Double and half angle identites help

**hacker804** · July 31st 2012, 01:00 AM

Hi.I am having trouble solving the following quesions:
Q1
$\frac{cos\alpha-sin\alpha}{cos\alpha+sin\alpha}=sec2\alpha-tan2\alpha$
Q2
$\frac{sin2\alpha}{1+cos2\alpha}=tan2\alpha$
Q3
$\sqrt{\frac{1+sin\alpha}{1-sin\alpha}} = \frac{sin\frac{a}{2}+cos\frac{a}{2}}{sin\frac{a}{2 }-cos\frac{a}{2}}$

Thanks

**Prove It** · July 31st 2012, 01:38 AM

Originally Posted by hacker804

Hi.I am having trouble solving the following quesions:
Q1
$\frac{cos\alpha-sin\alpha}{cos\alpha+sin\alpha}=sec2\alpha-tan2\alpha$
Q2
$\frac{sin2\alpha}{1+cos2\alpha}=tan2\alpha$
Q3
$\sqrt{\frac{1+sin\alpha}{1-sin\alpha}} = \frac{sin\frac{a}{2}+cos\frac{a}{2}}{sin\frac{a}{2 }-cos\frac{a}{2}}$

Thanks

Q1

$\displaystyle \begin{align*} \sec{2\alpha} - \tan{2\alpha} &\equiv \frac{1}{\cos{2\alpha}} - \frac{\sin{2\alpha}}{\cos{2\alpha}} \\ &\equiv \frac{1 - \sin{2\alpha}}{\cos{2\alpha}} \\ &\equiv \frac{1 - 2\sin{\alpha}\cos{\alpha}}{\cos^2{\alpha} - \sin^2{\alpha}} \\ &\equiv \frac{\cos^2{\alpha} - 2\sin{\alpha}\cos{\alpha} + \sin^2{\alpha}}{\cos^2{\alpha} - \sin^2{\alpha}} \\ &\equiv \frac{\left(\cos{\alpha} - \sin{\alpha}\right)^2 }{\left(\cos{\alpha} - \sin{\alpha}\right)\left(\cos{\alpha} + \sin{\alpha}\right)} \\ &\equiv \frac{\cos{\alpha} - \sin{\alpha}}{\cos{\alpha} + \sin{\alpha}} \end{align*}$

**Prove It** · July 31st 2012, 01:50 AM

Originally Posted by hacker804

Hi.I am having trouble solving the following quesions:
Q1
$\frac{cos\alpha-sin\alpha}{cos\alpha+sin\alpha}=sec2\alpha-tan2\alpha$
Q2
$\frac{sin2\alpha}{1+cos2\alpha}=tan2\alpha$
Q3
$\sqrt{\frac{1+sin\alpha}{1-sin\alpha}} = \frac{sin\frac{a}{2}+cos\frac{a}{2}}{sin\frac{a}{2 }-cos\frac{a}{2}}$

Thanks

Q2 is wrong, it should be obvious that $\displaystyle \begin{align*} \tan{2\alpha} \equiv \frac{\sin{2\alpha}}{\cos{2\alpha}} \end{align*}$ , NOT $\displaystyle \begin{align*} \frac{\sin{2\alpha}}{1 + \cos{2\alpha}} \end{align*}$ .

Q3

$\displaystyle \begin{align*} \sqrt{\frac{1 + \sin{\alpha}}{1 - \sin{\alpha}}} &\equiv \sqrt{\frac{1 + \sin{\left[2\left(\frac{\alpha}{2}\right)\right]}}{1 - \sin{\left[2\left(\frac{\alpha}{2}\right)\right]}}} \\ &\equiv \sqrt{ \frac{ 1 + 2\sin{ \left( \frac{\alpha}{2} \right) } \cos{ \left( \frac{\alpha}{2} \right) } }{ 1 - 2\sin{ \left( \frac{\alpha}{2} \right) }\cos{ \left( \frac{\alpha}{2} \right) } } } \\ &\equiv \sqrt{ \frac{ \sin^2{ \left( \frac{\alpha}{2} \right) } + 2\sin{ \left( \frac{\alpha}{2} \right) } \cos{ \left( \frac{\alpha}{2} \right) } + \cos^2{ \left( \frac{\alpha}{2} \right) } }{ \sin^2{ \left( \frac{\alpha}{2} \right) } - 2\sin{ \left( \frac{\alpha}{2} \right) }\cos{ \left(\frac{\alpha}{2} \right)} + \cos^2{ \left(\frac{\alpha}{2}\right) } } } \end{align*}$

$\displaystyle \begin{align*} &\equiv \sqrt{ \frac{ \left[ \sin{\left(\frac{\alpha}{2}\right)} + \cos{\left(\frac{\alpha}{2}\right)} \right]^2 }{ \left[ \sin{\left(\frac{\alpha}{2}\right)} - \cos{\left(\frac{\alpha}{2}\right)} \right]^2 } } \\ &\equiv \frac{ \sin{\left(\frac{\alpha}{2}\right)} + \cos{\left(\frac{\alpha}{2}\right)} }{ \sin{\left(\frac{\alpha}{2}\right)} - \cos{\left(\frac{\alpha}{2}\right)} } \end{align*}$

**hacker804** · July 31st 2012, 03:31 AM

$\frac{cos^{2}\alpha-2sin\alpha cos\alpha+sin^{2}\alpha}{cos^{2}\alpha-sin^{2}\alpha}$

Explain this please.How did you complete the square?

**Prove It** · July 31st 2012, 03:39 AM

Originally Posted by hacker804

$\frac{cos^{2}\alpha-2sin\alpha cos\alpha+sin^{2}\alpha}{cos^{2}\alpha-sin^{2}\alpha}$

Explain this please.How did you complete the square?

Surely you know that $\displaystyle \begin{align*} \cos^2{\alpha} + \sin^2{\alpha} \equiv 1 \end{align*}$ ...

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