SOLUTION: Solve the following trigonometric equation for 1π/6≤β<7π/4:sin(β)=−0.493
Specific Solution(s) with 2 decimals :
can someone tell me how to do thi
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-> SOLUTION: Solve the following trigonometric equation for 1π/6≤β<7π/4:sin(β)=−0.493
Specific Solution(s) with 2 decimals :
can someone tell me how to do thi
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Question 1117560: Solve the following trigonometric equation for 1π/6≤β<7π/4:sin(β)=−0.493
Specific Solution(s) with 2 decimals :
can someone tell me how to do this question ? i really need to know how to do it because this one might go to the final exam, please help me ! Found 2 solutions by greenestamps, ikleyn:Answer by greenestamps(13200) (Show Source):
By far the easiest way to solve the problem is to graph y=sin(x) and y = -0.493 on a graphing calculator over the specified interval and find where the graphs intersect.
But most likely that solution method will not be acceptable on an exam....
On the other hand, it would be helpful, if allowed, to find the solution that way so you can check the answer you get using your knowledge of the behavior of the sine function.
So here is a graph of the sine function from -pi to +2pi, along with a graph of y=-0.493:
On the specified interval, there is clearly a solution at about x = 3.5; in fact, my graphing calculator gives the value as x = 3.6571272. And it looks as if there might be a solution at about x=6; but it turns out that that intersection is just outside the specified interval.
So your job is to find the single solution at about x = 3.5.
Clearly you need to use your calculator to find an angle x for which sin(x)=-0.493. Most (all?) calculators will show to several decimal places.
You can see that on the above graph -- the first intersection point to the left of x=0. Your task is to use that value to find the solution at about x = 3.5, using what you know about the sine function.
There are many ways you could do that. Perhaps another tutor will also respond to your question and use a different method than I'm going to use. That could be helpful to you; perhaps you would find a different method than mine to be more to your liking.
So here is the reasoning I would use to find the answer.
(1) The value of x at about 3.5 where sin(x) = -0.493 is exactly half a period (pi) past the point where sin(x) = +0.493.
(2) The value of x where sin(x) = +0.493 is the opposite of the value where sin(x) = -0.493.
So I will find my answer by taking the angle my calculator gives me, multiplying it by -1 to get the opposite of that angle, and then adding pi:
Hooray! That agrees with the answer I got by graphing.
1. You should understand that the solutions are in quadrants QIII and QIV.
2. Calculate the given endpoints of the solution interval
= = 0.523; = = 5.495.
3. Find arcsin(0.493) = 0.516 radians (using your calculator). It is in quadrant QI.
4. You should understand that the solution to sin(b) = - 0.493 is units (half the period of sine) ahead in QIII:
= arcsin(0.493) + = 0.516 + 3.14 = 3.656.
It is still in the assigned limits. Thus you just found the solution in QIII.
5. Now you need to find (and to check) the solution in QIV.
There are two ways to find it.
a) You can find arcsin(-0.493), using your calculator. You will get
= -0.516 radians.
Add = 6.28 (one rotation about the unit circle) to it to get QIV.
So, your new value for is -0.516 + 6.28 = 5.764.
But it is greater than the limit 5.495 of the given domain, so this solution does not work.
b) Or, knowing the behavior of sine function, you can find the distance from to :
= = 1.054
and then to add this value 1.054 to : = 5.764 ( ! the SAME value as you found above ! )
It is your candidate for ; but it fails since it is greater than the upper boundary of 5.495.
Your analysis is completed. You just found the unique solution = 3.656.
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So, your algorithm again:
- Find the boundaries of the domain.
- Think in what quadrants the solution should be.
- Evaluate the solution using your calculator.
- Check if it satisfies the given constraints.
Surely, you should know the behavior / (how the plots look like) for basic trigonometry functions.
