Respuesta :

Danutz98

[tex]\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 6 +\sqrt 2}{4}[/tex]

kudzordzifrancis

Answer:

[tex]\cos (15)=\frac{\sqrt{6}+\sqrt{2}}{4}[/tex]

Step-by-step explanation:

We want to find the exact value of

[tex]\cos 15\degree[/tex]

We rewrite this expression using compound angles  

[tex]\cos (45\degree-30\degree)[/tex]

Recall that:

[tex]\cos (A-B)=\cos A\cos B+\sin A\sin B[/tex]

We apply this property to obtain:

[tex]\cos (45-30)=\cos 45\cos 30+\sin 45\sin 30[/tex]

[tex]\cos (15)=\frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2} \times \frac{1}{2}[/tex]

[tex]\cos (15)=\frac{\sqrt{6}+\sqrt{2}}{4}[/tex]

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