$\sin{(A+B)}$ $\,=\,$ $\sin{A}\cos{B}$ $+$ $\cos{A}\sin{B}$

The expansion of $\sin{(A+B)}$ identity is actually derived in geometrical approach by constructing a right angled triangle, in which its angle is divided as two different angles.

- $\small \Delta EDF$ is a right angled triangle and draw line from $\small D$ to divide the angle of $\small \Delta EDF$ as two different angles $\small A$ and $\small B$. The line intersects side $\small \overline{EF}$ at point $\small G$.
- Draw a line from point $\small G$ to side $\small \overline{FD}$ but it should be perpendicular to side $\small \overline{DG}$ and the line intersects the side $\small \overline{FD}$ at point $\small H$.
- Draw a perpendicular to side $\small \overline{ED}$ from point $\small H$ and it intersects the side $\small \overline{DG}$ at point $\small K$ and also intersects the side $\small \overline{ED}$ at point $\small J$.
- Finally, draw a perpendicular line to side $\small \overline{HJ}$ from $\small G$ and it intersects the side $\small \overline{HJ}$ at point $\small K$. Similarly, draw a perpendicular line from $\small I$ to side $\small \overline{EF}$ and intersects the side $\small \overline{EF}$ at point $\small L$.

$A+B$ is angle of the $\Delta JDH$ and it is useful to express sin of sum of angles as a ratio of lengths of the opposite side to hypotenuse.

$(1) \,\,\,\,\,\,$ $\overline{HJ}$ is opposite side

$(2) \,\,\,\,\,\,$ $\overline{DH}$ is hypotenuse

The lengths of them are written as $HJ$ and $DH$ respectively in mathematics.

$\sin{(A+B)} = \dfrac{HJ}{DH}$

The side $\overline{KG}$ is perpendicularly intersected the side $\overline{HJ}$ at point $K$. Therefore, the length of the side $\overline{HJ}$ can be written as sum of the lengths of the sides $\overline{HK}$ and $\overline{KJ}$.

$HJ = HK+KJ$

Now, replace the length of the side $\overline{HJ}$ in the equivalent expansion of $\sin{(A+B)}$ law.

$\implies$ $\sin{(A+B)} = \dfrac{HK+KJ}{DH}$

$\implies$ $\sin{(A+B)} = \dfrac{HK}{DH}+\dfrac{KJ}{DH}$

$\overline{KJ}$ and $\overline{GE}$ are parallel lines and also equal because of the perpendicular intersection of another set of parallel lines $\overline{KG}$ and $\overline{JE}$.

$KJ = GE$

Now, substitute the length of the side $\overline{KJ}$ by the length of the side $\overline{GE}$.

$\implies$ $\sin{(A+B)} = \dfrac{HK}{DH}+\dfrac{GE}{DH}$

$\overline{GE}$ is opposite side of right angled triangle $EDG$ and $A$ is an angle of that triangle. So, write the length of the side $\overline{GE}$ in trigonometric function form.

$\sin{A} = \dfrac{GE}{DG}$

$\implies GE = DG\sin{A}$

Now, replace the length of the opposite side $\overline{GE}$ by its equivalent value.

$\sin{(A+B)} = \dfrac{HK}{DH}+\dfrac{DG\sin{A}}{DH}$

$\overline{HK}$ is a side of right angled triangle $KHG$ but its angle is unknown. So, it is not possible to express the length of side $\overline{HK}$ in trigonometric form. This problem can be overcome by some geometrical analysis.

$\Delta EDG$ and $\Delta LIG$ are right angled triangles and moreover, they are similar triangles. Therefore, their corresponding angles should be equal geometrically. In this case, $\angle EDG$ and $\angle LIG$ are corresponding angles but $\angle EDG = A$.

$\angle LIG = \angle EDG = A$

$\overline{KG}$ and $\overline{IL}$ are parallel lines and $\overline{IG}$ is their transversal. $\angle LIG$ and $\angle IGK$ are interior alternate angles and they are equal geometrically.

$\angle IGK = \angle LIG = A$

The side $\overline{HG}$ is drawn perpendicularly to the side $\overline{DG}$. So, the sum of $\angle IGK$ and $\angle KGH$ should be $90^\circ$.

$\angle IGK + \angle KGH = 90^\circ$

$\implies A + \angle KGH = 90^\circ$

$\implies \angle KGH = 90^\circ-A$

$\Delta KHG$ is a right angled triangle. In this triangle, $\angle HKG = 90^\circ$, $\angle KGH = 90^\circ-A$ but $\angle GHK$ is unknown. However, it can be calculated by using sum of angles in a triangle rule.

$\angle GHK + \angle KGH + \angle HKG$ $\,=\,$ $180^\circ$

$\implies \angle GHK + 90^\circ-A + 90^\circ$ $\,=\,$ $180^\circ$

$\implies \angle GHK + 180^\circ-A$ $\,=\,$ $180^\circ$

$\implies \angle GHK$ $\,=\,$ $180^\circ-180^\circ+A$

$\implies \angle GHK \,=\, A$

$\cos{A} = \dfrac{HK}{HG}$

$\implies HK = HG\cos{A}$

Replace the length of the side $\overline{HK}$ by its equivalent value.

$\implies \sin{(A+B)}$ $\,=\,$ $\dfrac{HG\cos{A}}{DH}+\dfrac{DG\sin{A}}{DH}$

$\implies \sin{(A+B)}$ $\,=\,$ $\dfrac{HG}{DH} \times \cos{A} +\dfrac{DG}{DH} \times \sin{A}$

In the expansion of $\sin(A+B)$ rule, $HG$, $DG$ and $DH$ are remaining factors in the trigonometric expression. Actually, the three of them are lengths of the sides of the $\Delta GDH$ and $B$ is the angle of this right angled triangle. Now express each ratio of lengths of the sides by its respective trigonometric function.

$\sin{B} \,=\, \dfrac{HG}{DH}$

$\cos{B} \,=\, \dfrac{DG}{DH}$

Replace the ratio of lengths of the sides by the corresponding trigonometric function in the expansion of $\sin{(A+B)}$ rule.

$\implies \sin{(A+B)}$ $\,=\,$ $\sin{B}\times\cos{A}+\cos{B}\times\sin{A}$

$\implies \sin{(A+B)}$ $\,=\,$ $\sin{B}\cos{A}+\cos{B}\sin{A}$

$\implies \sin{(A+B)}$ $\,=\,$ $\cos{A}\sin{B}+\sin{A}\cos{B}$

$\,\,\, \therefore \,\,\,\,\,\, \sin{(A+B)}$ $\,=\,$ $\sin{A}\cos{B}+\cos{A}\sin{B}$

This trigonometric identity is used to expand sin of sum of two angles in terms of sine and cosine of those two angles.

List of most recently solved mathematics problems.

Jul 04, 2018

Limit (Calculus)

Evaluate $\displaystyle \large \lim_{x \,\to\, \tan^{-1}{3}} \normalsize {\dfrac{\tan^2{x}-2\tan{x}-3}{\tan^2{x}-4\tan{x}+3}}$

Jun 23, 2018

Limit (Calculus)

Evaluate $\displaystyle \large \lim_{x \to 0} \normalsize \dfrac{e^{x^2}-\cos{x}}{x^2}$

Jun 22, 2018

Integral Calculus

Evaluate $\displaystyle \int \dfrac{1+\cos{4x}}{\cot{x}-\tan{x}} dx$

Jun 21, 2018

Limit

Evaluate $\displaystyle \large \lim_{x \to \infty} \normalsize {\sqrt{x^2+x+1}-\sqrt{x^2+1}}$

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.