Solving trigonometric functions Multiple choice HSC (2 Viewers)

[BlueGas]

There should be two solutions for . You were meant to do 180º + (-63)º, and 360º + (-63)º for that.

So answer is that there are three solutions.

Alternatively, just sketch graphs of the two functions in the given domain. We can see then that the horizontal line intersects that of in two places in the given domain, and the line of touches the graph of exactly once in the given domain. Hence answer is 3.

But....but...but... the answers in success one book says there are two solutions.

 

[Ekman]

There should be two solutions for . You were meant to do 180º + (-63)º, and 360º + (-63)º for that.

So answer is that there are three solutions.

Alternatively, just sketch graphs of the two functions in the given domain. We can see then that the horizontal line intersects that of in two places in the given domain, and the line of touches the graph of exactly once in the given domain. Hence answer is 3.

Actually the answer is 2, believe me I said the same thing last year , but x=pi/2 does not exist as tan(pi/2) = undefined, so the solution you get from the (sinx-1) is ignored. So there is only 2 solutions that come from the (tanx + 2)

 

[InteGrand]

Ah right my bad.

But the method used by BlueGas had a mistake in the tan part, so it was a coincidence he got the right answer.

 

[BlueGas]

Ah right my bad.

But the method used by BlueGas had a mistake in the tan part, so it was a coincidence he got the right answer.

The mistake was 180 - (-63), but when I wrote 360 - (-63) that's right, isn't it?

 

[InteGrand]

The mistake was 180 - (-63), but when I wrote 360 - (-63) that's right, isn't it?

There's two solutions to the tan equation in the given domain.

Using the ASTC method, we're meant to find the acute (positive) angle, and then go to the quadrants.

So we'd use tan^-1 (2) (≈ +63.43º) as the angle, and then go to the ASTC diagram.

 

[Ekman]

The mistake was 180 - (-63), but when I wrote 360 - (-63) that's right, isn't it?

Well both of those answers are wrong. The way I do these questions is that I treat it as if its tanx = 2, then find the angle x, and since tan is negative in the second and fourth quadrant I do 180-x and 360-x. So in this question, tan(63)=2, so the solution to tanx + 2=0 is 180-63 and 360-63

 

[BlueGas]

There's two solutions to the tan equation in the given domain.

Using the ASTC method, we're meant to find the acute (positive) angle, and then go to the quadrants.

So we'd use tan^-1 (2) (≈ +63.43º) as the angle, and then go to the ASTC diagram.

Ohhh so you can do 0+x, or 180-x, or 180 +x or 360-x unless x is positive? And if it's negative you look at the negative quadrants but you take the positive of x?

 

[BlueGas]

Man I hate these sort of questions, so annoying.

 

[BlueGas]

Actually the answer is 2, believe me I said the same thing last year , but x=pi/2 does not exist as tan(pi/2) = undefined, so the solution you get from the (sinx-1) is ignored. So there is only 2 solutions that come from the (tanx + 2)

But why would you sub x=90 from sin into tanx?

 

[InteGrand]

But why would you sub x=90 from sin into tanx?

We say that the original equation's LHS is undefined at x = 90º since tan x is undefined there. So we can't use x = 90º.

 

[BlueGas]

We say that the original equation's LHS is undefined at x = 90º since tan x is undefined there. So we can't use x = 90º.

So what are the two solutions again?

 

[InteGrand]

So what are the two solutions again?

We don't need to know them for the question. We just need to know how many there are.

But they were: and .

 

[BlueGas]

Quick question, when inversing a trig function, is the answer in radians mode in or degrees?

 

[VBN2470]

Quick question, when inversing a trig function, is the answer in radians mode in or degrees?

Radians.

 

[InteGrand]

Quick question, when inversing a trig function, is the answer in radians mode in or degrees?

It depends on your calculator mode. If your calculator is set in degrees mode, the output will be in degrees. So when you got that output of ~63 when typing , that was in degrees, as your calculator was in degrees mode. If you set your calculator to radians mode, you'll get ~1.107 for your calculator's output when typing in .

 

[milkytea99]

There should be two solutions for . You were meant to do 180º + (-63)º, and 360º + (-63)º for that.

So answer is that there are three solutions.

Alternatively, just sketch graphs of the two functions in the given domain. We can see then that the horizontal line intersects that of in two places in the given domain, and the line of touches the graph of exactly once in the given domain. Hence answer is 3.

Isn't the answer 3?
As there's one solution for sine and 2 solutions for tan

 

[InteGrand]

Isn't the answer 3?
As there's one solution for sine and 2 solutions for tan

That was the trick of this question – the solution for the sine doesn't count, because the LHS is undefined there (as tan is undefined).

 

[milkytea99]

Can't see the pics

 

[Speed6]

BlueGas, see there thick past HSC questions book you are using, what is it called?

 

[Crisium]

BlueGas, see there thick past HSC questions book you are using, what is it called?

Excel Success One HSC Mathematics Past HSC Papers and Worked Answers