1. ## trig cont.

if 0<a<pi/2

prove that tana = root all (1-cos2a)/(1+cos2a)
( the square root is above the entire right hand side)

and

if 4tan(a-b) = 3tana, prove that tanb = sin2a/(7+cos2a)

thanks for any help, i really appreciate it ! =D

2. Originally Posted by flyinhigh123
if 0<a<pi/2

prove that tana = root all (1-cos2a)/(1+cos2a)
( the square root is above the entire right hand side)
Hi

Use the formula $\displaystyle \cos 2a = 1 - 2 \sin^2a = 2 \cos^2a - 1$

3. Originally Posted by flyinhigh123
if 4tan(a-b) = 3tana, prove that tanb = sin2a/(7+cos2a)

thanks for any help, i really appreciate it ! =D
\displaystyle \begin{aligned} 4\tan(a - b) &= 3\tan a \\ 4\left(\frac{\tan a - \tan b}{1 + \tan a \tan b}\right) &= 3\tan a \\ 4(\tan a - \tan b) &= 3\tan a(1 + \tan a \tan b) \\ 4\tan a - 4\tan b &= 3\tan a + 3\tan^2 a \tan b \\ \tan a &= 3\tan^2 a \tan b + 4 \tan b \\ \tan a &= \tan b(3\tan^2 a + 4) \\ \tan b &= \frac{\tan a}{3\tan^2 a + 4} \end{aligned}

Now we need to prove that
$\displaystyle \frac{\tan a}{3\tan^2 a + 4} = \frac{\sin 2a}{7 + \cos 2a}$.

Starting from the left side:
$\displaystyle \frac{\tan a}{3\tan^2 a + 4}$

\displaystyle \begin{aligned} &= \frac{\tan a}{3\tan^2 a + 4} \cdot \frac{2\cos^2 a}{2\cos^2 a} \\ &= \frac{2\sin a \cos a}{6\sin^2 a + 8\cos^2 a} \\ &= \frac{\sin 2a}{6\sin^2 a + 6\cos^2 a + 2\cos^2 a} \end{aligned}

\displaystyle \begin{aligned} &= \frac{\sin 2a}{6 + 2\cos^2 a} \\ &= \frac{\sin 2a}{7 + 2\cos^2 a - 1} \\ &= \frac{\sin 2a}{7 + \cos 2a} \end{aligned}

01

4. Hello, flyinhigh123!

Prove: .$\displaystyle \tan A \:=\: \sqrt{\frac{1-\cos2A}{1+\cos2A}}$
We need these two identities: .$\displaystyle \begin{array}{ccccccc}\sin^2\theta &=& \dfrac{1-\cos2\theta}{2} & \Rightarrow & \sin\theta &=& \sqrt{\dfrac{1-\cos2\theta}{2}} \\ \\[-3mm] \cos^2\theta &=& \dfrac{1+\cos2\theta}{2} & \Rightarrow & \cos\theta &=& \sqrt{\dfrac{1+\cos2\theta}{2}} \end{array}$

We have: .$\displaystyle \tan A \;=\;\frac{\sin A}{\cos A} \;=\;\frac{\sqrt{\frac{1-\cos2A}{2}} }{\sqrt{\frac{1+\cos2A}{2}}} \;=\; \sqrt{\frac{\frac{1-\cos2A}{2}} {\frac{1+\cos2A}{2}}}$

. . Therefore: .$\displaystyle \boxed{\tan A \;=\;\sqrt{\frac{1-\cos2A}{1+\cos2A}}}$

If $\displaystyle 4\tan(A-B) \:=\: 3\tan A$, prove that: .$\displaystyle \tan B \:=\:\frac{\sin2A}{7+\cos2A}$
We are given: .$\displaystyle 4\tan(A-B) \:=\:3\tan A$

. . $\displaystyle 4\left(\frac{\tan A - \tan B}{1 + \tan A\tan B}\right) \;=\;3\tan A$

. . $\displaystyle 4(\tan A - \tan B) \;=\;3\tan A(1 + \tan A\tan B)$

. . $\displaystyle 4\tan A - 4\tan B \;=\;3\tan A + 3\tan^2\!A\tan B$

. . $\displaystyle 3\tan^2\!A\tan B + 4\tan B \;=\;\tan A$

. . $\displaystyle \tan B(3\tan^2\!A + 4) \;=\;\tan A$

. . $\displaystyle \tan B \;=\;\frac{\tan A}{3\tan^2\!A + 4} \;=\; \frac{\frac{\sin A}{\cos A}} {3\frac{\sin^2\!A}{\cos^2\!A} + 4}$

$\displaystyle \text{Multiply by }\frac{\cos^2\!A}{\cos^2\!A}\!:\quad\tan B \;=\; \frac{\sin A\cos A}{3\sin^2\!A + 4\cos^2\!A} \;=\;\frac{\overbrace{\sin A\cos A}^{\text{This is }\frac{1}{2}\sin2A}}{3(1-\cos^2\!A) + 4\cos^2\!A}$

. . . . $\displaystyle \tan B \;=\;\frac{\frac{1}{2}\sin2A}{3 + \cos^2\!A} \;=\;\frac{\frac{1}{2}\sin2A}{3 + \frac{1+\cos2A}{2}}$

Multiply by $\displaystyle \frac{2}{2}\!:\quad \tan B \;=\; \frac{\sin2A}{6 + 1 + \cos2A} \quad\Rightarrow\quad \boxed{\tan B \;=\;\frac{\sin2A}{7 + \cos2A}}$

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### squarerrot 1-cos2A/1 cos2A=tanA

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