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Question: Evaluate the value of \(\sin {60^ \circ }\cos {30^ \circ } + \cos {60^ \circ }\sin {30^ \circ }\)....

Evaluate the value of sin⁡60∘cos⁡30∘+cos⁡60∘sin⁡30∘\sin {60^ \circ }\cos {30^ \circ } + \cos {60^ \circ }\sin {30^ \circ }.

Explanation

Solution

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.

Complete step-by-step answer:
Given,
sin⁡60∘cos⁡30∘+cos⁡60∘sin⁡30∘\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

θ\theta 0∘{0^ \circ }30∘{30^ \circ }45∘{45^ \circ }60∘{60^ \circ }90∘{90^ \circ }
sin⁡θ\sin \theta 0012\dfrac{1}{2}12\dfrac{1}{{\sqrt 2 }}32\dfrac{{\sqrt 3 }}{2}11
cos⁡θ\cos \theta 1132\dfrac{{\sqrt 3 }}{2}12\dfrac{1}{{\sqrt 2 }}12\dfrac{1}{2}00

From the above table we get
sin⁡60∘=32\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2} and sin⁡30∘=12\sin {30^ \circ } = \dfrac{1}{2}
cos⁡30∘=32\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2} and cos⁡60∘=12\cos {60^ \circ } = \dfrac{1}{2}
Substituting these values in the relation we get
⇒sin⁡60∘cos⁡30∘+cos⁡60∘sin⁡30∘ ⇒32×32+12×12   \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 3×3=(3)2=3\sqrt 3 \times \sqrt 3 = {(\sqrt 3 )^2} = 3
Similarly in denominator ,we have 2×2=(2)2=42 \times 2 = {(2)^2} = 4
=34+14= \dfrac{3}{4} + \dfrac{1}{4}
Taking 4 as L.C.M. , we get
=44 =1  = \dfrac{4}{4} \\\ = 1 \\\
So , the value of sin⁡60∘cos⁡30∘+cos⁡60∘sin⁡30∘\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∘cos⁡30∘+cos⁡60∘sin⁡30∘\sin {60^ \circ }\cos {30^ \circ } + \cos {60^ \circ }\sin {30^ \circ }
This is of the form sin⁡Acos⁡B+cos⁡Asin⁡B\sin A\cos B + \cos A\sin B
And we know sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B
So our equation sin⁡60∘cos⁡30∘+cos⁡60∘sin⁡30∘\sin {60^ \circ }\cos {30^ \circ } + \cos {60^ \circ }\sin {30^ \circ }would become
sin⁡(60+30)∘=sin⁡90∘\sin {(60 + 30)^ \circ } = \sin {90^ \circ }
And we know the value of sin⁡90∘=1\sin {90^ \circ } = 1from the table of trigonometric values for different angles.
Therefore , the value of sin⁡60∘cos⁡30∘+cos⁡60∘sin⁡30∘\sin {60^ \circ }\cos {30^ \circ } + \cos {60^ \circ }\sin {30^ \circ } is 1.