Question

TIf 1 degree=0.017 radian, then the approximate value of $\sin ({{46}^{\circ }})$ will be
a.0.7194
b. $\dfrac{0.017}{\sqrt{2}}$
c. $\dfrac{1.017}{\sqrt{2}}$
d.None of these

Answer 1

Hint:In this question, we are asked to find $\sin ({{46}^{\circ }})$ and are given the value of 1 degree in terms of radians. As the value of $\sin ({{45}^{\circ }})=\dfrac{1}{\sqrt{2}}$ , we can express $\sin ({{46}^{\circ }})$ in terms of $\sin ({{45}^{\circ }})$ by using the formula for sine of a sum of two angles. Thereafter, as the value of ${{1}^{\circ }}$ is very small, we can use the small angle approximation of sine and cosine functions, that is for $x\approx 0$ , $\sin (x)\approx x$ and $\cos (x)\approx 1$ where x is in radians and use the values from it in the equations to obtain our desired answer.

Complete step-by-step answer:
We know that the formula for the sine of a sum of angles is given by
$\sin (a+b)=\sin a\cos b+\cos a\sin b................(1.1)$
Therefore, taking $a={{45}^{\circ }}$ and $b={{1}^{\circ }}$ in equation (1.1), we obtain
$\sin \left( {{46}^{\circ }} \right)=\sin \left( {{45}^{\circ }} \right)\cos \left( {{1}^{\circ }} \right)+\cos \left( {{1}^{\circ }} \right)\sin \left( {{45}^{\circ }} \right)...............(1.2)$
However, we know that the value of $\sin \left( {{45}^{\circ }} \right)=\cos \left( {{45}^{\circ }} \right)=\dfrac{1}{\sqrt{2}}$ . Therefore, using these values in equation (1.2), we get
$\begin{aligned} & \sin \left( {{46}^{\circ }} \right)=\sin \left( {{45}^{\circ }} \right)\cos \left( {{1}^{\circ }} \right)+\cos \left( {{45}^{\circ }} \right)\sin \left( {{1}^{\circ }} \right) \\\ & =\dfrac{1}{\sqrt{2}}\times \cos \left( {{1}^{\circ }} \right)+\dfrac{1}{\sqrt{2}}\times \sin \left( {{1}^{\circ }} \right) \\\ & =\dfrac{1}{\sqrt{2}}\left( \cos \left( {{1}^{\circ }} \right)+\sin \left( {{1}^{\circ }} \right) \right).........................(1.3) \\\ \end{aligned}$
Now, we note that as ${{1}^{\circ }}\ll {{45}^{\circ }}$ and as ${{1}^{\circ }}$ is very close to ${{0}^{\circ }}$ , we can use the small angle approximation of sine and cosine which states that for $x\approx 0$ ,
$\begin{aligned} & \sin x\approx x \\\ & \cos x\approx \cos \left( {{0}^{\circ }} \right)\approx 1...........................(1.4) \\\ \end{aligned}$
Where x is given in radians. Therefore, taking the approximation in (1.4) with $x={{1}^{\circ }}$ , as ${{1}^{\circ }}$ is equivalent to 0.017 radians as given in the question, we get
$\begin{aligned} & \sin \left( {{1}^{\circ }} \right)\approx 0.017 \\\ & \cos \left( {{1}^{\circ }} \right)\approx 1............................(1.4a) \\\ \end{aligned}$
Using it in equation (1.3), we obtain
$\begin{aligned} & \sin \left( {{46}^{\circ }} \right)=\dfrac{1}{\sqrt{2}}\left( \cos \left( {{1}^{\circ }} \right)+\sin \left( {{1}^{\circ }} \right) \right) \\\ & \approx \dfrac{1}{\sqrt{2}}\left( 1+0.017 \right) \\\ & =\dfrac{1.017}{\sqrt{2}}.........................(1.5) \\\ \end{aligned}$
Which matches option (c) given in the question. Therefore, the required answer is option (c).

Note: We should note that we should convert the value of x in radians before using it in equation (1.4), therefore, we cannot write $\sin \left( {{1}^{\circ }} \right)\approx 1$ because here the angle is in degrees and not in radians. Also, we should be careful to use the correct sign between the terms in equation (1.1), there should be a positive sign in case of sine of a sum of angles whereas the sign is negative if we expand the cosine of a sum of angles.