# Math Help - trigo problem no. 11

1. ## trigo problem no. 11

p180 ex8b q5
prove the following identities:

2. Just solve for $\frac x2$. So $y:=\frac x2$ then $\frac{\sin{2y}-\cos{y}}{1-\cos{2y}-\sin{y}}=\frac{\cos{y}}{\sin{y}}$. And solving that is quite straight forward...

3. Originally Posted by james_bond
Just solve for $\frac x2$. So $y:=\frac x2$ then $\frac{\sin{2y}-\cos{y}}{1-\cos{2y}-\sin{y}}=\frac{\cos{y}}{\sin{y}}$. And solving that is quite straight forward...
Just to clarify:

Replace $\frac{x}{2}$ with y. Note that this means you replace x with 2y.

Then you're required to prove the identity $\frac{\sin{2y}-\cos{y}}{1-\cos{2y}-\sin{y}}=\frac{\cos{y}}{\sin{y}}$.

And this is, if not quite straight forward, at least more straight forward.

4. OK, so step by step:
$\frac{\sin{2y}-\cos{y}}{1-\cos{2y}-\sin{y}}=\frac{\cos{y}}{\sin{y}}\Longleftrightarro w\frac{2\sin y\cos y-\cos y}{1-(\cos^2y-\sin^2y)-\sin y}=\frac{\cos y}{\sin y}\Longleftrightarrow$
$\Longleftrightarrow\frac{2\sin y\cos y-\cos y}{1-\cos^2y+\sin^2y-\sin y}=\frac{\cos y}{\sin y}\Longleftrightarrow\frac{2\sin y\cos y-\cos y}{\sin^2y+\sin^2y-\sin y}=\frac{\cos y}{\sin y}\Longleftrightarrow$ $\frac{2\sin y\cos y-\cos y}{2\sin^2y-\sin y}=\frac{\cos y}{\sin y}\Longleftrightarrow\frac{\cos y \left(2\sin y-1\right)}{\sin y \left(2\sin y-1\right)}=\frac{\cos y}{\sin y}\Longleftrightarrow 0=0$
which is true.

5. i think of another method, but seems slow

6. Hello, afeasfaerw23231233!

. . Incredibly, it produces the desired result!

Let $t = \tan\frac{x}{2}$

Then we have: . $\frac{\dfrac{2t}{1+t^2} - \dfrac{1}{1+t^2} }{1 - \dfrac{1-t^2}{1+t^2} - \dfrac{t}{1+t^2}}$ .
. . . This is not correct!

Do you know the basis for this strange substitution?
In case you don't, I'll explain it.

We let: . $t \:=\:\tan\frac{x}{2}$

So we have: . $\tan\frac{x}{2} \:=\:\frac{t}{1} \:=\:\frac{opp}{adj}$

That is, angle $\frac{x}{2}$ is in a right triangle with: $opp = t,\;adj = 1$
. . Using Pythagorus, we have: . $hyp = \sqrt{1+t^2}$

Then: . $\boxed{\sin\frac{x}{2} \:=\:\frac{t}{\sqrt{1+t^2}}},\quad\boxed{\cos\frac {x}{2} \:=\:\frac{1}{\sqrt{1+t^2}}}$

We also have: . $\sin x \;=\;2\sin\frac{x}{2}\cos\frac{x}{2}\;=\;2\frac{t} {\sqrt{1+t^2}}\frac{1}{\sqrt{1+t^2}}\quad\Rightarr ow\quad\boxed{ \sin x \;=\;\frac{2t}{1+t^2}}$

and: . $\cos x \;=\;\cos^2\!\frac{x}{2} \,- \,\sin^2\!\frac{x}{2} \;=\;\left(\frac{1}{\sqrt{1+t^2}}\right)^2 - \left(\frac{t}{\sqrt{1+t^2}}\right)^2\quad\Rightar row\quad\boxed{ \cos x \;=\;\frac{1-t^2}{1+t^2}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Back to the problem . . .

We have: . $\frac{\sin x - \cos\frac{x}{2}}{1 - \cos x - \sin\frac{x}{2}}$

Substitute: . $\frac{\dfrac{2t}{1+t^2} - {\color{red}\dfrac{1}{\sqrt{1+t^2}}}} {1 - \dfrac{1-t^2}{1+t^2} - {\color{red}\dfrac{t}{\sqrt{1+t^2}}}}$ . .

Multiply top and bottom by $(1+t^2)\sqrt{1+t^2}$

. . $\frac{2t\sqrt{1+t^2} - (1 + t^2)}{(1+t^2)\sqrt{1+t^2} - (1-t^2)\sqrt{1+t^2} - t(1+t^2)} \;=\;\frac{2t\sqrt{1+t^2} - 1 - t^2}{2t^2\sqrt{1+t^2} - t - t^3}$

Factor and reduce: . $\frac{(2t\sqrt{1+t^2} - 1 - t^2)}{t(2t\sqrt{1+t^2} - 1 - t^2)} \;=\;\frac{1}{t}$

Therefore, we have: . $\frac{1}{t} \;=\;\frac{1}{\tan\frac{x}{2}} \;=\;\cot\frac{x}{2}$ . . . . There!