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2 years ago

(6x + 14) - (9x + 5)

Mathematics

1 answer:

Gre4nikov [31]2 years ago

6 0

Answer:

Step-by-step explanation:

(6x + 14) - (9x + 5) =

6x + 14 - 9x - 5 =

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Use a half-angle identity to find the exact value

Tatiana [17]

Given:

$\cos 15^{\circ}$

To find:

The exact value of cos 15°.

Solution:

$$\cos 15^{\circ}=\cos\frac{ 30^{\circ}}{2}$

Using half-angle identity:

$$\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos (x)}{2}}$

$$\cos \frac{30^{\circ}}{2}=\sqrt{\frac{1+\cos \left(30^{\circ}\right)}{2}}$

Using the trigonometric identity: $\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

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

Let us first solve the fraction in the numerator.

$$=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}$

Using fraction rule: $\frac{\frac{a}{b} }{c}=\frac{a}{b \cdot c}$

$$=\sqrt{\frac {2+\sqrt{3}}{4}}$

Apply radical rule: $\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

$$=\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}$

Using $\sqrt{4} =2$ :

$$=\frac{\sqrt{2+\sqrt{3}}}{2}$

$$\cos 15^\circ=\frac{\sqrt{2+\sqrt{3}}}{2}$

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