Q115 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
$\text { Value of } \cos 105^{\circ} \text { is }$
- A$\dfrac{(\sqrt{3}+1)}{2 \sqrt{2}}$
- B$-\dfrac{(1-\sqrt{3})}{2 \sqrt{2}}$
- C$-\dfrac{(\sqrt{3}+1)}{2 \sqrt{2}}$
- Correct answerD$-\dfrac{(\sqrt{3}-1)}{2 \sqrt{2}}$
Solution
The value of $\cos 105^{\circ}$ can be calculated using the cosine addition formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$.
We express $105^{\circ}$ as $60^{\circ} + 45^{\circ}$.
$\cos 105^{\circ} = \cos(60^{\circ} + 45^{\circ}) = \cos 60^{\circ} \cos 45^{\circ} - \sin 60^{\circ} \sin 45^{\circ}$.
Substituting the known trigonometric values:
$\cos 60^{\circ} = \dfrac{1}{2}$, $\cos 45^{\circ} = \dfrac{1}{\sqrt{2}}$, $\sin 60^{\circ} = \dfrac{\sqrt{3}}{2}$, $\sin 45^{\circ} = \dfrac{1}{\sqrt{2}}$.
$\cos 105^{\circ} = \left(\dfrac{1}{2}\right) \left(\dfrac{1}{\sqrt{2}}\right) - \left(\dfrac{\sqrt{3}}{2}\right) \left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{1}{2\sqrt{2}} - \dfrac{\sqrt{3}}{2\sqrt{2}} = \dfrac{1 - \sqrt{3}}{2\sqrt{2}}$.
This can be rewritten as $-\dfrac{\sqrt{3} - 1}{2\sqrt{2}}$.
Answer: $-\dfrac{(\sqrt{3}-1)}{2 \sqrt{2}}$
We express $105^{\circ}$ as $60^{\circ} + 45^{\circ}$.
$\cos 105^{\circ} = \cos(60^{\circ} + 45^{\circ}) = \cos 60^{\circ} \cos 45^{\circ} - \sin 60^{\circ} \sin 45^{\circ}$.
Substituting the known trigonometric values:
$\cos 60^{\circ} = \dfrac{1}{2}$, $\cos 45^{\circ} = \dfrac{1}{\sqrt{2}}$, $\sin 60^{\circ} = \dfrac{\sqrt{3}}{2}$, $\sin 45^{\circ} = \dfrac{1}{\sqrt{2}}$.
$\cos 105^{\circ} = \left(\dfrac{1}{2}\right) \left(\dfrac{1}{\sqrt{2}}\right) - \left(\dfrac{\sqrt{3}}{2}\right) \left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{1}{2\sqrt{2}} - \dfrac{\sqrt{3}}{2\sqrt{2}} = \dfrac{1 - \sqrt{3}}{2\sqrt{2}}$.
This can be rewritten as $-\dfrac{\sqrt{3} - 1}{2\sqrt{2}}$.
Answer: $-\dfrac{(\sqrt{3}-1)}{2 \sqrt{2}}$
Hence, the correct answer is option: 'D'.