac + bc) (a² + b² − c²) / (2abc)
[(ab + ac) (b² + c² − a²) + (ab + bc) (a² + c² − b²) + (ac + bc) (a² + b² − c²)] / (2abc)
We have to distribute, which is messy. To keep things neat, let's do this one at a time. First, let's look at the a² terms.
-a² (ab + ac) + a² (ab + bc) + a² (ac + bc)
a² (-ab − ac + ab + bc + ac + bc)
2a²bc
Repeating for the b² terms:
b² (ab + ac) − b² (ab + bc) + b² (ac + bc)
b² (ab + ac − ab − bc + ac + bc)
2ab²c
And the c² terms:
c² (ab + ac) + c² (ab + bc) − c² (ac + bc)
c² (ab + ac + ab + bc − ac − bc)
2abc²
Substituting:
(2a²bc + 2ab²c + 2abc²) / (2abc)
2abc (a + b + c) / (2abc)
a + b + c
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