Double Angle Formulas

vanbeersj

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Aug 6, 2008
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I have the following question. I always have difficulties getting started with word problems....

The cross section of a radio wave reflector is defined by x = cos2b, y=sinb. Find the relation between x and y by eliminating b.

I'm not sure where to start.
 
Solve x=cos(2b)\displaystyle x=cos(2b) for b and sub into y=sin(b)\displaystyle y=sin(b)

y will then be in terms of x.
 
Hello, vanbeersj!

The cross section of a radio wave reflector is defined by:   x=cos2θ[1]y=sinθ[2]\displaystyle \text{The cross section of a radio wave reflector is defined by: }\;\begin{array}{cccc} x &=& \cos2\theta & [1] \\ y&=&\sin\theta & [2]\end{array}

Find the relation between x and y by eliminating θ.\displaystyle \text{Find the relation between }x\text{ and }y\text{ by eliminating }\theta.

We know that:   sin2 ⁣θ+cos2 ⁣θ=1\displaystyle \text{We know that: }\;\sin^2\!\theta + \cos^2\!\theta \:=\:1

Substitute [2]:   y2+cos2 ⁣θ=1cos2 ⁣θ=1y2cosθ=1y2    [3]\displaystyle \text{Substitute [2]: }\;y^2 + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \cos^2\!\theta \:=\:1-y^2 \quad\Rightarrow\quad \cos\theta \:=\:\sqrt{1-y^2}\;\;[3]


From [1], we have:   x  =  cos2θx  =  cos2 ⁣θsin2 ⁣θ\displaystyle \text{From [1], we have: }\;x \;=\;\cos2\theta \quad\Rightarrow\quad x \;=\;\cos^2\!\theta - \sin^2\!\theta

Substitute [2] and [3]:   x  =  (1y2)2(y)2  =  1y2y2\displaystyle \text{Substitute [2] and [3]: }\;x \;=\;(\sqrt{1-y^2})^2 - (y)^2 \;=\;1-y^2 - y^2

Therefore:   x  =  12y2\displaystyle \text{Therefore: }\;\boxed{x \;=\;1-2y^2}

 
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