Question

Find the value of $\sin {22^ \circ }\cos {38^ \circ } + \cos {22^ \circ }\sin {38^ \circ }$
A. $\dfrac{1}{2}$
B. $\dfrac{{ - \sqrt 3 }}{2}$
C. $\dfrac{{\sqrt 3 }}{2}$
D. $\dfrac{1}{{\sqrt 2 }}$

Answer 1

We multiply and divide the given equation by 2. Form two separate equations relating to the formulas of trigonometry i.e. $2\sin A\cos B$ and $2\cos A\sin B$. Use the formulas and solve the equation.

• $2\sin A\cos B = \sin (A + B)\cos (A - B)$ and $2\cos A\sin B = \sin (A + B) - \sin (A - B)$

Complete step-by-step answer:
We are given the equation $\sin {22^ \circ }\cos {38^ \circ } + \cos {22^ \circ }\sin {38^ \circ }$
Multiply and divide the given equation by 2
$\Rightarrow \sin {22^ \circ }\cos {38^ \circ } + \cos {22^ \circ }\sin {38^ \circ } = \dfrac{1}{2}\left[ {2\left( {\sin {{22}^ \circ }\cos {{38}^ \circ } + \cos {{22}^ \circ }\sin {{38}^ \circ }} \right)} \right]$
Multiply 2 to each term of RHS
$\Rightarrow \sin {22^ \circ }\cos {38^ \circ } + \cos {22^ \circ }\sin {38^ \circ } = \dfrac{1}{2}\left[ {\left( {2\sin {{22}^ \circ }\cos {{38}^ \circ }} \right) + \left( {2\cos {{22}^ \circ }\sin {{38}^ \circ }} \right)} \right]$...............… (1)
We know $2\sin A\cos B = \sin (A + B)\cos (A - B)$and$2\cos A\sin B = \sin (A + B) - \sin (A - B)$
Here $A = {22^ \circ },B = {38^ \circ }$
$\Rightarrow 2\sin {22^ \circ }\cos {38^ \circ } = \sin ({22^ \circ } + {38^ \circ }) + \sin ({22^ \circ } - {38^ \circ })$and
$2\cos {22^ \circ }\sin {38^ \circ } = \sin ({22^ \circ } + {38^ \circ }) - \sin ({22^ \circ } - {38^ \circ })$
Substitute the values obtained in RHS of equation (1)
$\Rightarrow \sin {22^ \circ }\cos {38^ \circ } + \cos {22^ \circ }\sin {38^ \circ } = \dfrac{1}{2}\left[ {\sin ({{22}^ \circ } + {{38}^ \circ }) + \sin ({{22}^ \circ } - {{38}^ \circ }) + \sin ({{22}^ \circ } + {{38}^ \circ }) - \sin ({{22}^ \circ } - {{38}^ \circ })} \right]$
Cancel terms having same magnitudes and opposite signs
$\Rightarrow \sin {22^ \circ }\cos {38^ \circ } + \cos {22^ \circ }\sin {38^ \circ } = \dfrac{1}{2}\left[ {\sin \left( {{{22}^ \circ } + {{38}^ \circ }} \right) + \sin \left( {{{22}^ \circ } + {{38}^ \circ }} \right)} \right]$
Add like terms in RHS of the equation
$\Rightarrow \sin {22^ \circ }\cos {38^ \circ } + \cos {22^ \circ }\sin {38^ \circ } = \dfrac{1}{2}\left[ {2\sin ({{22}^ \circ } + {{38}^ \circ })} \right]$
Add the angles in the bracket in RHS of the equation
$\Rightarrow \sin {22^ \circ }\cos {38^ \circ } + \cos {22^ \circ }\sin {38^ \circ } = \dfrac{1}{2}\left[ {2\sin {{60}^ \circ }} \right]$
Cancel same factors from numerator and denominator in RHS of the equation
$\Rightarrow \sin {22^ \circ }\cos {38^ \circ } + \cos {22^ \circ }\sin {38^ \circ } = \sin {60^ \circ }$
We know the value of $\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$
$\Rightarrow \sin {22^ \circ }\cos {38^ \circ } + \cos {22^ \circ }\sin {38^ \circ } = \dfrac{{\sqrt 3 }}{2}$

$\therefore$Option C is correct.

Note: Students many times try to find the value of the functions from the calculator and directly substitute those values in the equation. Students are advised not to proceed in this way as the value can be found very easily using trigonometric formulas. Many students don’t remember the value of sine function directly; they are advised to take the help of the table which gives us the values of some trigonometric functions at common angles like ${0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ }$ is

ANGLEFUNCTION	${0^ \circ }$	${30^ \circ }$	${45^ \circ }$	${60^ \circ }$	${90^ \circ }$
Sin	0	$\dfrac{1}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{{\sqrt 3 }}{2}$	1
Cos	1	$\dfrac{{\sqrt 3 }}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{1}{2}$	0
Tan	0	$\dfrac{1}{{\sqrt 3 }}$	1	$\sqrt 3$	Not defined

Students can solve this question by carefully observing the given equation and comparing it to the general trigonometric formula of $\sin (A + B)$.
We are given $\sin {22^ \circ }\cos {38^ \circ } + \cos {22^ \circ }\sin {38^ \circ }$
We know the identity $\sin (A + B) = \sin A\cos B + \cos A\sin B$
If we put the values $A = {22^ \circ },B = {38^ \circ }$in the formula
$\Rightarrow \sin {22^ \circ }\cos {38^ \circ } + \cos {22^ \circ }\sin {38^ \circ } = \sin ({22^ \circ } + {38^ \circ })$
Add the angles in the bracket in RHS of the equation
$\Rightarrow \sin {22^ \circ }\cos {38^ \circ } + \cos {22^ \circ }\sin {38^ \circ } = \sin {60^ \circ }$
Substitute the value of $\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$
$\Rightarrow \sin {22^ \circ }\cos {38^ \circ } + \cos {22^ \circ }\sin {38^ \circ } = \dfrac{{\sqrt 3 }}{2}$
$\therefore$Option C is correct.