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Math Help - Double-Angle Formulas

  1. #1
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    Double-Angle Formulas

    Hello,

    Can someone please help me with these problems?

    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
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  2. #2
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    Hello l flipboi l
    Quote Originally Posted by l flipboi l View Post
    Hello,

    Can someone please help me with these problems?

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

    2. You are right: \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 \tan x < 0 means that x must be in QII. Using Pythagoras, then:
    \tan x = -\frac{1}{\sqrt{15}} and \cos x = -\frac{\sqrt{15}}{4}
    Again, the standard double-angle formula will give you the results you need.

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello l flipboi l
    Are you familiar with this diagram that tells you when the various trig functions are positive?
    \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 \tan x = -\frac43 and x is in QII, \sin x = \frac45 and \cos x = \frac35 (using Pythagoras). You should be able to find the sine, cosine and tangent of 2x using the standard double-angle formulae.

    2. You are right: \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 \tan x < 0 means that x must be in QII. Using Pythagoras, then:
    \tan x = -\frac{1}{\sqrt{15}} and \cos x = -\frac{\sqrt{15}}{4}
    Again, the standard double-angle formula will give you the results you need.

    Grandad
    Thanks Grandad!
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