Solutions to Homework Problems from Section 7.5 of Stewart

Transcription

1 Solutions to Homewk Problems from Section 7.5 of Stewart The solutions of the basic equations cos k, sin k, tan k are as follows:. If k is a number such that k, then the solutions of the equation cos k are arccos k n ( arccos k 36 n if we are using degrees).. If k is a number such that k, then the solutions of the equation sin k are arcsin k n arcsin k n. 3. If k is any real number, then the solutions of the equation tan k are arctan k n. It is actually not a good idea to try to memize the above three facts. It is better to rely on your knowledge of the unit circle.. To solve the equation cos, we write this equation as cos observe that the solutions are 3 n. 3. To solve the equation sin 3, we write this equation as observe that the solutions are sin 3 3 n 3 n. 5. To solve 4cos, we write this as cos /4, finally as cos /. The terminal points on the unit circle that have first codinate equal to / crespond to /3, /3. The solutions are thus 3 n 3 n. 7. If sec, then sec which implies that cos / that cos /. Using similar reasoning to that in problem 5, we obtain the solutions 4 n

2 3 4 n. 9. If cos sin, then either cos sin /. The solutions of cos are n the solutions of sin / are 6 n 7 6 n. All of these are solutions of the iginal equation.. If tan 3 cos, then either tan 3 cos, but the latter is not possible because is not in the range of the cosine function. Thus we must have tan 3. Noting that arctan 3 /3, we obtain the solutions 3 n. 3. To solve cossin cos, we write this equation as cos sin note that all solutions of this equation must satisfy either cos (because it is not possible that sin ). Therefe, the solutions are n. 5. The equation 4cos 4cos can be written as cos. Thus its solutions must satisfy cos cos /. The solutions of this equation are thus 3 n. 7. The equation sin sin 3 can be written as sin sin 3 as sin 3 sin. The only solutions are those f which sin (because it is not possible that sin 3). The solutions are thus n. 9. The equation sin 4 cos can be written as sin cos 4 cos cos (using the Pythagean identity) as 4 cos as cos 3. However it is clearly not possible that cos 3 so this equation has no solutions. (The graphs of y sin y 4 cos are shown below. Note that they do not intersect

3 anywhere.) To solve sin3, we write this equation as sin3. This equation has solutions (let s write them in degrees f a change) n n. Thus the solutions are n 5 n. 3. If cos /, then cos / which means that 36 n which means that 7 n. 5. To solve tan 5 9tan, we write this equation as tan tan 4 9, better yet, write it as tan tan 3 tan 3. There are thus three possibilities: solutions of tan which are 8 n, solutions of tan 3, which (since arctan 3 6 ) are 6 8 n, solutions of tan 3 which (since arctan 3 6 ) are 6 8 n. (Note that the equation tan 3 has no solutions. Why?)

4 7. To solve 4sincos sin cos, we write this equation as sin cos cos then as sin cos. The two possibilities are thus that sin / that cos /. The fmer equation has solutions 3 36 n 5 36 n the latter equation has solutions 36 n. 9. To solve cos sin, we write this equation as cos 4 (Recall the identity cos cos sin.) The solutions of cos 4 are given by n. Thus the solutions of this equation are.5 9 n. 3. We can write cos3 as cos3 /. We are only seeking solutions such that these are obtained when 3 6. The solutions are thus given by 3 3, 5 3, 7 3, 3, 3 3, 7 3 9, 5 9, 7 9, 9, 3 9, 7 9. Here is a graph of y cos3 on the interval,. Note that it crosses the line y si times Graphs of y cos 3 y 33. We want to find all solutions of sintan tan sin that lie in the interval,. Here are the graphs of y sintan tan y sin on that interval:

5 It looks like the equation has four solutions. How can we find them? Let s write the equation as tan sin sin then as tan sin sin finally as tan sin. From this we see that there are two possibilities: either tan sin /. The solutions of tan that lie in the interval, are 3 /4 7 /4. The solutions of sin / that lie in the interval, are /6 5 /6. These are our four solutions! Note that this seems to agree with what we see in the above graph. 35. The equation tan 3cot can be written as tan 3cot as tan 3 tan as tan 3. The solutions must thus satisfy tan 3. The only such numbers in the interval, are 3, 3, 4 3, 5 3. Are these all solutions? Let us graph the function y tan 3cot on the interval, :

6 Graph of y tan 3cot It indeed appears to cross the ais four times. 37. We would like to solve the equation tan 3 sec 3. First let s graph y tan 3 y sec 3 on the interval, : It looks like there are many points of intersection. Can you count them? To solve the equation tan 3 sec 3 we square both sides to obtain tan 3 sec 3 tan 3 tan 3 sec 3. Since tan 3 sec 3 (a Pythagean identity), we simply obtain tan 3 tan 3. The values of that satisfy this equation f which 3 6 are 3,,,3,4,5. The possible solutions f which are thus

7 , 3, 3,, 4 3, 5 3. Note that I said possible solutions. The only ones that actually wk when we plug them back into the iginal equation are, /3, 4 /3. F eample, if we try, we obtain tan 3 but sec 3 cos 3 which shows that is not a solution. Why did we find some etra solutions? The answer is that this is always a possibility when we attempt to solve an equation by squaring both sides. Now here is the really surprising part: When we first looked at the graphs, it looked like they had many points of intersection. However, there are only three! Since the graphs lie very close to each other at certain places, it just looks like they intersect but they really don t. This is an eample of why we can t always rely on graphical analysis. Looking at the graphs again, do you now see the three points of intersection? (Note that is actually a fourth point of intersection but we are not counting it since we were asked only to find solutions in the interval,.) 59. The equation sin sin3 can be written as sin cos. The solutions must thus satisfy either sin cos. The solutions of sin are 36 n 8 36 n 8 n 9 8 n. The solutions of cos are 9 36 n. 6. Since cos 4 cos cos 3 cos, the equation cos 4 cos cos can be written as cos 3 cos cos as cos 3 cos cos as cos cos 3. The solutions of cos are given by 9 36 n

8 the solutions of cos 3 / are given by n n. Thus all solutions of the equation cos 4 cos cos are given by 9 36 n n. Here are graphs of y cos 4 cos y cos on the interval,36 : The eight points of intersection that we see in this picture are