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Transcript

4.2.1 Trigonometric equations Welcome. If you want to know how long it takes to rise 34 meters while in the London Eye starting from the rightmost point, you have to solve an equation involving the sine of this duration. Typically, equations involving angles include trigonometric functions as sines and cosines. In this video I will show you how to solve such equations. Let us begin with the most basic kind of equation. Consider the angle x which satisfies sin(x) = frac12. You can look in the table of the sine function and see that pi/6 or 30 degrees is a solution. You can also find this solution from your calculator by using the sin inverse button. This angle must thus be pi/6 right? What else could it be? Well, consider the triangle on the slide. That is not a 30 degree angle, but its sine is also equal to 1/2. Let us look at the graph of sin(x). We add a horizontal line at y=1/2. The [click] intersections of this line with the graph of sin(x) are the solutions to the equation sin(x)=1/2. We see that there are many solutions. Pi/6 is just one solution. How can you find the other solutions, knowing pi/6 is a solution? You can use the properties of the graph of the sine function to find these others. The simplest property is that the sine is periodic, so the graph remains the same if we shift it a distance 2 pi horizontally. The red points of intersection are obtained by adding a multiple of 2pi to the original solution. So we have 13pi/6 and 25 pi/6, and to the left -‐11pi/6, and so on. The green point is obtained by using the symmetry of the graph of the sine upon reflecting in the line x=pi/2. This reflection sends x to pi-‐x, so it is pi-‐pi/6=5pi/6. The other green solutions can subsequently be found by adding multiples of 2pi to 5 pi/6. You can see in the graph that we have now found all solutions. We find infinitely many solutions to this equation. How can you describe all these solutions without using an infinite amount of ink? You can do this by giving a formula which describes all solutions. For example, [click] you can write x=pi/6 + 2pi k or x=5pi/6 + 2pi k, where k is an integer, either positive or negative. By plugging in all integer values of k you obtain all solutions of the equation. For k=0 you obtain pi/6 and 5pi/6, for k=1 you obtain 13pi/6 and 17 pi/6, etc. You have now seen how to solve a basic equation sin(x)= constant. A similar method works for cos(x) =constant: Find one solution using a table or calculator, and shift and reflect this solution to get the other ones. Other equations involving trigonometric functions first have to be simplified to this form. There are two basic ways to do this. The first method is to use a substitution. Consider the example cos(x)^2-‐cos(x)=3/4. We can replace cos(x) by a new variable p. Using this substitution the equation simplifies to p^2-‐p=3/4. You should make sure no stray x's remain in your equation: You need to have an equation involving only p's and no x's, as we do here. We can solve this equation for p as p=-‐1/2 or p=3/2. Thus we find that x satisfies cos(x)=-‐1/2 or 3/2. We now have two equations of the basic form. The second equation has no solutions as a cosine is always between -‐1 and 1. For the first equation we need to solve cos(x) = -‐1/2. Let us consider the graph and table of cos(x). There are no negative values in our table, but the cos^{-‐1} button of our calculator will still work. We can do better than using our calculator though. We can expand our basic table to x-‐values from 0 to pi, and we know that the cosine takes negative values from pi/2 to pi. Another symmetry of the graph shows that those values mirror the values of the cosine from 0 to pi/2, but with a minus sign. Thus we see that cos(pi/3)=1/2 implies that cos(pi-‐ pi/3) =-‐1/2. This gives our first answer. Reflecting 2 pi/3 in the y-‐axis gives the green solution -‐2pi/3. The remaining solutions are 2 pi shifts of plus minus 2pi/3. All solutions can thus be expressed as 2pi/3 + 2 pi k or -‐2pi /3 + 2pi k. Let us go back to our original example and use these solutions. As a final step we check that these are valid by plugging these in our equation, using that cos(x)=-‐1/2 for all of them. Warning! If you want to use a substitution a common mistake is to use a substitution that leaves some x's remaining in the resulting equation. For example if we use p=cos(x) in the equation x cos(x) = 1, to get x p =1 we have an x remaining. This must not happen: The final equation should only contain the new variable. In the previous example we have used a substitution to simplify our equation to the basic form cos(x)=c. A second method for simplifying is to use trigonometric identities. Consider the example cos(x) sin(x)=1/4. Recall that sin(2x)=2cos(x)sin(x). Thus we recognize the result of the doubling formula for the sine in our equation. Using this, the equation simplifies to sin(2x)=1/2. We can use that we already solved sin(x)=1/2 to obtain 2x=pi/6 or 5pi/6 + 2pi k. Dividing by 2 gives our final solution. In summary there are two important aspects to solving equations involving trigonometric functions: To solve a more complex equation you need to first simplify it to a basic equation. Use a substitution whenever this is possible. Remember that no x's are allowed in the resulting equation. Otherwise you should look for a simplifying trigonometric identity. We reviewed them in week 2. Be on the lookout for sin^2+cos^2 and the results of doubling formulas and addition formulas as you want to go from complicated expressions to easier one. Sometimes you may even need to combine both methods. To solve a basic equation sin(x)=c or cos(x) = c you first find a single solution by looking in the table or using your calculator, and subsequently use the graph of the function to obtain all other solutions. Try these techniques yourself today!