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# Evaluate the value of $\sin {60^ \circ }\cos {30^ \circ } + \cos {60^ \circ }\sin {30^ \circ }$.

Last updated date: 13th Jun 2024
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Hint: Start by writing down the given equation . Try to recall all the values of trigonometric ratios for corresponding angles and substitute in the given equation. Simplify the equation in order to get the desired value for the given equation.

Given,
$\sin {60^ \circ }\cos {30^ \circ } + \cos {60^ \circ }\sin {30^ \circ }$
Step by step complete solution
We know the values of following trigonometric ratios
 $\theta$ ${0^ \circ }$ ${30^ \circ }$ ${45^ \circ }$ ${60^ \circ }$ ${90^ \circ }$ $\sin \theta$ $0$ $\dfrac{1}{2}$ $\dfrac{1}{{\sqrt 2 }}$ $\dfrac{{\sqrt 3 }}{2}$ $1$ $\cos \theta$ $1$ $\dfrac{{\sqrt 3 }}{2}$ $\dfrac{1}{{\sqrt 2 }}$ $\dfrac{1}{2}$ $0$

From the above table we get
$\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$ and $\sin {30^ \circ } = \dfrac{1}{2}$
$\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$ and $\cos {60^ \circ } = \dfrac{1}{2}$
Substituting these values in the relation we get
$\Rightarrow \sin {60^ \circ }\cos {30^ \circ } + \cos {60^ \circ }\sin {30^ \circ } \\ \Rightarrow \dfrac{{\sqrt 3 }}{2} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{2} \times \dfrac{1}{2} \\ \\$
We know $\sqrt 3 \times \sqrt 3 = {(\sqrt 3 )^2} = 3$
Similarly in denominator ,we have $2 \times 2 = {(2)^2} = 4$
$= \dfrac{3}{4} + \dfrac{1}{4}$
Taking 4 as L.C.M. , we get
$= \dfrac{4}{4} \\ = 1 \\$
So , the value of $\sin {60^ \circ }\cos {30^ \circ } + \cos {60^ \circ }\sin {30^ \circ }$ is 1.

Note: Students must remember all the trigonometric values of different angles , which are used more often. Attention must be given while substituting the values , keeping in mind the quadrant system, here we had all the positive values only , But one might get different quadrant values as well.
Alternative method:-
We can simplify the given trigonometric equation by using trigonometric formulas
We have $\sin {60^ \circ }\cos {30^ \circ } + \cos {60^ \circ }\sin {30^ \circ }$
This is of the form $\sin A\cos B + \cos A\sin B$
And we know $\sin (A + B) = \sin A\cos B + \cos A\sin B$
So our equation $\sin {60^ \circ }\cos {30^ \circ } + \cos {60^ \circ }\sin {30^ \circ }$would become
$\sin {(60 + 30)^ \circ } = \sin {90^ \circ }$
And we know the value of $\sin {90^ \circ } = 1$from the table of trigonometric values for different angles.
Therefore , the value of $\sin {60^ \circ }\cos {30^ \circ } + \cos {60^ \circ }\sin {30^ \circ }$ is 1.