Help please! (double angle formulas)

S2Krazy · #1 $Old$ October 30th, 2009, 11:04 PM

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

Grandad · #2 $Old$ October 31st, 2009, 10:00 AM

Hello S2Krazy

Quote:

Originally Posted by S2Krazy $View Post$

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:

$\tan2A = \frac{2\tan A}{1-\tan^2A}$ and $\cos2A= 2\cos^2A-1$

When the various trig functions are positive. Here's the diagram:

$\begin{array}{c|c}S & A\\ \hline T & C \end{array}$

(Remember ACTS starting in QI and going clockwise.)

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

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

Using $\tan2A = \frac{2\tan A}{1-\tan^2A}$ :

$\tan 2A = \frac{2\times \frac{5}{12}}{1-(\frac{5}{12})^2} = ... = \frac{120}{119}$

For # 4, x is in QIV, so only cosine is positive; sine and tangent are negative. So $\cos x = \frac{2}{\sqrt5}\Rightarrow\sin x = - \frac{1}{\sqrt5}$ (Pythagoras) $\Rightarrow \tan x = -\frac12$

$\Rightarrow \tan2x=...=-\frac43$ (You can check this out.)

$\Rightarrow \cot2x = -\frac34$

For #5, all functions are positive in QI. So $\cos\theta = \frac{12}{13}$ (Pythagoras)

$\Rightarrow \cos2\theta = 2\cos^2\theta - 1 = ... = \frac{119}{169}$ (Check my working)

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

Grandad

Soroban · #3 $Old$ October 31st, 2009, 10:24 AM

Hello, S2Krazy!

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

Quote:

Let $\sin A = -\frac{5}{13}$ .with $A$ in Q3. . Find $\tan(2A)$

We need the identity: . $\tan(2A) \:=\:\frac{2\tan A}{1 - \tan^2\!A}$

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

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

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

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

Quote:

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

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

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

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

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

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

Quote:

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

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

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

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

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

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