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

Prove sin⁡(A+B)+sin⁡(A−B)=2sin⁡Acos⁡B\sin (A + B) + \sin (A - B) = 2\sin A\cos B.

Explanation

Solution

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⁡θ=cos⁡(90−θ)\sin \theta = \cos (90 - \theta ). The cosine of any acute angle is equal to the sine of its complement-cos⁡θ=sin⁡(90−θ)\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:
⇒sin⁡(A+B)+sin⁡(A−B)\Rightarrow \sin (A + B) + \sin (A - B)
Now we will use the theorem-
Theorem One:
⇒sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\Rightarrow \sin (A + B) = \sin A\cos B + \cos A\sin B
Theorem Two:
⇒sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\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⁡Acos⁡B+cos⁡Asin⁡B)+(sin⁡Acos⁡B−cos⁡Asin⁡B)\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⁡Acos⁡B+cos⁡Asin⁡B+sin⁡Acos⁡B−cos⁡Asin⁡B\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⁡Acos⁡B\sin (A + B) + \sin (A - B)= \sin AcosB + \sin A\cos B
Taking the common factor, we get,
sin⁡(A+B)+sin⁡(A−B)=2sin⁡Acos⁡B=RHS\sin (A + B) + \sin (A - B)= 2\sin A\cos B =RHS

Hence it is proved that sin⁡(A+B)+sin⁡(A−B)=2sin⁡Acos⁡B\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, AA, equals the cosine of the other acute angle, BB.