[Inverse Trig] Why does the solution only take into consideration 1 case? (1 Viewer)

[okeydonkey]

Hi I am a bit confused by the solution of the following question. In the solution, it automatically assumes that tan(alpha) is negative but if sin(alpha) is negative then alpha can either be in the third quadrant or the fourth quadrant which means that tan(alpha) can either be positive or negative. So shouldn't we have two answers instead. Now when I think about that, if we do that for sin inverse then we need to do the same for cos inverse which means we should totally have 4 cases which is too much. Could someone please explain this to me? So if we have sin inverse, cos inverse or tan inverse as negative in questions similar to this, what rules should I follow to figure out the signs of other trig functions?

Thank you so much for helping me!

 

[TheShy]

The range of sin inverse is from pi/2 to -pi/2 so that would be in the first and 4rth quadrant. cos inverse range is from 0 to pi so 1st or 2nd quadrant, but since the cos is positive should be in 1st quadrant. Is this what you were asking?

 

[okeydonkey]

The range of sin inverse is from pi/2 to -pi/2 so that would be in the first and 4rth quadrant. cos inverse range is from 0 to pi so 1st or 2nd quadrant, but since the cos is positive should be in 1st quadrant. Is this what you were asking?

I guess so. How about tan inverse? Is it also from pi/2 to -pi/2 like sine inver

 

[TheShy]

I guess so. How about tan inverse? Is it also from pi/2 to -pi/2 like sine inver

Yeh tan inverse is from pi/2 to -pi/2
cos inverse is from 0 to pi
sin inverse is from pi/2 to -pi/2

 

[Drongoski]

See diagram in solution you provided. I think I can explain some of the confusion.



So our answers are the same. But the provided solution got it "the wrong way". The alpha in the triangle is not the correct inverse sin of (-3/5) because it is a negative angle, and a triangle cannot show a negative angle. The alpha in the triangle is the "related angle of -alpha", and you need to use this in the triangle, which is helpful in working out the trig ratios. But you must be aware, in this particular case, the inverse is -alpha. I have many worked solutions to this type of questions, always pointing out that the angles in the triangles may not be the actual inverse angles.
inverse sine of (3/5) is approx 37 deg. So inverse sine of(-3/5) is approx -37 deg. Inverse cosine of (2/3) is approx 48 deg. So

 

[CM_Tutor]

Another way is to start by remembering that the inverse sine function is odd, and so has the property that:



Cpnsequently, the problem we have to solve simplifies:



I can now take:



Now, is an acute angle in a right-angled triangle with hypoteneuse 3 and adjacent side 2 (or in any triangle similar to it). Using Pythagoras' Theorem, the third side is given by:



from which I can immediately deduce that:



And I can also take:



Now, is an acute angle in a right-angled triangle with hypoteneuse 5 and opposite side 3 (or any triangle similar to it). Using Pythagoras' Theorem, the third side is given by:



from which I can immediately deduce that:



Now, the original expression can be easily calculated:

 

[Velocifire]

Call me crazy, but don't you have to rationalise the denominator?

Cause I had that problem in a test with similar numbers but got marked down for that.

 

[CM_Tutor]

Hi I am a bit confused by the solution of the following question. In the solution, it automatically assumes that tan(alpha) is negative but if sin(alpha) is negative then alpha can either be in the third quadrant or the fourth quadrant which means that tan(alpha) can either be positive or negative. So shouldn't we have two answers instead. Now when I think about that, if we do that for sin inverse then we need to do the same for cos inverse which means we should totally have 4 cases which is too much. Could someone please explain this to me? So if we have sin inverse, cos inverse or tan inverse as negative in questions similar to this, what rules should I follow to figure out the signs of other trig functions?

Thank you so much for helping me!
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Okeydonkey, you asked: So shouldn't we have two answers instead?

Leaving aside the working from above showing a single answer, the answer to your question is definitely no. An expression like has a single value. It is the angle found in either the first or fourth quadrants where the sine function has the value 0.5. There are, of course, many angles with sine of 0.5, but only one such angle in those quadrants and restricted the range of outputs possible from the inverse sine function.

So, the original questions gives two single angles, seeks their difference, then asks for the tan of the resulting angle. There being only one difference and thus only one angle whose tan we seek, there can only be one answer.

Asides:

• It is poor practice to leave an answer as a fraction with an irrational denominator, and so the answer should be rationalised.
• If drawing triangles as the answer you gave does, confusion is avoided if they are put on a quadrants diagram. Representing should show an angle in quadrant four and a hypoteneuse drawn from the origin to (4, -3). The triangle may have sides of length / size 3, 4, and 5, but the coordinate relating to the opposite side place the point in quadrant 4 (with positive x and negative y, explaining the sine and tan being negative while the cosine is positive). Drawing this way, rather than as just plain right-angled triangles, makes the implications of the minus sign clear. I encourage everyone to draw the triangles for a question like this on a quadrants diagram to handle the signs correctly.

 

[CM_Tutor]

Call me crazy, but don't you have to rationalise the denominator?

Cause I had that problem in a test with similar numbers but got marked down for that.

We cross-posted, I agree that it is poor form to leave the denominator unrationalised and I would not give the answer full marks left as it was (and, I admit, that I did) in an exam

 

[Velocifire]

Okeydonkey, you asked: So shouldn't we have two answers instead?

Leaving aside the working from above showing a single answer, the answer to your question is definitely no. An expression like has a single value. It is the angle found in either the first or fourth quadrants where the sine function has the value 0.5. There are, of course, many angles with sine of 0.5, but only one such angle in those quadrants and restricted the range of outputs possible from the inverse sine function.

So, the original questions gives two single angles, seeks their difference, then asks for the tan of the resulting angle. There being only one difference and thus only one angle whose tan we seek, there can only be one answer.

Asides:

• It is poor practice to leave an answer as a fraction with an irrational denominator, and so the answer should be rationalised.
• If drawing triangles as the answer you gave does, confusion is avoided if they are put on a quadrants diagram. Representing should show an angle in quadrant four and a hypoteneuse drawn from the origin to (4, -3). The triangle may have sides of length / size 3, 4, and 5, but the coordinate relating to the opposite side place the point in quadrant 4 (with positive x and negative y, explaining the sine and tan being negative while the cosine is positive). Drawing this way, rather than as just plain right-angled triangles, makes the implications of the minus sign clear. I encourage everyone to draw the triangles for a question like this on a quadrants diagram to handle the signs correctly.

Really, I spent quite a lot of time on my paper yearning to find one. It was towards the back anyway.

I just hate losing marks for silly mistakes, got 3/4 at least.

 

[CM_Tutor]

 

[Trebla]

Call me crazy, but don't you have to rationalise the denominator?

Cause I had that problem in a test with similar numbers but got marked down for that.

We cross-posted, I agree that it is poor form to leave the denominator unrationalised and I would not give the answer full marks left as it was (and, I admit, that I did) in an exam

In the actual HSC exam itself, you will not be penalised for having an irrational denominator as the long as the answer is correct and the question didn’t explicitly ask for it.

 

[Velocifire]

In the actual HSC exam itself, you will not be penalised for having an irrational denominator as the long as the answer is correct and the question didn’t explicitly ask for it.

Fair enough. But our teacher is urging to scrape marks of people lol

 

[Trebla]

Fair enough. But our teacher is urging to scrape marks of people lol

Yeah unfortunately some teachers make up their own extra pedantic criteria and take off marks for the most trivial things, which is inconsistent to how the HSC exams are marked. You just have to follow their rules for internal assessments.

Generally speaking for computation questions, you can actually get away with leaving answers unsimplified in the HSC exam (though it you’re going to use it later on you’re better off simplifying it). Ironically, this was a hack that my school teacher taught me.

 

[CM_Tutor]

Yeah unfortunately some teachers make up their own extra pedantic criteria and take off marks for the most trivial things, which is inconsistent to how the HSC exams are marked. You just have to follow their rules for internal assessments.

Generally speaking for computation questions, you can actually get away with leaving answers unsimplified in the HSC exam (though it you’re going to use it later on you’re better off simplifying it). Ironically, this was a hack that my school teacher taught me.

I agree with Trebla that standards and criteria are applied variably. It's inevitable if the candidature at the school does not match the distribution / characteristics of the whole HSC candidature in the exam - and remember that the purpose of an exam is to rank students in order from highest to lowest, so variations that make this easier in a given context are reasonable. Having said that, the potential for the problem to become large is much higher if the criteria you are used to are looser than those applied in the HSC... or, (as was demonstrated a few years ago, when variable criteria crept into the HSC itself). Plus, in this case, the stricter criterion has a sensible rationale.

I cannot tell just by looking if



is rational or not, but I can if I rationalise the denominator. It is the same in Extension 2 with complex numbers, a fraction should not be left with a non-real denominator.

I also have a problem with a question that asks that you show your answer is rational and the answer is left as



or as



or as



even though all of these are rational (unless I've made a mistake in my mental arithmetic). It's pretty clear the first will be rational when expanded, and the second just needs a common factor cancelled, while the third is not obviously rational (IMO)... but are any of these acceptable if followed by "which is rational" full stop?

There are habits and practices that are arbitrary, and there are habits and practices that make sense / are just good to do. I think this one is a habit worth developing.

 

[Trebla]

I agree with Trebla that standards and criteria are applied variably. It's inevitable if the candidature at the school does not match the distribution / characteristics of the whole HSC candidature in the exam - and remember that the purpose of an exam is to rank students in order from highest to lowest, so variations that make this easier in a given context are reasonable. Having said that, the potential for the problem to become large is much higher if the criteria you are used to are looser than those applied in the HSC... or, (as was demonstrated a few years ago, when variable criteria crept into the HSC itself). Plus, in this case, the stricter criterion has a sensible rationale.

I cannot tell just by looking if



is rational or not, but I can if I rationalise the denominator. It is the same in Extension 2 with complex numbers, a fraction should not be left with a non-real denominator.

I also have a problem with a question that asks that you show your answer is rational and the answer is left as



or as



or as



even though all of these are rational (unless I've made a mistake in my mental arithmetic). It's pretty clear the first will be rational when expanded, and the second just needs a common factor cancelled, while the third is not obviously rational (IMO)... but are any of these acceptable if followed by "which is rational" full stop?

There are habits and practices that are arbitrary, and there are habits and practices that make sense / are just good to do. I think this one is a habit worth developing.

The examples you provide are specific scenarios where the rationality of a denominator is important for purposes of the question. Usually this requirement would be made pretty clear in the question itself.

However, if it’s a straight computation (like the original question above) where the rationality of a denominator is irrelevant to the purpose of the question, then it makes no sense to require it in the marking criteria. A simple example - find the exact value of . It makes no sense to pedantically deduct marks if you write and only give full marks to ...given these are the equivalent answers. The rationality of the denominator is completely irrelevant to the point of the question so that type of criteria is a poor way to discriminate between someone who actually knows the material versus someone who does not.

In the HSC exam itself the criteria is more thoughtfully set out with the purpose of the question in mind and done in a way to differentiate stronger and weaker candidates with respect to how they demonstrate their knowledge of the course. Therefore, how you got the answer is far more important than what format the answer itself takes. So when it comes to straight computation questions in the HSC exam (which don't ask for the denominator to be rational in the final answer) you shouldn't be wasting your time trying rationalise the denominator of the final answer because you've already demonstrated how you got to the answer and won't earn any extra marks for additional simplification.

 

[Trebla]

Hi I am a bit confused by the solution of the following question. In the solution, it automatically assumes that tan(alpha) is negative but if sin(alpha) is negative then alpha can either be in the third quadrant or the fourth quadrant which means that tan(alpha) can either be positive or negative. So shouldn't we have two answers instead. Now when I think about that, if we do that for sin inverse then we need to do the same for cos inverse which means we should totally have 4 cases which is too much. Could someone please explain this to me? So if we have sin inverse, cos inverse or tan inverse as negative in questions similar to this, what rules should I follow to figure out the signs of other trig functions?

Since there are only numbers involved, when you plug this in your calculator there should only ever be one answer with this type of question (it's also a way to check if you have the right exact answer as well).

 

[Velocifire]

ALL STATIONS TO CENTRAL

Me sitting on a train: how do you remember the quadrants
Train: all stations to central
Me: thank you train... wait WHAT ALL STOPS?

 

[okeydonkey]

Thank you everyone!

 

[5uckerberg]

ALL STATIONS TO CENTRAL

Me sitting on a train: how do you remember the quadrants
Train: all stations to central
Me: thank you train... wait WHAT ALL STOPS?

I got another acronym for this one, ALL STATES TO CANBERRA