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STUDY GUIDE: Trigonometry |
INTRODUCTION |
| HERE'S a different approach |
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THE MAIN FUNCTIONS: SIN, COS & TAN |
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Consider the right triangle above
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WHY WE CARE |
If we know the angle of a right triangle and just one of the other sides we can find the lengths of the other sides. Check this out! Let's say that angle Θ above is actually 30 degrees. And we know that the length of side a = 10 meters. Using the definition above, we know that sin(30 degrees) = length of the side of the triangle opposite angle Θ divided by the length of the hypotenuse. Writing that using math terms, we get: sin(Θ) = a/c. If we plug in the angle we know (30 degrees) and the length of side a, we get: sine(30 degrees) = 10 meters/(length of hypotenuse) plugging in sine(30) into our calculator: we get a value of .5 Substituting above: .5 = 10 meters/(length of hypotenuse) substituting the letter c for the (length of the hypotenuse) we get .5 = 10 meters/c and solving for c we get .5c = 10 meters c= 10 meters/.5 c=20 meters In physics, we frequently talk about the magnitude of a vector. That magnitude is frequently the hypotenuse of a triangle and often symbolized as R. As such, we can often calculate the x component of that vector by Rcos(Θ) while the y component of the vector is calculated by Rsin(Θ). Where Θ is the angle between the vector and the x axis. |
INVERSE FUNCTIONS |
Once again the terms are somewhat more intimidating than the functions. Inverse trig functions simply relate known sides to angles. In physics we sometimes know the sides of a triangle but the angle between them is unknown. For example. In the example above. If we knew the side opposite angle Θ was 10, and we knew the hypotenuse was 20 then we would write the information we know like this: sin(Θ) = 10/20 sin(Θ) = .5 Our thinking would be: "There is an angle Θ where the ratio of the side opposite angle Θ to the hypotenuse is equal to .5" using the inv button on our calculator together with the sin button we type in: (.5)(inv)(sin) AND VOILA!!! Out pops 30 degrees!! |
TRIG IN PHYSICS |
| We almost *always* draw figures when we solve problems in physics. Throughout this term you will find references to angles and triangles and it is critical that you place those triangles in the correct quadrant of the coordinate system. |
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| Consider the cartesian coordinate plane system shown above with quadrants labeled as I, II, III and IV. |
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| X values are always positive in quadrants I and IV while Y values are always positive in quadrants I and II. |
| Triangles drawn with x components in those quadrants will behave exactly the same way. |