Question

Which of the following numbers is/are rational?

• A. $\sin \left({15}^{°}\right)$
• B. $\cos \left({15}^{°}\right)$
• C. $\sin \left({15}^{°}\right)\cos \left({15}^{°}\right)$
• D. $\sin \left({15}^{°}\right)\cos \left({75}^{°}\right)$

Answer 1

The correct option is C

$\sin \left({15}^{°}\right)\cos \left({15}^{°}\right)$

Explanation for the correct answer:

Rational numbers are the numbers that are written in the form of fractions and the denominator should not be equal to zero.

$\sin \left({15}^{°}\right)=\sin \left({45}^{°}-{30}^{°}\right)$

$\sin \left({15}^{°}\right)=\sin \left({45}^{°}\right)\cos \left({30}^{°}\right)-\cos \left({45}^{°}\right)\sin \left({30}^{°}\right)$ $\left[\sin \left(\mathrm{A}-\mathrm{B}\right)=\sin \left(\mathrm{A}\right)\cos \left(\mathrm{B}\right)-\cos \left(\mathrm{A}\right)\sin \left(\mathrm{B}\right)\right]$

$\sin \left({15}^{°}\right)=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}$ [Putting in the values]

$\sin \left({15}^{°}\right)=\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}$

$\sin \left({15}^{°}\right)=\frac{\sqrt{3}-1}{2\sqrt{2}}$

Similarly, the value of $\cos \left({15}^{°}\right)=\frac{\sqrt{3}+1}{2\sqrt{2}}$.

$\sin \left({15}^{°}\right)\cos \left({15}^{°}\right)$

$=\frac{\sqrt{3}-1}{2\sqrt{2}}\times \frac{\sqrt{3}+1}{2\sqrt{2}}$

$=\frac{{\left(\sqrt{3}\right)}^2-1^2}{{\left(2\sqrt{2}\right)}^2}$ $\left[\because {\mathrm{a}}^2-{\mathrm{b}}^2=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right]$

$=\frac{3-1}{8}$

$=\frac{2}{8}$

$=\frac{1}{4}$, which is a rational number.

Explanation for incorrect options:

a) $\sin \left({15}^{°}\right)$

The value of $\sin \left({15}^{°}\right)$ is $\frac{\sqrt{3}-1}{2\sqrt{2}}$which is not rational.

b) $\cos \left({15}^{°}\right)$

The value of $\cos \left({15}^{°}\right)$ is $\frac{\sqrt{3}+1}{2\sqrt{2}}$which is not rational.

d)$\sin \left({15}^{°}\right)\cos \left({75}^{°}\right)$

$\cos \left({75}^{°}\right)=\cos \left({45}^{°}+{30}^{°}\right)$

$\cos \left({75}^{°}\right)=\cos \left({45}^{°}\right)\cos \left({30}^{°}\right)-\sin \left({45}^{°}\right)\sin \left({30}^{°}\right)\left[\because \cos \left(\mathrm{A}+\mathrm{B}\right)=\cos \left(\mathrm{A}\right)\cos \left(\mathrm{B}\right)-\sin \left(\mathrm{A}\right)\sin \left(\mathrm{B}\right)\right]$

$\cos \left({75}^{°}\right)=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}$ [Putting in the values]

$\cos \left({75}^{°}\right)=\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}$

$\cos \left({75}^{°}\right)=\frac{\sqrt{3}-1}{2\sqrt{2}}$

$\sin \left({15}^{°}\right)\cos \left({75}^{°}\right)$

$=\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)$

$=\frac{{\left(\sqrt{3}\right)}^2+1^2-2\left(\sqrt{3}\right)\left(1\right)}{\left(4\right)\left(2\right)};\left[{\left(\mathrm{a}-\mathrm{b}\right)}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}\right]$

$=\frac{3+1-2\sqrt{3}}{8}$

$=\frac{4-2\sqrt{3}}{8}$

$=\frac{2-\sqrt{3}}{4}$ ; [Dividing numerator and denominator by $2$]

The value of $\sin \left({15}^{°}\right)\cos \left({75}^{°}\right)$ is $\frac{2-\sqrt{3}}{4}$which is not rational.

Hence, the correct option is c) $\sin \left({15}^{°}\right)\cos \left({15}^{°}\right)$.