Question

Which of the following is correct?
a.$\sin {{1}^{\circ }}<\sin 1$
b.$\sin {{1}^{\circ }}>\sin 1$
c.$\sin {{1}^{\circ }}=\sin 1$
d.$\sin {{2}^{\circ }}=\sin 2$

Answer 1

Hint: Here $\sin 1$, mentions the angle in radians. Thus with the help of a circular sector, establish the relation between degree and radius. Thus find the value of $\sin {{1}^{\circ }}$ and $\sin 1$. Compare them and find who is greatest with the same relations compare $\sin {{2}^{\circ }}$ and $\sin 2$.

Complete step-by-step answer:

First let us compare between $\sin {{1}^{\circ }}$ and $\sin 1$.
Hence, $\sin 1$ means the sine of 1 radius. It can be represented by placing C in power of that angle, i.e. ${{1}^{C}}$.
We can define one radian as the angle in degrees for which the radius of a circular sector is equal to its length. We told that,
radius = length of a circular sector.
We now that length of circular sector $=\dfrac{\theta }{{{360}^{\circ }}}\times 2\pi r$
Thus we can say that,
$r=\dfrac{\theta }{{{360}^{\circ }}}\times 2\pi r$
Thus cancelling like terms we get,
$\theta =\dfrac{{{180}^{\circ }}}{\pi }$, which is 1 radian.
The approximate value of 1 radian is i.e. ${{1}^{C}}={{57.3}^{\circ }}$
We know that sine is an increasing function in the first quadrant.

& \therefore \sin {{1}^{\circ }}=0.017 \\\ & \sin {{1}^{C}}=0.841 \\\ \end{aligned}$$ Hence from the above we can see the value of sine in degrees and radians, which is correct to three decimals. Here, $$\sin {{1}^{\circ }}$$ is greater than $$\sin {{1}^{\circ }}$$ i.e. $$\sin {{1}^{\circ }}<\sin 1$$. $$\therefore $$ Thus from the options given we can say that, $$\sin {{1}^{\circ }}<\sin 1$$. We got, $${{1}^{C}}={{57.3}^{\circ }}$$ $$\therefore {{2}^{C}}=2\times {{57.3}^{\circ }}={{114.6}^{\circ }}$$ Thus the value of $$\sin {{2}^{\circ }}=0.0348$$ $$\sin {{2}^{C}}=0.909$$ By comparing the options we can say that only $$\sin {{1}^{\circ }}<\sin 1$$ is correct. $$\therefore $$ Option (a) is the correct answer. Note: If we compare between $$\sin {{1}^{\circ }}$$ and $$\sin {{2}^{\circ }}$$ degree, we get that $$\sin {{2}^{\circ }}$$ is greater than $$\sin {{1}^{\circ }}$$. $$\sin {{1}^{\circ }}<\sin {{2}^{\circ }}$$ i.e. 0.017 < 0.0348, similarly comparing between $$\sin 1$$ and $$\sin 2$$ radius, we can say that $$\sin 2$$ is greater than $$\sin 1$$. $$\therefore \sin 1<\sin 2$$ i.e. 0.841 < 0.909