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Question: State whether the following are true or false. Justify your answer. \(\sin \left( A+B \right)=\sin...

State whether the following are true or false. Justify your answer.
sin⁡(A+B)=sin⁡A+sin⁡B\sin \left( A+B \right)=\sin A+\sin B.

Explanation

Solution

Hint:To state the above statement as true or false. Take any combination of angle of A and B and substitute in the given equation sin⁡(A+B)=sin⁡A+sin⁡B\sin \left( A+B \right)=\sin A+\sin B and then see whether the combination of angles satisfies this equation or not.

Complete step-by-step answer:
The equation given in the question is:
sin⁡(A+B)=sin⁡A+sin⁡B\sin \left( A+B \right)=\sin A+\sin B ……… Eq. (1)
We have to state whether this equation is true or false.
Let us take a combination of angle A and angle B.
∠A=30∘&∠B=60∘\angle A={{30}^{\circ }}\And \angle B={{60}^{\circ }}
Substitute these values of angles in the eq. (1).
sin⁡(30∘+60∘)=sin⁡30∘+sin⁡60∘\sin \left( {{30}^{\circ }}+{{60}^{\circ }} \right)=\sin {{30}^{\circ }}+\sin {{60}^{\circ }}
L.H.S of the above equation yields:
sin⁡90∘=1\sin {{90}^{\circ }}=1
R.H.S of the above equation yields:
sin⁡30∘+sin⁡60∘ =12+32 =3+12 \begin{aligned} & \sin {{30}^{\circ }}+\sin {{60}^{\circ }} \\\ & =\dfrac{1}{2}+\dfrac{\sqrt{3}}{2} \\\ & =\dfrac{\sqrt{3}+1}{2} \\\ \end{aligned}
From the above calculation we have found that,
L.H.S = 1
R.H.S=3+12\text{R}\text{.H}\text{.S}=\dfrac{\sqrt{3}+1}{2}
From the above solution we can see that L.H.S ≠ R.H.S. Hence, this equation sin⁡(A+B)=sin⁡A+sin⁡B\sin \left( A+B \right)=\sin A+\sin B is false.

Note: You can check the truth value of this equation sin⁡(A+B)=sin⁡A+sin⁡B\sin \left( A+B \right)=\sin A+\sin B by taking different angles also.
Like if we take A=450 and B=450A={{45}^{0}}\text{ and B=4}{{\text{5}}^{0}} and then substituting these values of A and B in the given equation sin⁡(A+B)=sin⁡A+sin⁡B\sin \left( A+B \right)=\sin A+\sin B we get,
sin⁡(45∘+45∘)=sin⁡45∘+sin⁡45∘\sin \left( {{45}^{\circ }}+{{45}^{\circ }} \right)=\sin {{45}^{\circ }}+\sin {{45}^{\circ }}
Solving L.H.S of the given equation:
sin⁡(45∘+45∘) =sin⁡90∘=1 \begin{aligned} & \sin \left( {{45}^{\circ }}+{{45}^{\circ }} \right) \\\ & =\sin {{90}^{\circ }}=1 \\\ \end{aligned}
Solving R.H.S of the given equation:
sin⁡45∘+sin⁡45∘ =2sin⁡45∘ =2(12)=2 \begin{aligned} & \sin {{45}^{\circ }}+\sin {{45}^{\circ }} \\\ & =2\sin {{45}^{\circ }} \\\ & =2\left( \dfrac{1}{\sqrt{2}} \right)=\sqrt{2} \\\ \end{aligned}
From the above calculations we have found that:
L.H.S = 1
R.H.S=2R.H.S=\sqrt{2}
From the above solution we can see that L.H.S ≠ R.H.S. Hence, this equation sin⁡(A+B)=sin⁡A+sin⁡B\sin \left( A+B \right)=\sin A+\sin B is false.
Hence, this combination of angles of A and B also states that the given equation is false.We can also check by using the formula of sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B.Comparing the formula with given equation we can say that the statement is false.