1. ## Help please! (double angle formulas)

Im so confused! im sure these r the correct answers!! but it keeps saying no! is it really worng or is is the programs error!? (some times hte program does make errors -_-) thxs for the help

2. Hello S2Krazy
Originally Posted by S2Krazy
Im so confused! im sure these r the correct answers!! but it keeps saying no! is it really worng or is is the programs error!? (some times hte program does make errors -_-) thxs for the help
What you need to know:

• Double-angle formulae. These are the ones you need here:

$\displaystyle \tan2A = \frac{2\tan A}{1-\tan^2A}$ and $\displaystyle \cos2A= 2\cos^2A-1$
• When the various trig functions are positive. Here's the diagram:

$\displaystyle \begin{array}{c|c}S & A\\ \hline T & C \end{array}$
(Remember ACTS starting in QI and going clockwise.)

• $\displaystyle \cot A = \frac{1}{\tan A}$ and $\displaystyle \sec A = \frac{1}{\cos A}$

So for #3, A is in QIII, so only tangent is positive: sine and cosine are both negative. So if $\displaystyle \sin A = -\frac{5}{13}, \cos A = -\frac{12}{13}$ (Pythagoras) and $\displaystyle \tan A = \frac{5}{12}$.

Using $\displaystyle \tan2A = \frac{2\tan A}{1-\tan^2A}$:
$\displaystyle \tan 2A = \frac{2\times \frac{5}{12}}{1-(\frac{5}{12})^2} = ... = \frac{120}{119}$
For # 4, x is in QIV, so only cosine is positive; sine and tangent are negative. So $\displaystyle \cos x = \frac{2}{\sqrt5}\Rightarrow\sin x = - \frac{1}{\sqrt5}$ (Pythagoras) $\displaystyle \Rightarrow \tan x = -\frac12$
$\displaystyle \Rightarrow \tan2x=...=-\frac43$ (You can check this out.)

$\displaystyle \Rightarrow \cot2x = -\frac34$
For #5, all functions are positive in QI. So $\displaystyle \cos\theta = \frac{12}{13}$ (Pythagoras)
$\displaystyle \Rightarrow \cos2\theta = 2\cos^2\theta - 1 = ... = \frac{119}{169}$ (Check my working)

$\displaystyle \Rightarrow \sec2\theta = \frac{169}{119}$

3. Hello, S2Krazy!

Sorry, you made errors on all of them . . .

Let $\displaystyle \sin A = -\frac{5}{13}$ .with $\displaystyle A$ in Q3. . Find $\displaystyle \tan(2A)$
We need the identity: .$\displaystyle \tan(2A) \:=\:\frac{2\tan A}{1 - \tan^2\!A}$

We have: .$\displaystyle \sin A \:=\:-\frac{5}{13} \;=\;\frac{opp}{hyp}$

So:. $\displaystyle opp = -5,\;hyp = 13$
. . Pythagorus says: .$\displaystyle adj = \pm 12$
Since $\displaystyle A$ is in Q3, $\displaystyle adj = -12$

. . Hence: .$\displaystyle \tan A \:=\:\frac{opp}{adj} \:=\:\frac{-5}{-12} \:=\:\frac{5}{12}$

Therefore: .$\displaystyle \tan2A \;=\;\frac{2(\frac{5}{12})}{1 - (\frac{5}{12})^2} \;=\;\frac{120}{119}$

Let $\displaystyle \cos x = \frac{2}{\sqrt{5}}$ .with $\displaystyle x$ in Q4. . Find $\displaystyle \cot(2x)$

We have: .$\displaystyle \cos x \:=\:\frac{2}{\sqrt{5}} \:=\:\frac{adj}{hyp}$

So: .$\displaystyle adj = 2,\;\;hyp = \sqrt{5}$
. . Pythagorus says: .$\displaystyle opp \,=\,\pm1$
Since $\displaystyle x$ is in Q4: .$\displaystyle opp = -1$

. . Hence: .$\displaystyle \tan x \:=\:\frac{opp}{adj} \:=\:-\frac{1}{2}$

Then: .$\displaystyle \tan(2x) \:=\:\frac{2(-\frac{1}{2})}{1 - (-\frac{1}{2})^2} \;=\;-\frac{4}{3}$

Therefore: .$\displaystyle \cot2x \:=\:-\frac{3}{4}$

Let $\displaystyle \tan\theta = \frac{5}{12}$ .with $\displaystyle \theta$ in Q1. . Find $\displaystyle \sec2\theta$

We have: .$\displaystyle \tan\theta \:=\:\frac{5}{12}\:=\:\frac{opp}{adj}$
So: .$\displaystyle opp = 5,\;adj = 12$
. . Pythagorus says: .$\displaystyle hyp = 13$

Hence: .$\displaystyle \cos\theta \:=\:\frac{adj}{hyp} \:=\:\frac{12}{13}$

Then: .$\displaystyle \cos2\theta \;=\;2\cos^2\theta - 1 \;=\;2\left(\frac{12}{13}\right)^2 - 1 \;=\;\frac{119}{169}$

Therefore: .$\displaystyle \sec2\theta \;=\;\frac{169}{119}$

Edit: Too fast for me, Grandad . . . Nice job, too!
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