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Linearising a cosine curve

  • Thread starter Maluras
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  • #1
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Homework Statement



For my physics EEI, I have developed the formula: g-forces=√(391.88-337.12 cosθ)/9.8

I need to linearise the graph into the form y=mx+c.

I'm not sure where just the angle is the independent or cos of the angle.

Homework Equations



y=k√(x) can be graphed as y vs. √(x) to linearise

The Attempt at a Solution



The square root of a negative is undefined, therefore √(cosθ) is unable to be done for all values
 

Answers and Replies

  • #2
SammyS
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Homework Statement



For my physics EEI, I have developed the formula: g-forces=√(391.88-337.12 cosθ)/9.8

I need to linearise the graph into the form y=mx+c.

I'm not sure where just the angle is the independent or cos of the angle.

Homework Equations



y=k√(x) can be graphed as y vs. √(x) to linearise

The Attempt at a Solution



The square root of a negative is undefined, therefore √(cosθ) is unable to be done for all values
Hello Maluras. Welcome to PF !

Is (391.88-337.12 cosθ ) ever negative?

-1 ≤ cos(θ) ≤ 1

Multiply by -337.12 then add 391.88 .

What's your conclusion regarding the range of (391.88-337.12 cosθ ) ?
 

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