Double arc trigonometric functions

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In the study of trigonometric functions, there are often problems involving double arches. Therefore, knowing the specific formulas of the sine, cosine It is tangent this type of arc is fundamental in simplifying many calculations.

Consider any arc of measure $\dpi{120} \alpha$ , the double arc is the arc of measure $\dpi{120} 2\alpha$ . In this way, we want to obtain sine formulas of $\dpi{120} 2\alpha$ , cosine of $\dpi{120} 2\alpha$ and tangent of $\dpi{120} 2\alpha$ .

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These formulas can be obtained from the two-arc addition formulas:

$\dpi{120} \mathbf{sen(\boldsymbol{\alpha + \beta}) sin\, \boldsymbol{\alpha} \cdot cos\, \boldsymbol{\beta} + sin\, \boldsymbol{\beta} \cdot cos\, \boldsymbol{\alpha}}$

$\dpi{120} \mathbf{cos(\boldsymbol{\alpha + \beta}) cos\, \boldsymbol{\alpha} \cdot cos\, \boldsymbol{\beta} - sen\, \boldsymbol{\beta} \cdot sen\, \boldsymbol{\alpha}}$

$\dpi{120} \mathbf{tan(\boldsymbol{\alpha + \beta}) \frac{sen(\boldsymbol{\alpha + \beta})}{cos(\boldsymbol{\alpha + \beta})} \frac{tan\, \boldsymbol{\alpha} + tan\, \boldsymbol{\beta}}{1 - tan\, \boldsymbol{\alpha} \cdot tan\, \boldsymbol{\beta}}}$

Remember the use of these formulas from an example where we obtain the sine of 75° from the sine and cosine of remarkable angles 30° and 45°.

$\dpi{120} \mathrm{sen (75^{\circ})sen (30^{\circ} + 45^{\circ}) sin\, 30^{\circ}\cdot cos\, 45^{ \circ} +sen\, 45^{\circ}\cdot cos\, 30^{\circ}}$

$\dpi{120} \mathrm{ \frac{1}{2}\cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt {3}}{2} }$

$\dpi{120} \mathrm{ \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} }$

$\dpi{120} \mathrm{ \frac{\sqrt{2} + \sqrt{6} }{4} }$

$\dpi{120} 0.96$

Now, let's see how the formulas of the double arc trigonometric functions.

Trigonometric functions of double arcs

Given an arc of measure $\dpi{120} \alpha$ , the double arc is the arc of measure $\dpi{120} 2\alpha$ . Since $\dpi{120} 2\alpha \alpha + \alpha$ , we can use the formulas for adding two arcs to get the formulas for the double arc.

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$\dpi{120} \mathbf{sen (2\boldsymbol{\alpha})sen(\boldsymbol{\alpha + \alpha}) sin\, \boldsymbol{\alpha} \cdot cos\, \boldsymbol{\alpha} + sen\, \boldsymbol{\alpha} \cdot cos\, \boldsymbol{\alpha}}$

$\dpi{120} \mathbf{ 2. (sen\, \boldsymbol{\alpha} \cdot cos\, \boldsymbol{\alpha}) }$

Therefore, the double arc sine is obtained by the following formula:

$\dpi{120} \mathbf{sen (2\boldsymbol{\alpha}) 2. (sen\, \boldsymbol{\alpha} \cdot cos\, \boldsymbol{\alpha}) }$

Now, see that:

$\dpi{120} \mathbf{cos (2\boldsymbol{\alpha})cos(\boldsymbol{\alpha + \alpha}) cos\, \boldsymbol{\alpha} \cdot cos\, \boldsymbol{\alpha} - sen\, \boldsymbol{\alpha} \cdot sen\, \boldsymbol{\alpha}}$

$\dpi{120} \mathbf{ cos^2\, \boldsymbol{\alpha} - sin^2\, \boldsymbol{\alpha} }$

Therefore, the double arc cosine is obtained by the following formula:

$\dpi{120} \mathbf{cos (2\boldsymbol{\alpha}) cos^2\, \boldsymbol{\alpha} - sin^2\, \boldsymbol{\alpha} }$

Regarding the tangent, we have:

$\dpi{120} \mathbf{tan (2\boldsymbol{\alpha})tan(\boldsymbol{\alpha + \alpha}) \frac{tan\, \boldsymbol{\alpha} + tan\, \boldsymbol{\alpha}}{1 - tan\, \boldsymbol{\alpha} \cdot tan\, \boldsymbol{\alpha}}}$