1. ## more trig questions

Hi
I am having trouble solving the following:

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

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

P.S

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

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

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

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

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

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

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

$\displaystyle = \cot{\theta}$.

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

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

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

P.S
If $\displaystyle \sin{2\theta} = 2\sin{\theta}\cos{\theta}$

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle 4t = 2$

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

Now let $\displaystyle t = \tan{\frac{x}{2}}$ and solve for $\displaystyle x$.

4. can you explain to me how you got this$\displaystyle 2\cos^2{\theta} - 1$ and $\displaystyle 1 - 2\sin^2{\theta}$

5. The Pythagorean Identity.

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

So $\displaystyle \sin^2{\theta} = 1 - \cos^2{\theta}$ and $\displaystyle \cos^2{\theta} = 1 - \sin^2{\theta}$.

6. ok thanks i understand now.

7. can you explain to me how you got this?$\displaystyle = 2\left(\frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}\right)\left(\frac{1}{\sqrt{1 + \tan^2{\theta}}}\right)$

8. $\displaystyle \sin{\theta} = \frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}$ and $\displaystyle \cos{\theta} = \frac{1}{\sqrt{1 + \tan^2{\theta}}}$.

These are also easily proven using the Pythagorean Identity.