Math / Trigonometry

QuestionDado $\cos \alpha=\frac{\sqrt{3}}{2}$ , calcular $\operatorname{Sen} 2 \alpha, \operatorname{Cos} 2 \alpha, \operatorname{Tg} 2 \alpha$

Studdy Solution

STEP 1

Suposiciones
1. Se nos da que $\cos \alpha = \frac{\sqrt{3}}{2}$ .
2. Queremos calcular $\operatorname{Sen} 2\alpha$ , $\operatorname{Cos} 2\alpha$ , y $\operatorname{Tg} 2\alpha$ .
3. Utilizaremos las identidades trigonométricas de ángulos dobles: - $\operatorname{Sen} 2\alpha = 2 \sin \alpha \cos \alpha$ - $\operatorname{Cos} 2\alpha = \cos^2 \alpha - \sin^2 \alpha$ - $\operatorname{Tg} 2\alpha = \frac{\operatorname{Sen} 2\alpha}{\operatorname{Cos} 2\alpha}$

STEP 2

Calcular $\sin \alpha$ usando la identidad trigonométrica $\sin^2 \alpha + \cos^2 \alpha = 1$ .
$\sin^2 \alpha = 1 - \cos^2 \alpha$

STEP 3

Sustituir $\cos \alpha = \frac{\sqrt{3}}{2}$ en la ecuación.
$\sin^2 \alpha = 1 - \left(\frac{\sqrt{3}}{2}\right)^2$

STEP 4

Calcular $\cos^2 \alpha$ .
$\cos^2 \alpha = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$

STEP 5

Calcular $\sin^2 \alpha$ .
$\sin^2 \alpha = 1 - \frac{3}{4} = \frac{1}{4}$

STEP 6

Determinar $\sin \alpha$ . Dado que $\sin^2 \alpha = \frac{1}{4}$ , entonces $\sin \alpha = \pm \frac{1}{2}$ .

STEP 7

Calcular $\operatorname{Sen} 2\alpha$ usando la identidad $\operatorname{Sen} 2\alpha = 2 \sin \alpha \cos \alpha$ .
$\operatorname{Sen} 2\alpha = 2 \left(\pm \frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right)$

STEP 8

Calcular $\operatorname{Sen} 2\alpha$ .
$\operatorname{Sen} 2\alpha = \pm \frac{\sqrt{3}}{2}$

STEP 9

Calcular $\operatorname{Cos} 2\alpha$ usando la identidad $\operatorname{Cos} 2\alpha = \cos^2 \alpha - \sin^2 \alpha$ .
$\operatorname{Cos} 2\alpha = \frac{3}{4} - \frac{1}{4}$

STEP 10

Calcular $\operatorname{Cos} 2\alpha$ .
$\operatorname{Cos} 2\alpha = \frac{2}{4} = \frac{1}{2}$

STEP 11

Calcular $\operatorname{Tg} 2\alpha$ usando la identidad $\operatorname{Tg} 2\alpha = \frac{\operatorname{Sen} 2\alpha}{\operatorname{Cos} 2\alpha}$ .
$\operatorname{Tg} 2\alpha = \frac{\pm \frac{\sqrt{3}}{2}}{\frac{1}{2}}$

STEP 12

Calcular $\operatorname{Tg} 2\alpha$ .
$\operatorname{Tg} 2\alpha = \pm \sqrt{3}$
La solución es:
- $\operatorname{Sen} 2\alpha = \pm \frac{\sqrt{3}}{2}$ - $\operatorname{Cos} 2\alpha = \frac{1}{2}$ - $\operatorname{Tg} 2\alpha = \pm \sqrt{3}$

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QuestionDado $\cos \alpha=\frac{\sqrt{3}}{2}$ , calcular $\operatorname{Sen} 2 \alpha, \operatorname{Cos} 2 \alpha, \operatorname{Tg} 2 \alpha$

Studdy Solution

STEP 1

STEP 2

Calcular $\sin \alpha$ usando la identidad trigonométrica $\sin^2 \alpha + \cos^2 \alpha = 1$ .
$\sin^2 \alpha = 1 - \cos^2 \alpha$

STEP 3

Sustituir $\cos \alpha = \frac{\sqrt{3}}{2}$ en la ecuación.
$\sin^2 \alpha = 1 - \left(\frac{\sqrt{3}}{2}\right)^2$

STEP 4

Calcular $\cos^2 \alpha$ .
$\cos^2 \alpha = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$

STEP 5

Calcular $\sin^2 \alpha$ .
$\sin^2 \alpha = 1 - \frac{3}{4} = \frac{1}{4}$

STEP 6

Determinar $\sin \alpha$ . Dado que $\sin^2 \alpha = \frac{1}{4}$ , entonces $\sin \alpha = \pm \frac{1}{2}$ .

STEP 7

Calcular $\operatorname{Sen} 2\alpha$ usando la identidad $\operatorname{Sen} 2\alpha = 2 \sin \alpha \cos \alpha$ .
$\operatorname{Sen} 2\alpha = 2 \left(\pm \frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right)$

STEP 8

Calcular $\operatorname{Sen} 2\alpha$ .
$\operatorname{Sen} 2\alpha = \pm \frac{\sqrt{3}}{2}$

STEP 9

Calcular $\operatorname{Cos} 2\alpha$ usando la identidad $\operatorname{Cos} 2\alpha = \cos^2 \alpha - \sin^2 \alpha$ .
$\operatorname{Cos} 2\alpha = \frac{3}{4} - \frac{1}{4}$

STEP 10

Calcular $\operatorname{Cos} 2\alpha$ .
$\operatorname{Cos} 2\alpha = \frac{2}{4} = \frac{1}{2}$

STEP 11

Calcular $\operatorname{Tg} 2\alpha$ usando la identidad $\operatorname{Tg} 2\alpha = \frac{\operatorname{Sen} 2\alpha}{\operatorname{Cos} 2\alpha}$ .
$\operatorname{Tg} 2\alpha = \frac{\pm \frac{\sqrt{3}}{2}}{\frac{1}{2}}$

STEP 12

Calcular $\operatorname{Tg} 2\alpha$ .
$\operatorname{Tg} 2\alpha = \pm \sqrt{3}$
La solución es:
- $\operatorname{Sen} 2\alpha = \pm \frac{\sqrt{3}}{2}$ - $\operatorname{Cos} 2\alpha = \frac{1}{2}$ - $\operatorname{Tg} 2\alpha = \pm \sqrt{3}$

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QuestionDado cos⁡α=32\cos \alpha=\frac{\sqrt{3}}{2}cosα=23​​, calcular Sen⁡2α,Cos⁡2α,Tg⁡2α\operatorname{Sen} 2 \alpha, \operatorname{Cos} 2 \alpha, \operatorname{Tg} 2 \alphaSen2α,Cos2α,Tg2α

Studdy Solution

STEP 1

STEP 2

Calcular sin⁡α\sin \alphasinα usando la identidad trigonométrica sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1sin2α+cos2α=1.sin⁡2α=1−cos⁡2α\sin^2 \alpha = 1 - \cos^2 \alphasin2α=1−cos2α

STEP 3

Sustituir cos⁡α=32\cos \alpha = \frac{\sqrt{3}}{2}cosα=23​​ en la ecuación.sin⁡2α=1−(32)2\sin^2 \alpha = 1 - \left(\frac{\sqrt{3}}{2}\right)^2sin2α=1−(23​​)2

STEP 4

Calcular cos⁡2α\cos^2 \alphacos2α.cos⁡2α=(32)2=34\cos^2 \alpha = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}cos2α=(23​​)2=43​

STEP 5

Calcular sin⁡2α\sin^2 \alphasin2α.sin⁡2α=1−34=14\sin^2 \alpha = 1 - \frac{3}{4} = \frac{1}{4}sin2α=1−43​=41​

STEP 6

Determinar sin⁡α\sin \alphasinα. Dado que sin⁡2α=14\sin^2 \alpha = \frac{1}{4}sin2α=41​, entonces sin⁡α=±12\sin \alpha = \pm \frac{1}{2}sinα=±21​.

STEP 7

STEP 8

Calcular Sen⁡2α\operatorname{Sen} 2\alphaSen2α.Sen⁡2α=±32\operatorname{Sen} 2\alpha = \pm \frac{\sqrt{3}}{2}Sen2α=±23​​

STEP 9

Calcular Cos⁡2α\operatorname{Cos} 2\alphaCos2α usando la identidad Cos⁡2α=cos⁡2α−sin⁡2α\operatorname{Cos} 2\alpha = \cos^2 \alpha - \sin^2 \alphaCos2α=cos2α−sin2α.Cos⁡2α=34−14\operatorname{Cos} 2\alpha = \frac{3}{4} - \frac{1}{4}Cos2α=43​−41​

STEP 10

Calcular Cos⁡2α\operatorname{Cos} 2\alphaCos2α.Cos⁡2α=24=12\operatorname{Cos} 2\alpha = \frac{2}{4} = \frac{1}{2}Cos2α=42​=21​

STEP 11

STEP 12

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QuestionDado $\cos \alpha=\frac{\sqrt{3}}{2}$ , calcular $\operatorname{Sen} 2 \alpha, \operatorname{Cos} 2 \alpha, \operatorname{Tg} 2 \alpha$

Calcular $\sin \alpha$ usando la identidad trigonométrica $\sin^2 \alpha + \cos^2 \alpha = 1$ .
$\sin^2 \alpha = 1 - \cos^2 \alpha$

Sustituir $\cos \alpha = \frac{\sqrt{3}}{2}$ en la ecuación.
$\sin^2 \alpha = 1 - \left(\frac{\sqrt{3}}{2}\right)^2$

Calcular $\cos^2 \alpha$ .
$\cos^2 \alpha = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$

Calcular $\sin^2 \alpha$ .
$\sin^2 \alpha = 1 - \frac{3}{4} = \frac{1}{4}$

Determinar $\sin \alpha$ . Dado que $\sin^2 \alpha = \frac{1}{4}$ , entonces $\sin \alpha = \pm \frac{1}{2}$ .

Calcular $\operatorname{Sen} 2\alpha$ .
$\operatorname{Sen} 2\alpha = \pm \frac{\sqrt{3}}{2}$

Calcular $\operatorname{Cos} 2\alpha$ usando la identidad $\operatorname{Cos} 2\alpha = \cos^2 \alpha - \sin^2 \alpha$ .
$\operatorname{Cos} 2\alpha = \frac{3}{4} - \frac{1}{4}$

Calcular $\operatorname{Cos} 2\alpha$ .
$\operatorname{Cos} 2\alpha = \frac{2}{4} = \frac{1}{2}$