For any triangle abc note down the sine and cos theorems ( sina/a= sinb/b
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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

answer: hey there,

the answer you are looking for is a..

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