Trig Help (1 Viewer)

[Lith_30]

How would is solve a question like this.

 

[username_2]

How would is solve a question like this.
View attachment 33616

Well uh have you tried auxiliary angle method where you let: R sin(x + a) = cos(x) + 3sin(x)

R = sqrt(3^2+1^2) = sqrt(10)
a = cos^-1(1/sqrt(10)) = 1.249

Hence, you can solve the equation:
sqrt(10) sin(x+1.249) = 1

And i trust you can do it from here.

 

[A1La5]

The first approaches that come to mind for me here are T formulae and aux angle but those approaches are out of question as this was posted in the Advanced forum.

Square both sides of the equation and use the Pythagorean identity to get a quadratic in terms of sine or cosine. Solve said quadratic as normal, but be sure to disregard any solutions that are outside the range or . Also pay attention to the given domain, and substitute the values into the original equation to see if it matches up (squaring an equation provides extra solutions).

 

[5uckerberg]

How would is solve a question like this.
View attachment 33616

Well uh have you tried auxiliary angle method where you let: R sin(x + a) = cos(x) + 3sin(x)

R = sqrt(3^2+1^2) = sqrt(10)
a = cos^-1(1/sqrt(10)) = 1.249

Hence, you can solve the equation:
sqrt(10) sin(x+1.249) = 1

And i trust you can do it from here.

Um, do Mathematics Advanced students learn the Auxillary angle method? Did this come from an Extension I textbook?
Actually I have a way out of this square both sides in doing so we have . Next step is to bring everything on the LHS and there we will have . there, . The rest is just 2U techniques.

 

[okayokay123]

This is an Extension 1 question not an Advanced question. you would have to used auxillary method, t formulae etc.

 

[Lith_30]

Well uh have you tried auxiliary angle method where you let: R sin(x + a) = cos(x) + 3sin(x)

R = sqrt(3^2+1^2) = sqrt(10)
a = cos^-1(1/sqrt(10)) = 1.249

Hence, you can solve the equation:
sqrt(10) sin(x+1.249) = 1

And i trust you can do it from here.

I have heard about the auxiliary method but I haven't learnt it. Also, thankyou!

Did this come from an Extension I textbook?

Not really sure, this was just part of the homework given to me by my advanced teacher. I probably will ask her were she got this.

 

[ExtremelyBoredUser]

I have heard about the auxiliary method but I haven't learnt it. Also, thankyou!


Not really sure, this was just part of the homework given to me by my advanced teacher. I probably will ask her were she got this.

The only way I can think of inside the syllabus is by graphing the function and finding the values by just finding the points that are at 1.

 

[A1La5]

Not really sure, this was just part of the homework given to me by my advanced teacher. I probably will ask her were she got this.

The question in your original post is Question 7A of Exercise 12.2 ["Further Solution of Trigonometric Equations"] from the New Senior Mathematics Advanced textbook. Example 10 from the same exercise tackles a similar style question:

 

[specificagent1]

this does not look like advance lol

 

[CM_Tutor]

Squaring then testing the resulting solutions (to exclude "solutions" created by the squaring) is an accepted Advanced method. The auxiliary angle method is the standard method for such equations but is not taught below MX1 level.