Some very useful Trigonometry formulas and Identities with proof are given here. Here we would also discuss some of the very important practical applications of these formulas in real life with examples.
Sine Formula
Let, In a △ABC, the length of sides: BC=a, CA=b, and AB=c then according to the Sine Formula
\frac{a}{Sin(A)}=\frac{b}{Sin(B)}=\frac{c}{Sin(C)}=2R.
Here R is the Circumradius of the △ABC.
Here is the proof of this formula.
![Trigonometry Formulas with Proof (Bonus tips and Cheat sheet 2023) 2 Sine formula proof](https://logicxonomy.com/wp-content/uploads/2022/12/Sine-1024x576.webp)
Draw AD⟂BC, the height AD=h
In △BDA, Sin(B)=\frac{h}{c}.
h=c.Sin(B)
Similarly, In △CDA
Sin(C)=\frac{h}{b}.
h=b.Sin(C)
So, c.Sin(B)=b.Sin(C)
\frac{b}{Sin(B)}=\frac{c}{Sin(C)} ……………(i)
Now, Draw BE⟂AC, the height BE=k
\frac{a}{Sin(A)}=\frac{c}{Sin(C)} ……………(ii)
From the equation (i) and (ii):
\frac{a}{Sin(A)}=\frac{b}{Sin(b)}=\frac{c}{Sin(C)} ……………(iii)
![Trigonometry Formulas with Proof (Bonus tips and Cheat sheet 2023) 3 Circumcircle of the triangle](https://logicxonomy.com/wp-content/uploads/2022/12/Circum-1024x576.webp)
Draw a Circumcircle of △ABC. Here O is the Circumcenter of the triangle.
The length of Circumradius=R
So, the length of the diameter, BD=2R
∠BAC=∠BDC=∠A (Angles, subtended by the same arc at the circumference in the same segment)
∠BCD=90° (Angle made by the diameter at the Circumference)
Now in △BCD, according to Sine Rule:
\frac{BC}{Sin(BDC)}=\frac{BD}{Sin(90°)}.
\frac{a}{Sin(A)}=\frac{2R}{1}.
\frac{a}{Sin(A)}=2R …………(iv)
From equation (iii) and (iv):
\frac{a}{Sin(A)}=\frac{b}{Sin(B)}=\frac{c}{Sin(C)}=2R.
![Trigonometry Formulas with Proof (Bonus tips and Cheat sheet 2023) 4 Online Class Logicxonomy](https://logicxonomy.com/wp-content/uploads/2023/03/20230314_070946_0000-1024x1024.webp)
Cosine Formula
Let, In an △ABC, the length of sides: BC=a, CA=b, and AB=c. BD is perpendicular drawn from vertex B to side AC, here h is the height of the triangle.
Then, According to Cosine Formula:
a2=b2+c2-2bc×Cos(A)
b2=c2+a2-2ac×Cos(B)
c2=a2+b2-2ab×Cos(C)
Here is the proof of this formula.
![Trigonometry Formulas with Proof (Bonus tips and Cheat sheet 2023) 5 Trigonometry Formulas](https://logicxonomy.com/wp-content/uploads/2022/12/Cosine-1024x576.webp)
Apply Pythagoras Theorem in △BDC
a2=h2+DC2
R.H.S.
=h2+(AC-AD)2
=h2+AC2+AD2-2.AC.AD
=(h2+AD2)+b2-2.b.AD
=c2+b2-2.b.AD
Note: h2+AD2=c2 (Pythagoras Theorem)
Note: In △ADB
Cos(A)=\frac{AD}{AB}=\frac{AD}{c}.
AD=c.Cos(A)
So, a2=b2+c2-2.bc.Cos(A)
Similarly, we can prove other combinations also.
In △ADB, Cos(A)=\frac{AD}{c}.
AD=c.Cos(A) ………(i)
In △CDB, Cos(C)=\frac{CD}{a}.
CD=a.Cos(C) ………(ii)
Add both equations:
AD+CD=c.Cos(A)+a.Cos(C)
AC=c.Cos(A)+a.Cos(C)
b=c.Cos(A)+a.Cos(C) ……… (iii)
Similarly, we can prove:
a=b.Cos(C)+c.Cos(B) ……… (iv)
c=a.Cos(B)+b.Cos(A) ……… (v)
The Stewart’s and the Apollonius’s Theorem Proof: Click Here
More Trigonometry Formulas and Concepts are coming soon…