Hi all,

 

This is my first post in this section of the forums!

 

I was doing some past papers and I am stuck in this question. Need help any1.I am not very good at maths! I am doing A’ levels by the way.

 

 

Q. Solve the equation for angle in the range -180 ≤ Ф ≤ 180.

 

4 Cos2 Ф + 5 Sin Ф = 3

 

 

My solution is:

 

4 (1- Sin 2 Ф) + 5 Sin Ф -3 = 0.

 

4 – 4 Sin 2 Ф + 5 sin Ф -3 = 0.

 

-4 Sin 2 Ф + 5 sin Ф = 0.

 

Assuming sin Ф = x, then

 

 

-4x2 + 5x + 1 = 0.

 

Using quadratic formula:

 

x = -b -+ √ b2 – 4ac

2a

 

x = - 5 -+ √ 5 -4(-4)(1)

-8

 

So x = -5 – 4.58

-8

 

x = 1.197

 

OR

 

x = -5 + 4.58

-8

 

x = 0.0525

 

Now I have 2 values of x which I am going to replace in sin Ф = x

 

So Ф = Sin -1 1.197 ( calculator says Ma error)

 

We have one x value left which I am going to put now:

 

Ф = sin -1 0.0525

= 3.009 or 3.01 ( 3 significant figures and 2 decimal places).

 

I know that sin is always positive in the first and second quadrant but I have never managed to get the answers! Anyone can tell me what to do now or what mistake have I made?

 

I’ll really appreciate it thnx!

- On b² you forgot to square and put 5 instead of 5²=25.

- I hope you understand why your calculator gives an error for [math]\sin^{-1} 1.197 [/math].

- [math]\sin^{-1} x [/math] does not map on the whole range -180° <= phi <= 180°. You might (you do) miss solutions. Not exactly sure how you get the 2nd solution (you did make a sketch/plot to see that there's two solutions, did you?). I'd try by reexpressing x=cos(phi - 90°) rather than x=sin(phi).

 

Sidenote:

- The meaning of "quadrant" you probably meant is incompatible with my understanding of the term. Quadrant often refers to the four areas divided by the axes of a plot. Incidently, the 1st and 2nd quadrant are those where the value of the function is/would be positive

- On b² you forgot to square and put 5 instead of 5²=25.

- I hope you understand why your calculator gives an error for .

- does not map on the whole range -180° <= phi <= 180°. You might (you do) miss solutions. Not exactly sure how you get the 2nd solution (you did make a sketch/plot to see that there's two solutions, did you?). I'd try by reexpressing x=cos(phi - 90°) rather than x=sin(phi).

 

On b² you forgot to square and put 5 instead of 5²=25.

 

I think then x = 0.175 or -1.425. SO if I take the inverse of tan the phi will be 9.926 and -54.94 respectively.

 

- I hope you understand why your calculator gives an error for .

 

I do yes!

does not map on the whole range -180° <= phi <= 180°. You might (you do) miss solutions. Not exactly sure how you get the 2nd solution (you did make a sketch/plot to see that there's two solutions, did you?). I'd try by reexpressing x=cos(phi - 90°) rather than x=sin(phi).

 

 

I have no clue whatsoever what it means or to do next or how to find the answers out .

I think then [when squaring the 5] x = 0.175 or -1.425.

When the term under the the square root increases, the lower result (for x) must decrease and the higher result must increase. Therefore, these values (or your previous ones) cannot be correct. Simply redo the calculation starting from plugging values into the quadratic forumula and don't try to patch the erronous calculation.

 

I have no clue whatsoever what it means

Assuming with "it" you mean my comment about the range of arcsin:

 

Completely independently of your homework:

sin(0°) = 0,

sin(180°) = 0.

So what does arcsin(0) equal to? 0° or 180°? Point to be learned: [math]\sin^{-1}[/math] is not the inverse of sin, except for special cases that do not hold true here.

 

or to do next or how to find the answers out .

The problem is that what I said already was on the borderline of what I consider appropriate for helping with homework. Finding that more than one (the one you get from arcsin(x)) solution exists and finding the other solutions imho is exactly the point of the problem that requires a bit of creativity and understanding of mathematics beyond rearranging equations. Because of this and to avoid possibly unnecessary confusion (my idea is probably not the only way, it's just the approach that popped up into my mind, first) I will not give further advice on that point.

 

Just to make one important point clear and to reduce confusion: You understand that the way you approached the problem is correct and that if you don't screw up the numbers when applying the quadratic formula you will get 1 correct solution, are you? So you have solved at least part of the problem.

Just to make one important point clear and to reduce confusion: You understand that the way you approached the problem is correct and that if you don't screw up the numbers when applying the quadratic formula you will get 1 correct solution, are you? So you have solved at least part of the problem

 

Yes, you are right. I am weak in maths and can't understand things unless I am face to face with whoever is instructing me.

 

 

The problem is that what I said already was on the borderline of what I consider appropriate for helping with homework.

 

Sorry if i gave the wrong impression but this is no homework. I didn't ask to work things out for me, I know I have to do it myself. I only wanted help! however I thank you for your help! I really appreciate it.

So what do you need it for? If you just need an approximate solution, you could for example just solve it graphically and read out the solutions from the graph. Anyways, your path is fine. You'll just possibly miss one solution when taking the arcsin without paying attention.

uh-huh, alright then thnx very much. I'll just try to practice some more questions and see what happens. Thanks again!

So the equation is:

[math]4cos^2\Phi + 5sin\Phi - 3 = 0[math]

 

as [math]cos^2\Phi = 1 - sin^2\Phi[/math] (like you posted) ten the equation would look like this:

 

[math]4sin^2\Phi - 5sin\Phi + 1[/math] and then use the quadratic formula and you get:

 

[math]x_{1/2}=\frac{5 \pm \sqrt{25 - 16}}{8}[/math]

 

[math]x_{1/2}=\frac{5\pm 3}{8}[/math]

 

and the solutions of quadratic equation are [math]x_1=\frac{1}{4} ; x_2=1[/math]

 

so you got:

 

[math]sin\Phi=\frac{1}{4}[/math] and [math] sin\Phi=1[/math]

 

Now let's check them into equation.

 

Case 1 [math]sin\Phi=1/4[math]

 

[math]4sin^2\Phi - 5sin\Phi +1 = 0[/math]

 

[math]4\times (\frac{1}{4})^2 - 5\times \frac{1}{5} + 1 = 0[/math]

 

[math]\frac{1}{4} - \frac{5}{4} + 1 = 0[/math]

 

[math]0=0[/math] works out!

 

Case 2 [math] sin\Phi=1[/math]

 

[math]4sin^2\Phi -5sin\Phi + 1 = 0[/math]

[math]4\times 1 - 5\times 1 + 1 = 0[/math]

[math]0=0[/math]

 

works out too.

 

And please LaTeXise next math post.

 

Cheers,

Shade

... then the equation would look like this:

[math]4sin^2\Phi - 5sin\Phi + 1[/math]

That is not an equation.

 

 

Now let's check them into equation.

 

Case 1 [math]sin\Phi=1/4[/math]

 

[math]4sin^2\Phi - 5sin\Phi +1 = 0[/math] ...

I strongly advice checking results with the original equation. Otherwise, you might end up checking with an erronous or incomplete one.

That is not an equation.

I seem to have forgotten =0, but I do believe that you are smart enough to understand that by your own.

Sure, but then the equation wrong. T'was more a hint that errors often come from the parts where you worked sloppily.

 

I know Atheist, and there is just one mistake on the whole equation (it's that damn +1 at the end that should have been -1), and screws up my calculation and makes me sad:doh:!