# more trig questions

• Jan 7th 2010, 02:49 PM
Paymemoney
more trig questions
Hi
I am having trouble solving the following:

1)show that $\frac{1+sin{2\theta}+cos{2\theta}}{1+sin{2\theta}-cos{2\theta}}=cot{\theta}$

2)Given $t=tan{\frac{x}{2}}$, prove that $sin(x)=\frac{2t}{1+t^2}$
Hence solve the equation $2sin(x)-cos(x)=1$

P.S
• Jan 7th 2010, 03:07 PM
Prove It
Quote:

Originally Posted by Paymemoney
Hi
I am having trouble solving the following:

1)show that $\frac{1+sin{2\theta}+cos{2\theta}}{1+sin{2\theta}-cos{2\theta}}=cot{\theta}$

2)Given $t=tan{\frac{x}{2}}$, prove that $sin(x)=\frac{2t}{1+t^2}$
Hence solve the equation $2sin(x)-cos(x)=1$

P.S

1) $\frac{1 + \sin{2\theta} + \cos{2\theta}}{1 + \sin{2\theta} - \cos{2\theta}} = \frac{1 + 2\sin{\theta}\cos{\theta} + \cos^2{\theta} - \sin^2{\theta}}{1 + 2\sin{\theta}\cos{\theta} - (\cos^2{\theta} - \sin^2{\theta})}$

$= \frac{1 + 2\sin{\theta}\cos{\theta} + 2\cos^2{\theta} - 1}{1 + 2\sin{\theta}\cos{\theta} - (1 - 2\sin^2{\theta})}$

$= \frac{2\cos{\theta}(\sin{\theta} + \cos{\theta})}{2\sin{\theta}(\sin{\theta} + \cos{\theta})}$

$= \frac{\cos{\theta}}{\sin{\theta}}$

$= \cot{\theta}$.
• Jan 7th 2010, 03:12 PM
Prove It
Quote:

Originally Posted by Paymemoney
Hi
I am having trouble solving the following:

1)show that $\frac{1+sin{2\theta}+cos{2\theta}}{1+sin{2\theta}-cos{2\theta}}=cot{\theta}$

2)Given $t=tan{\frac{x}{2}}$, prove that $sin(x)=\frac{2t}{1+t^2}$
Hence solve the equation $2sin(x)-cos(x)=1$

P.S

If $\sin{2\theta} = 2\sin{\theta}\cos{\theta}$

$= 2\left(\frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}\right)\left(\frac{1}{\sqrt{1 + \tan^2{\theta}}}\right)$

$= \frac{2\tan{\theta}}{1 + \tan^2{\theta}}$.

Now let $\theta = \frac{x}{2}$.

Now to solve $2\sin{x} - \cos{x} = 1$

$2\sin{x} - \sqrt{1 - \sin^2{x}} = 1$

$\frac{4t}{1 + t^2} - \sqrt{1 - \left(\frac{2t}{1 + t^2}\right)^2} = 1$

$\frac{4t}{1 + t^2} - \sqrt{\frac{(1 + t^2)^2 - 4t^2}{(1 + t^2)^2}} = 1$

$\frac{4t}{1 + t^2} - \sqrt{\frac{1 + 2t^2 + t^4 - 4t^2}{(1 + t^2)^2}} = 1$

$\frac{4t}{1 + t^2} - \sqrt{\frac{1 - 2t^2 + t^4}{(1 + t^2)^2}} = 1$

$\frac{4t}{1 + t^2} - \sqrt{\frac{(1 - t^2)^2}{(1 + t^2)^2}} = 1$

$\frac{4t}{1 + t^2} - \frac{1 - t^2}{1 + t^2} = 1$

$\frac{t^2 + 4t - 1}{1 + t^2} = 1$

$t^2 + 4t - 1 = 1 + t^2$

$4t = 2$

$t = \frac{1}{2}$.

Now let $t = \tan{\frac{x}{2}}$ and solve for $x$.
• Jan 7th 2010, 03:22 PM
Paymemoney
can you explain to me how you got this $
2\cos^2{\theta} - 1$
and $1 - 2\sin^2{\theta}$
• Jan 7th 2010, 03:27 PM
Prove It
The Pythagorean Identity.

$\sin^2{\theta} + \cos^2{\theta} = 1$

So $\sin^2{\theta} = 1 - \cos^2{\theta}$ and $\cos^2{\theta} = 1 - \sin^2{\theta}$.
• Jan 7th 2010, 03:28 PM
Paymemoney
ok thanks i understand now.
• Jan 7th 2010, 03:45 PM
Paymemoney
can you explain to me how you got this? $
= 2\left(\frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}\right)\left(\frac{1}{\sqrt{1 + \tan^2{\theta}}}\right)
$
• Jan 7th 2010, 03:56 PM
Prove It
$\sin{\theta} = \frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}$ and $\cos{\theta} = \frac{1}{\sqrt{1 + \tan^2{\theta}}}$.

These are also easily proven using the Pythagorean Identity.