Trig inequality help

Yatayyat

Hello everyone!

Here is my Q shown below

Screenshot 2020-03-15 at 21.00.14.png

I have had a go at all the parts but I'm struggling on how I should do part d) of this question

Any help would be really grateful, thanks!

Scroll to see replies

Reply 1

Yazomi

Original post by Yatayyat

Hello everyone!

Here is my Q shown below

Screenshot 2020-03-15 at 21.00.14.png

I have had a go at all the parts but I'm struggling on how I should do part d) of this question

Any help would be really grateful, thanks!

I think

You should sketch the graph of when cosx=1/2 or =-0.7... within the range they gave you, preferably on the same graph

Since y=cosx is similar to a quadratic you could do the same where you look on the graph and whichever part of it is above the X axis is your answer (remembering to put your inequality signs)

Reply 2

RDKGames

Study Forum Helper

Original post by Yatayyat

Hello everyone!

Here is my Q shown below

I have had a go at all the parts but I'm struggling on how I should do part d) of this question

Any help would be really grateful, thanks!

Maybe the factorisation isn't at all obvious, but consider

(x-\alpha)(x-\beta)

which expands into

x^2 - (\alpha + \beta)x + \alpha\beta = 0

.

The question is of exactly this form. Hence if you notice this you can immediately factorise into

(\cos x - \frac{1}{\sqrt{2}})(\cos x - \frac{1}{2}) \geq 0

So you should solve this inequality. We require that

\cos x \leq \dfrac{1}{\sqrt{2}}

*or*

\cos x \geq \dfrac{1}{2}

. So you seek x values so that this statement is true.

(edited 4 years ago)

Reply 3

Yatayyat

Original post by RDKGames

Maybe the factorisation isn't at all obvious, but consider

(x-\alpha)(x-\beta)

which expands into

x^2 - (\alpha + \beta)x + \alpha\beta = 0

.

The question is of exactly this form. Hence if you notice this you can immediately factorise into

(\cos x - \frac{1}{\sqrt{2}})(\cos x - \frac{1}{2}) \geq 0

So you should solve this inequality. We require that

\cos x \leq \dfrac{1}{\sqrt{2}}

*and*

\cos x \geq \dfrac{1}{2}

. So you seek x values so that this statement is true.

Thanks I hadn't seen it like that. Plus my calculator wouldn't give one of the solutions in exact form when using the quadratic formula, where 0.7017... = 1/sqrt(2)

I have tried to work out each of the smaller 2 trig inequalities and got this. I had to always refer back to the cos graph to help me get the solutions I need to include in the given interval.

Reply 4

mqb2766

Original post by Yatayyat

You want the union of the intervals where both factors are positive or both factors are negative?
Also you've written the intersection of two intervals which gives the range (45,60) degrees. This looks to be the opposite of what you want? Just stick some values into your original equation if you're unsure.

(edited 4 years ago)

Reply 5

Yatayyat

Original post by mqb2766

Yep my mistake, I see why I need to use the union here instead :smile:

Reply 6

mqb2766

Original post by Yatayyat

Yep my mistake, I see why I need to use the union here instead :smile:

And the sets are wrong, as they need to be both positive or both negative.

Reply 7

ISHxxxx

Is this year 12 or year 13 maths

Reply 8

Yatayyat

Original post by mqb2766

And the sets are wrong, as they need to be both positive or both negative.

Are they not both positive over here? What do you mean when you say a set is positive or negative?

Reply 9

Yatayyat

Original post by mqb2766

And the sets are wrong, as they need to be both positive or both negative.

All x values that lie in both of these sets do give a value of greater than or equal to zero. So here both my sets are positive? Plus I need to adjust the set accordingly in order to give all x values that lie in the given interval of the question

Reply 10

mqb2766

Original post by Yatayyat

Are they not both positive over here? What do you mean when you say a set is positive or negative?

You had the sets (0,60) and (45,90). Either plug a few values back into the quadratic or reason about the signs of the factors
(cos(x) - 0.7)(cos(x) - 0.5)
To reason how each set is defined - both terms are positive or (union) both terms are negative.

Tbh, its easier to just go back to a basic quadratic. You have a quadratic in z=cos(x). The z^2 coefficient is positive so the gra on is a bowl. The zero crossing points are in the interval of interest (0,1), so the graph is positive in the intervals at the ends and is negative in the centre interval. These 3 intervals do not overlap. Your original solution defined two overlapping intervals and took their intersection. This gave the interval in which the quadratic is nagative.

(edited 4 years ago)

Reply 11

Yatayyat

Original post by mqb2766

You had the sets (0,60) and (45,90). Either plug a few values back into the quadratic or reason about the signs of the factors
(cos(x) - 0.7)(cos(x) - 0.5)
To reason how each set is defined - both terms are positive or (union) both terms are negative.

Tbh, its easier to just go back to a basic quadratic. You have a quadratic in z=cos(x). The z^2 coefficient is positive so the gra on is a bowl. The zero crossing points are in the interval of interest (0,1), so the graph is positive in the intervals at the ends and is negative in the centre interval. Thread 3 intervals do not overlap. Your original solution defined two overlapping intervals and took their intersection. This gave the interval in which the quadratic is nagative.

Ok thanks, but I do seem to a get a value of zero or more when I plug in x values that belong in the set (0,60) and (45,90) back in the quadratic?

I have tried using Desmos to help me picture of what is going on here and I see this:

Screenshot 2020-03-17 at 08.57.25.png

Therefore my set should look something like this?

Reply 12

mqb2766

Original post by Yatayyat

Therefore my set should look something like this?

The intervals are wrong. A s mentioned above, a simple quadratic in z argument gives (0,45) where both factors are positive or (60,90) where both factors are negative. In either case, their product is positive.
I'm not sure how your thinking is going wrong.

Your working is quite verbose (long) at times. Great to see clear explanations, but I feel you sometimes miss the big picture? Sometimes a heuristic, quick and dirty approximate solution keeps you on track.

(edited 4 years ago)

Reply 13

Yatayyat

Original post by mqb2766

The intervals are wrong. A s mentioned above, a simple quadratic in a argument gives (0,45) where both factors are negative or (60,90) where both factors are positive. In either case, their product is positive.
I'm not sure how your thinking is going wrong.

Your working is quite verbose (long) at times. Great to see clear explanations, but I feel you sometimes miss the big picture? Sometimes a heuristic, quick and dirty approximate solution keeps you on track.

I understand that that interval is wrong, but I when I do key in the 2 inequalities to how it exactly is on Desmos, it does appear to include the region I have specified with my original interval before. This is my confusion.

For the bit about you saying that both factors must be either both positive or both negative. Isn't both my factors positive, as when I sub in x values that belong to the set I have given, they come to be zero or more, so therefore the product of those 2 factors will always equal zero or more which is what the inequality wants

Reply 14

mqb2766

Original post by Yatayyat

I'm not sure what you mean about desmos. In the final graph, the middle dark green interval is when the quadratic is <= 0. You want the two intervals on either side.

Your quadratic factories is
(z - 0.7)(z - 0.5)
Where z goes from 0 to 1. Can you sketch it? Almost without doing any analysis you should be able to say its
positive for
z < 0.5
and
z > 0.7
So the two intetvals to the left and right of the minimum and maximum roots.

Reply 15

Yatayyat

Original post by mqb2766

Is this what you mean?

I get these 2 intervals which I have circled in my working

Reply 16

Yatayyat

The question states that I must provide my sols in the range 0 to 90, how would I do this with the 2 intervals I have just found

Reply 17

mqb2766

Original post by Yatayyat

Is this what you mean?

I get these 2 intervals which I have circled in my working

So when z<0.5, you get the interval (60,90)
For z>0.7 you get the interval (0,45)
You keep the same inequality when taking acos(). Sketch the graph and notice it flips. That's your mistake.

Then take their union.

(edited 4 years ago)

Reply 18

Yatayyat

Original post by mqb2766

Yes this is specifically the part I am struggling to understand? I'm very confused with how you arrived at the interval (60,90) from z<0.5

Because from that I arrived at x<= 60 and not 60 <= x <= 90

Reply 19

mqb2766

Original post by Yatayyat

Yes this is specifically the part I am struggling to understand? I'm very confused with how you arrived at the interval (60,90) from z<0.5

Because from that I arrived at x<= 60 and not 60 <= x <= 90

Sketch z = cos(x), the question asks you to.
So when z<0.5, this is satisfied by x=90 as z=cos(90)=0 and all values back to x=60 as z=cos(60)=0.5.
Going back any further means z>0.5. So you want (60,90).
Similar for the other case.

(edited 4 years ago)