
What is the value of $\sin {{163}^{\circ }}\cos {{347}^{\circ }}+\sin {{73}^{\circ }}\sin {{167}^{\circ }}$?
A . 0
B . $\dfrac{1}{2}$
C . 1
D . None of these
Answer
162.9k+ views
Hint: In this question, we are given the expression $\sin {{163}^{\circ }}\cos {{347}^{\circ }}+\sin {{73}^{\circ }}\sin {{167}^{\circ }}$ and we have to find the value of expression. First we write the equations in the form of complimentary angles and then using the identity $\sin (A+B)$ and comparing and finding the values, we are able to get the value which is equal to $\sin {{163}^{\circ }}\cos {{347}^{\circ }}+\sin {{73}^{\circ }}\sin {{167}^{\circ }}$.
Formula Used:
Identity which is used to solve this question is as follow:-
$\sin (A+B)=\sin A\cos B+\cos A\sin B$
And complimentary angles are
$\sin ({{180}^{\circ }}-A)=\sin A$
And $\cos ({{360}^{\circ }}-A)=\cos A$
$\sin ({{90}^{\circ }}-A)=\cos A$
Complete Step- by- step Solution:
Given that $\sin {{163}^{\circ }}\cos {{347}^{\circ }}+\sin {{73}^{\circ }}\sin {{167}^{\circ }}$
We write the above equation as
$\sin ({{180}^{\circ }}-{{17}^{\circ }})\cos ({{360}^{\circ }}-{{13}^{\circ }})+\sin ({{90}^{\circ }}-{{17}^{\circ }})\sin ({{180}^{\circ }}-{{13}^{\circ }})$
We know that by the complimentary angle of trigonometric ratio the value of
$\sin ({{180}^{\circ }}-A)=\sin A$
And $\cos ({{360}^{\circ }}-A)=\cos A$
$\sin ({{90}^{\circ }}-A)=\cos A$
By simplifying the above equation, we get
$\sin {{17}^{\circ }}\cos {{13}^{\circ }}+\cos {{17}^{\circ }}\sin {{13}^{\circ }}$…………………………… (1)
We know the identity
$\sin (A+B)=\sin A\cos B+\cos A\sin B$…………………………… (2)
By comparing the equation (1) with equation (2), we get
$\sin {{17}^{\circ }}\cos {{13}^{\circ }}+\sin {{17}^{\circ }}\sin {{13}^{\circ }}$ = $\sin ({{17}^{\circ }}+{{13}^{\circ }})$
Which is equal to $\sin {{30}^{\circ }}$
And we know value of $\sin {{30}^{\circ }}$ = $\dfrac{1}{2}$
Hence, $\sin {{163}^{\circ }}\cos {{347}^{\circ }}+\sin {{73}^{\circ }}\sin {{167}^{\circ }}$ = $\dfrac{1}{2}$
Thus, Option ( B ) is correct.
Note: Whenever you get this type of question, important is to know about the complementary angles of trigonometric ratios and how to write the equation in the form of complementary angles. We know any two angle with their sum of ${{90}^{\circ }}$are called complementary. So the complement of any angle is that value which is obtained by subtracting it from ${{90}^{\circ }}$. According to trigonometric complimentary ratio theorem, trigonometric function of complementary angle is defined as another trigonometric function of the original angle. Then
$\sin ({{90}^{\circ }}-A)=\cos A$
$\cos ({{90}^{\circ }}-A)=\sin A$
$tan({{90}^{\circ }}-A)=\cot A$
$\cot ({{90}^{\circ }}-A)=\tan A$
$\sec ({{90}^{\circ }}-A)=\cos ecA$
$\cos ec({{90}^{\circ }}-A)=\sec A$
Formula Used:
Identity which is used to solve this question is as follow:-
$\sin (A+B)=\sin A\cos B+\cos A\sin B$
And complimentary angles are
$\sin ({{180}^{\circ }}-A)=\sin A$
And $\cos ({{360}^{\circ }}-A)=\cos A$
$\sin ({{90}^{\circ }}-A)=\cos A$
Complete Step- by- step Solution:
Given that $\sin {{163}^{\circ }}\cos {{347}^{\circ }}+\sin {{73}^{\circ }}\sin {{167}^{\circ }}$
We write the above equation as
$\sin ({{180}^{\circ }}-{{17}^{\circ }})\cos ({{360}^{\circ }}-{{13}^{\circ }})+\sin ({{90}^{\circ }}-{{17}^{\circ }})\sin ({{180}^{\circ }}-{{13}^{\circ }})$
We know that by the complimentary angle of trigonometric ratio the value of
$\sin ({{180}^{\circ }}-A)=\sin A$
And $\cos ({{360}^{\circ }}-A)=\cos A$
$\sin ({{90}^{\circ }}-A)=\cos A$
By simplifying the above equation, we get
$\sin {{17}^{\circ }}\cos {{13}^{\circ }}+\cos {{17}^{\circ }}\sin {{13}^{\circ }}$…………………………… (1)
We know the identity
$\sin (A+B)=\sin A\cos B+\cos A\sin B$…………………………… (2)
By comparing the equation (1) with equation (2), we get
$\sin {{17}^{\circ }}\cos {{13}^{\circ }}+\sin {{17}^{\circ }}\sin {{13}^{\circ }}$ = $\sin ({{17}^{\circ }}+{{13}^{\circ }})$
Which is equal to $\sin {{30}^{\circ }}$
And we know value of $\sin {{30}^{\circ }}$ = $\dfrac{1}{2}$
Hence, $\sin {{163}^{\circ }}\cos {{347}^{\circ }}+\sin {{73}^{\circ }}\sin {{167}^{\circ }}$ = $\dfrac{1}{2}$
Thus, Option ( B ) is correct.
Note: Whenever you get this type of question, important is to know about the complementary angles of trigonometric ratios and how to write the equation in the form of complementary angles. We know any two angle with their sum of ${{90}^{\circ }}$are called complementary. So the complement of any angle is that value which is obtained by subtracting it from ${{90}^{\circ }}$. According to trigonometric complimentary ratio theorem, trigonometric function of complementary angle is defined as another trigonometric function of the original angle. Then
$\sin ({{90}^{\circ }}-A)=\cos A$
$\cos ({{90}^{\circ }}-A)=\sin A$
$tan({{90}^{\circ }}-A)=\cot A$
$\cot ({{90}^{\circ }}-A)=\tan A$
$\sec ({{90}^{\circ }}-A)=\cos ecA$
$\cos ec({{90}^{\circ }}-A)=\sec A$
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