Trigonometric equations (1 Viewer)

[FOB all the way]

I am having real problems solving trig equations using asin@ + bcos@ = r (@ + &)

sorry for the bad symbols & is alpha

i keep geting every question wrong i think it is something small i am doing could someone please help with this question

4sin@ - cos@ + 3 = 0

the answers are 240 degrees 43 minutes and 327 degrees 21 minutes

thanks again

 

[Mark576]

Maybe you could show us your working, then we could more easily identify where exactly you're going wrong.

 

[ssglain]

I think you meant questions in the form of asin@ + bcos@ = Rsin(@ + &) or Rcos(@ + &)? I don't believe in categorising questions, but this is one of the few question-types that you can attack and succeed by following a set of steps every time. Basically, what you need to do is expand the RHS and then compare like terms on either side. Then you simply need to solve for R and & (taking -90 < & < 90).

Now, let's solve 4sin@ - cos@ + 3 = 0 (0 < @ < 360) to better illustrate what is known as the 'auxiliary angle method'.

Step 1.
Let 4sin@ - cos@ = Rsin(@ + &)
(The cos form would also do. Use whichever that suits your preference, or that of the examiner.)

Step 2.
Then 4sin@ - cos@ = Rsin@cos& + Rcos@sin&
Rcos& = 4 ...(1)
Rsin& = -1 ...(2)

Step 3.
(2)/(1) gives: tan& = -1/4 --> & = arctan(-1/4) = -14deg2min

Step 4.
(1)^2 + (2)^2 gives: R^2[(sin&)^2 + (cos&)^2] = 17 --> R^2 = 17 --> R = sqrt(17)

Now we've managed to establish that 4sin@ - cos@ = sqrt(17)sin(@ - 14deg2min)
Not a particularly nice example, is it? The numbers would be a lot nicer, but nevertheless...

Step 5.
Rewrite the original equation as sqrt(17)sin(@ - 14deg2min) = -3, which we will need to solve for @.
sin (@ - 14deg2min) = -0.7276
@ - 14deg2min = 226deg41min or 313deg19min
.: @ = 240deg43min or 327deg21min

So where did you go wrong? Please don't hesitate to ask if you need more help.

 

[FOB all the way]

thanks

i tried to do it like this

asin@ +bcos@ = rsin(@+&) where r= Square root of (a squared + b squared) And tan & = b/a

a= 4
b=1
r = Square root 17
tan & = 1/4
& = 14deg12min

4sin@ - cos@ = square root 17sin(@ + 14deg12min)
root17sin(@+14deg12min) = -3

base angle = 46deg41min

@+ 14deg02min = 226deg41min, and 313deg19min
@ = 212deg39min and 299deg17min

wrong i know

 

[ssglain]

Yeah, you got b wrong and it followed from there. However, you must show how your derived R and & every time. I know that Margaret Grove fiddles with R = sqrt(a^2 + b^2) and tan& = b/a in Maths In Focus, but it is never acceptable to quote these.

As you proceed through the course, you will see that this is just like, amongst many many other examples, parabola questions in which you are required to show how you obtained the equations of the tangent, normal, chord of contact, etc. or projectile motion questions in which you must use integration to show how you derived the equations of the horizonal/vertical velocity and displace, trajectory, etc.

A lot of people will relate memorising formulae to maths. Let me assure you that their presumption is immensely shallow, and will not get them far. Maths is about understanding. I don't believe that maths, especially at extension level, aims to teach "content" as such that can be memorised - rather, powerful tools and their applications.

Like I said before, the example we discussed is one of very few that can be solved by blindly following a set of steps without much understanding. So please don't expect to see "scaffolding" or "flow-charting" steps to work too often.

 

[FOB all the way]

thanks for your help