Question

What is the exact value of

Answer 1

Answer:

4th option

Step-by-step explanation:

using the addition formula for cosine

cos(a - b) = cosacosb + sinasinb

note that 15° = (45 - 30)° , then

cos15°

= cos(45 - 30)°

= cos45°cos30° + sin45°sin30°

= ( $\frac{\sqrt{2} }{2}$ × $\frac{\sqrt{3} }{2}$ ) + ( $\frac{\sqrt{2} }{2}$ × $\frac{1}{2}$ )

= $\frac{\sqrt{6} }{4}$ + $\frac{\sqrt{2} }{4}$

= $\frac{\sqrt{6}+\sqrt{2} }{4}$

= $\frac{\sqrt{2}+\sqrt{6} }{4}$

Answer 2

Answer:

$\cos 15^{\circ}= \dfrac{\sqrt{2}+\sqrt{6}}{4}$

Step-by-step explanation:

Trig Identity:

$\cos (A-B)=\cos A \cos B + \sin A \sin B$

$\begin{aligned}\implies \cos 15^{\circ}=\cos (45^{\circ}-30^{\circ}) & =\cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ}\\\\& = \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2}\\\\& = \dfrac{\sqrt{2}\sqrt{3}}{2 \cdot 2}+\dfrac{\sqrt{2} \cdot 1}{2 \cdot 2}\\\\& = \dfrac{\sqrt{2\cdot 3}}{4}+\dfrac{\sqrt{2}}{4}\\\\& = \dfrac{\sqrt{6}+\sqrt{2}}{4}\\\\& = \dfrac{\sqrt{2}+\sqrt{6}}{4}\end{aligned}$