Question

Evaluate the following expression:
$\sin \left( {{60}^{\circ }} \right)\cos \left( {{30}^{\circ }} \right)+\sin \left( {{30}^{\circ }} \right)\cos \left( {{60}^{\circ }} \right)$

Answer 1

Hint: We will use the direct values that are given in trigonometry. Some of them that are used here are sin(${{60}^{\circ }}$ ) as $\dfrac{\sqrt{3}}{2}$ and cos(${{30}^{\circ }}$ ) has the value with the value $\dfrac{\sqrt{3}}{2}$. Also, sin( ${{30}^{\circ }}$ ) has the value $\dfrac{1}{2}$ and cos( ${{60}^{\circ }}$ ) has the value equal to $\dfrac{1}{2}$. Moreover, we will apply the trigonometric formula given by sin(a + b) = sin(a)cos(b) + cos(a)sin(b). We will use this formula and use the values in order to solve the question.
Complete step-by-step answer:
Now, we will consider the expression $\sin \left( {{60}^{\circ }} \right)\cos \left( {{30}^{\circ }} \right)+\sin \left( {{30}^{\circ }} \right)\cos \left( {{60}^{\circ }} \right)...(i)$.
Now we will apply substitution here. We will substitute a as ${{60}^{\circ }}$ and b as ${{30}^{\circ }}$. Therefore our equation (i) changes to sin(a)cos(b) + cos(a)sin(b).
 Now at this step we will apply the formula given by sin(a + b) = sin(a)cos(b) + cos(a)sin(b). Therefore, we have a new expression given by sin(a + b).
As we know that the substitution a as ${{60}^{\circ }}$ and b as ${{30}^{\circ }}$ in equation (i) results into sin(a + b) as$\sin \left( {{60}^{\circ }}+{{30}^{\circ }} \right)$.
This further shows that after adding the degrees we have sin(a + b) as $\sin \left( {{90}^{\circ }} \right)$.
As we know that the value of $\sin \left( {{90}^{\circ }} \right)$ is equal to 1.
Therefore, we have the answer of the equation (i) as 1.
Hence, the value of the expression $\sin \left( {{60}^{\circ }} \right)\cos \left( {{30}^{\circ }} \right)+\sin \left( {{30}^{\circ }} \right)\cos \left( {{60}^{\circ }} \right)$ is equal to 1.

Note: Alternate method of doing this question is by substituting the trigonometric values at certain degrees in the expression $\sin \left( {{60}^{\circ }} \right)\cos \left( {{30}^{\circ }} \right)+\sin \left( {{30}^{\circ }} \right)\cos \left( {{60}^{\circ }} \right)$. This can be done as by substituting $\sin \left( {{60}^{\circ }} \right)=\dfrac{\sqrt{3}}{2}$ and $\cos \left( {{30}^{\circ }} \right)=\dfrac{\sqrt{3}}{2}$ in the first terms and substituting $\sin \left( {{30}^{\circ }} \right)=\dfrac{1}{2}$ and $\cos \left( {{60}^{\circ }} \right)=\dfrac{1}{2}$. Therefore, we get $\sin \left( {{60}^{\circ }} \right)\cos \left( {{30}^{\circ }} \right)+\sin \left( {{30}^{\circ }} \right)\cos \left( {{60}^{\circ }} \right)=\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \dfrac{1}{2}$.
This result into $\sin \left( {{60}^{\circ }} \right)\cos \left( {{30}^{\circ }} \right)+\sin \left( {{30}^{\circ }} \right)\cos \left( {{60}^{\circ }} \right)=\dfrac{3}{4}+\dfrac{1}{4}$ or $\sin \left( {{60}^{\circ }} \right)\cos \left( {{30}^{\circ }} \right)+\sin \left( {{30}^{\circ }} \right)\cos \left( {{60}^{\circ }} \right)=\dfrac{4}{4}$ resulting into 1.