What is the Value of Cos 15°? (Derivation & Formula)

• Fractional Value: The exact value of $\cos 15^\circ$ is $\frac{\sqrt{6} + \sqrt{2}}{4}$.
• Decimal Value: Approximately 0.9659.

To find $\cos 15^\circ$, we can write $15^\circ$ as $(45^\circ - 30^\circ)$. We then apply the trigonometric identity: $\cos(A - B) = \cos A \cos B + \sin A \sin B$

• Let $A = 45^\circ$ and $B = 30^\circ$
• $\cos(45^\circ - 30^\circ) = \cos 45^\circ \cdot \cos 30^\circ + \sin 45^\circ \cdot \sin 30^\circ$

Now, substitute the standard values:

• $\cos 45^\circ = \frac{1}{\sqrt{2}}$
• $\cos 30^\circ = \frac{\sqrt{3}}{2}$
• $\sin 45^\circ = \frac{1}{\sqrt{2}}$
• $\sin 30^\circ = \frac{1}{2}$

Plugging these in: $= (\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}} \cdot \frac{1}{2})$ $= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$ $= \frac{\sqrt{3} + 1}{2\sqrt{2}}$

To rationalize the denominator, multiply top and bottom by $\sqrt{2}$: $= \frac{(\sqrt{3} + 1) \cdot \sqrt{2}}{2\sqrt{2} \cdot \sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4}$.