How Can You Linearise a Cosine Curve for g-forces in Physics?

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The discussion revolves around linearizing the formula for g-forces, expressed as g-forces=√(391.88-337.12 cosθ)/9.8. The user seeks to convert this into the linear form y=mx+c but is uncertain whether to treat the angle or its cosine as the independent variable. A key point raised is the concern about the square root of negative values, specifically whether the expression (391.88-337.12 cosθ) can ever be negative. It is clarified that since the range of cos(θ) is between -1 and 1, the expression remains positive, allowing for successful linearization. Understanding this range is crucial for proceeding with the linearization process.
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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
 
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Maluras said:

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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