Trigonometric Identities & Vector Laws – Complete Notes

𝛼 ሺ𝑠−𝑏ሻሺ𝑠−𝑐ሻ
• Derive sin = ට
2 𝑏𝑐


To Proof :
Consider a triangle ABC , in which a ,b , c are the measure of sides
opposite to the angles 𝛼 , 𝛽 & 𝛾 respectively.
Using law of cosine
𝑏 2 + 𝑐 2 − 𝑎2
cos 𝛼 =
2𝑏𝑐
Subtracting both sides by 1
𝑏 2 + 𝑐 2 − 𝑎2
1 − cos 𝛼 = 1 −
2𝑏𝑐
2𝑏𝑐 − 𝑏 − 𝑐 2 + 𝑎2
2
1 − cos 𝛼 =
2𝑏𝑐
α
∴ 1 − cos α = 2 sin2
2
α 2𝑏𝑐 − 𝑏 2 − 𝑐 2 + 𝑎2
2
2 sin =
2 2𝑏𝑐

α 𝑎2 − 𝑏 2 + 2𝑏𝑐 − 𝑐 2
2 sin2 =
2 2𝑏𝑐

α 𝑎2 − ሺ𝑏 2 − 2𝑏𝑐 + 𝑐 2 ሻ
2
2 sin =
2 2𝑏𝑐
α 𝑎2 − ሺ𝑏 − 𝑐ሻ2
2
2 sin =
2 2𝑏𝑐

, α ሺ𝑎 + 𝑏 − 𝑐ሻሺ𝑎 − 𝑏 + 𝑐ሻ
2 sin2 =
2 2𝑏𝑐
∴ 𝑎 + 𝑏 + 𝑐 = 2𝑠 => 𝑎 + 𝑏 − 𝑐 = 2𝑠 − 2𝑐
∴ 𝑎 + 𝑏 + 𝑐 = 2𝑠 => 𝑎 − 𝑏 + 𝑐 = 2𝑠 − 2𝑏


α ሺ2𝑠 − 2𝑐ሻሺ2𝑠 − 2𝑏ሻ
sin2 =
2 2 × 2𝑏𝑐
α 4ሺ𝑠 − 𝑐ሻሺ𝑠 − 𝑏ሻ
sin2 =
2 4𝑏𝑐
α ሺ𝑠−𝑏ሻሺ𝑠−𝑐ሻ
sin = ට
2 𝑏𝑐

HENCE PROVED !



𝛽 ሺ𝑠−𝑎ሻሺ𝑠−𝑐ሻ
• Derive sin = ට
2 𝑎𝑐


To Proof :
Consider a triangle ABC , in which a ,b , c are the measure of sides
opposite to the angles 𝛼 , 𝛽 & 𝛾 respectively.
Using law of cosine
𝑎2 + 𝑐 2 − 𝑏 2
cos 𝛽 =
2𝑎𝑐
Subtracting both sides by 1
𝑎2 + 𝑐 2 − 𝑏 2
1 − cos 𝛽 = 1 −
2𝑎𝑐
2𝑎𝑐 − 𝑎 − 𝑐 2 + 𝑏 2
2
1 − cos 𝛽 =
2𝑎𝑐
𝛽
∴ 1 − cos 𝛽 = 2 sin2
2