# Double-Angle Formulas

• Nov 2nd 2009, 01:49 PM
l flipboi l
Double-Angle Formulas
Hello,

Find Sin2x, Cos 2x, and Tan2x for:

1. tan x= -4/3 quadrant II
2. csc x = 4, tan x<0

how do I start the problem?

for 1, is Sin -4/1 and Cos 3/1
and for 2 is sin 1/4?

Thanks
• Nov 3rd 2009, 07:01 AM
Hello l flipboi l
Quote:

Originally Posted by l flipboi l
Hello,

Find Sin2x, Cos 2x, and Tan2x for:

1. tan x= -4/3 quadrant II
2. csc x = 4, tan x<0

how do I start the problem?

for 1, is Sin -4/1 and Cos 3/1
and for 2 is sin 1/4?

Thanks

Are you familiar with this diagram that tells you when the various trig functions are positive?
$\displaystyle \begin{array}{c|c}S & A\\ \hline T & C \end{array}$
(Remember ACTS starting in QI and going clockwise.)

1. In QII, only sine is positive. So if $\displaystyle \tan x = -\frac43$ and $\displaystyle x$ is in QII, $\displaystyle \sin x = \frac45$ and $\displaystyle \cos x = \frac35$ (using Pythagoras). You should be able to find the sine, cosine and tangent of $\displaystyle 2x$ using the standard double-angle formulae.

2. You are right: $\displaystyle \csc x = \frac{1}{\sin x}= 4 \Rightarrow \sin x = \frac14$. Since sine is positive, we are either in QI or QII. So the fact that $\displaystyle \tan x < 0$ means that $\displaystyle x$ must be in QII. Using Pythagoras, then:
$\displaystyle \tan x = -\frac{1}{\sqrt{15}}$ and $\displaystyle \cos x = -\frac{\sqrt{15}}{4}$
Again, the standard double-angle formula will give you the results you need.

• Nov 3rd 2009, 10:15 AM
l flipboi l
Quote:

Hello l flipboi l
Are you familiar with this diagram that tells you when the various trig functions are positive?
$\displaystyle \begin{array}{c|c}S & A\\ \hline T & C \end{array}$
(Remember ACTS starting in QI and going clockwise.)

1. In QII, only sine is positive. So if $\displaystyle \tan x = -\frac43$ and $\displaystyle x$ is in QII, $\displaystyle \sin x = \frac45$ and $\displaystyle \cos x = \frac35$ (using Pythagoras). You should be able to find the sine, cosine and tangent of $\displaystyle 2x$ using the standard double-angle formulae.

2. You are right: $\displaystyle \csc x = \frac{1}{\sin x}= 4 \Rightarrow \sin x = \frac14$. Since sine is positive, we are either in QI or QII. So the fact that $\displaystyle \tan x < 0$ means that $\displaystyle x$ must be in QII. Using Pythagoras, then:
$\displaystyle \tan x = -\frac{1}{\sqrt{15}}$ and $\displaystyle \cos x = -\frac{\sqrt{15}}{4}$
Again, the standard double-angle formula will give you the results you need.