Sine of summation of two angles is equal to summation of, product of sine of one angle and cosine of second angle and product of cosine of one angle and sine of second angle, is called sine of sum of two angles. It is also called as sine of addition of two angles.

Trigonometric ratio sine mostly appears with angle of right angled triangle to represent the ratio of length of opposite side to length of hypotenuse but sometimes it appears with sum of two different angles. Sine of addition of two angles is expressed as sum and product form of those two angles in sine and cosine terms.

A triangle whose angle is split into two different angles, is essential to know how to prove this mathematical formula in trigonometry with geometric system. Follow below $7$ simple steps to construct required triangle. You can also know them in below HD video tutorial.

- $\Delta BAC$ is a right angled triangle and its angle is $\theta$.
- Draw a line from point $A$ towards side $\overline{BC}$ and it touches the side $\overline{BC}$ at a point $D$.
- The angle of right angled triangle theta $(\theta)$ is split into two angles and are named $x$ and $y$.
- Draw a line from point $D$ and it should be perpendicular to the line $\overline{AD}$. The extended line touches the side $\overline{AC}$ at a point $E$.
- Draw a line from point $E$ and it should be parallel to side $\overline{BC}$ but perpendicular to side $\overline{AB}$. It touches the side $\overline{AB}$ at a point $F$.
- The intersection of extended lines $\overline{EF}$ and $\overline{AD}$ intersect at a point $H$.
- Draw a line parallel to line segment $\overline{GD}$ from point $H$ and it touches the side $\overline{BC}$ at a point $I$.

According to triangle $\Delta EAF$

$$sin(x+y) = {EF \over AE}$$

The length of side $\overline{EF}$ is equal to the sum of lengths of line segments $\overline{EG}$ and $\overline{GF}$. In other words, $EF = EG + GF$ and replace $\overline{EF}$ with its equivalent value in numerator of above expression.

$$⇒ sin(x+y) = {EG + GF \over AE}$$

$\Delta BAD$ and $\Delta IHD$ are similar triangles. Therefore, $∠IHD = ∠BAD = x$. The angles $∠ IHD$ and $∠ HDG$ are vertically opposite angles. Therefore, $∠ HDG = ∠ IHD = x$. According to $\Delta DEH$, the angle $∠ HDE = 90^°$ but the line segment $\overline{DG}$ splits it into two angles $∠ HDG$ and $∠ GDE$.

$∠HDE = ∠HDG + ∠GDE$

$90^° = x + ∠GDE$

$∠GDE = 90^° – x$

The angle $∠GDE$ is $90^° -x$

Consider right angled triangle $\Delta GED$. The sum of three angles of any triangle is $180^°$ geometrically.

$∠GED + ∠EDG + ∠DGE = 180^°$

$∠GED + 90^°-x + 90^° = 180^°$

$∠GED + 180^°-x = 180^°$

$∠GDE = 180^°–180^° + x$

Geometrically, it is proved that the angle of right angled triangle $\Delta GED$ is also $x$.

According to $\Delta GED$,

$$cos x = {EG \over ED}$$

$∴ \, EG = ED.cos x$

Substitute the length of $EG$ in the numerator of the above expression.

$$⇒ sin(x+y) = {ED.cos x + GF \over AE}$$

The length of line segment $\overline{GF}$ is equal to length of line segment $\overline{BD}$. It means $GF = BD$ and substitute it in numerator of above expression.

$$⇒ sin(x+y) = {ED.cos x + BD \over AE}$$

According to $\Delta BAD$.

$$sin x = {BD \over AD}$$

$∴ \, BD = AD.sin x$

Substitute the length of $BD$ in the second part of the numerator of above expression.

$$⇒ sin(x+y) = {ED.cos x + AD.sin x \over AE}$$

$$⇒ sin(x+y) = {ED \over AE}.cos x + {AD \over AE}.sin x$$

According to $\Delta DAE$

$$siny = {ED \over AE}$$

$$cosy = {AD \over AE}$$

Now, substitute ratios in the expression to obtain the result.

$⇒ sin(x+y) = siny.cos x + cosy.sin x$

It can be written in simple remembering form in mathematics.

$sin(x+y) = sinx.cosy + cosx.siny$

Sine of sum of two angles is equal to summation of product of sine of first angle and cosine of second angle, and product of cosine of first angle and sine of second angle.

Mathematically, it is proved but verification of $sin(x+y) = sinx.cosy + cosx.siny$ is essential before use it as formula anywhere in mathematics. So, substitute two angles in this expression.

Take angles $x = 20^°$ and $y = 30^°$

Substitute them in left hand side of the expression.

$sin (20^° + 30^°)$

$⇒ \, sin50^° = 0.766$

Substitute the same angles in right hand side of the equation.

$sin20^°.cos30^° + cos20^°.sin30^° = (0.342).(0.866) + (0.9397).(0.5)$

$⇒ \, sin20^°.cos30^° + cos20^°.sin30^° = 0.766$

It is proved that the value of left hand side and the value of right hand side are equal. Therefore, $sin(x+y) = sinx.cosy + cosx.siny$ is true for these two angles. Not only these two angles, it is true for all the angles. Hence, it is called as a trigonometric identity.