Question

Prove \[\sin (A + B) + \sin (A - B) = 2\sin A\cos B\].

Answer 1

Hint: The given solution requires us to prove that both the sides of equations are equal using trigonometry rules. Trigonometry is a branch of mathematics that investigates the relationship between triangle side lengths and angles. Relationship between sine and cosine can be explained as follows: the sine of any acute angle is equal to the cosine of its complement- \[\sin \theta = \cos (90 - \theta )\]. The cosine of any acute angle is equal to the sine of its complement-\[\cos \theta = \sin (90 - \theta )\].

Complete step by step answer:
The given question requires us to use theorem to satisfy the conditions. A theorem is a statement that has been proven true by a rigorous proof with help of logical argument. We know with 100 percent certainty that a theorem is valid because it has been proven.
First, we will evaluate the left-hand side of the equation as follows:
\[ \Rightarrow \sin (A + B) + \sin (A - B)\]
Now we will use the theorem-
Theorem One:
\[ \Rightarrow \sin (A + B) = \sin A\cos B + \cos A\sin B\]
Theorem Two:
\[ \Rightarrow \sin (A - B) = \sin A\cos B - \cos A\sin B\]

Substituting the above theorems in the left-hand side of the equation, we get,
\[\sin (A + B) + \sin (A - B) = (\sin A\cos B + \cos A\sin B) + (\sin A\cos B - \cos A\sin B)\]
Opening up the brackets, we get,
\[ \sin (A + B) + \sin (A - B)= \sin A\cos B + \cos A\sin B + \sin A\cos B - \cos A\sin B\]
Subtracting in the above equation, we get,
\[ \sin (A + B) + \sin (A - B)= \sin AcosB + \sin A\cos B\]
Taking the common factor, we get,
\[ \sin (A + B) + \sin (A - B)= 2\sin A\cos B =RHS \]

Hence it is proved that \[\sin (A + B) + \sin (A - B) = 2\sin A\cos B\].

Note: Sine and Cosine are mainly used with reference to right-angled triangles. They are the ratio of the sides of the right-angled triangle. Sine is the ratio of side opposite to given angle and hypotenuse whereas cosine/cos is the ratio of side adjacent to given angle and hypotenuse. Sine and cosine are called "co-functions' ', where the sine (or cosine) function of any acute angle equals its cofunction of the angle's complement. In a right triangle, the sine of one acute angle, \[A\], equals the cosine of the other acute angle, \[B\].