• Best of luck to the class of 2020 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • Looking for HSC notes and resources?
    Check out our Notes & Resources page

Further Trig Questions (1 Viewer)

bujolover

Active Member
Joined
Jan 5, 2017
Messages
154
Gender
Undisclosed
HSC
2018
Hi guys,

1.

(a) ***I do not need help with this part*** Consider the statement cos[(π/2) + A] = +/- sinA. For which sign is this statement true for all A? For which A is the statement true for both signs?

Answer: Negative, A = 0, nπ (n is an integer)

(b) Taking A = 5θ in (a), write down the value of B such that -sin5θ = cosB. Hence find the least value of θ between 0 and 2π such that sin5θ + cos8θ = 0.
I actually haven't gotten any answers for this ^^, as I have little idea of how to do the question.

2. Hence, or otherwise, solve 2cosθ + 2cos(θ + (π/3)) = 3, for 0 < θ < 2π. (This is the second part of the question. The first part was to express the expression on the LHS in the form Rcos(θ + α), where R>0 and α is between 0 and 90 degrees (not including them). The answer was 2√3cos(θ + (π/6)).) I keep getting 5π/3, and I don't know where I'm going wrong (I'm sure I'm wrong).

Thanks in advance! :)

P.S. π is actually pi, just before anyone asks, to prevent people wasting time asking this.
 
Last edited:

integral95

Well-Known Member
Joined
Dec 16, 2012
Messages
780
Gender
Male
HSC
2013
Welp I can't seem to use the "hence" method. But for otherwise I just did like


 

integral95

Well-Known Member
Joined
Dec 16, 2012
Messages
780
Gender
Male
HSC
2013
Hi guys,

1.

(a) ***I do not need help with this part*** Consider the statement cos[(π/2) + A] = +/- sinA. For which sign is this statement true for all A? For which A is the statement true for both signs?

Answer: Negative, A = 0, nπ (n is an integer)

(b) Taking A = 5θ in (a), write down the value of B such that -sin5θ = cosB. Hence find the least value of θ between 0 and 2π such that sin5θ + cos8θ = 0.
I actually haven't gotten any answers for this ^^, as I have little idea of how to do the question.

2. Hence, or otherwise, solve 2cosθ + 2cos(θ + (π/3)) = 3, for 0 < θ < 2π. (This is the second part of the question. The first part was to express the expression on the LHS in the form Rcos(θ + α), where R>0 and α is between 0 and 90 degrees. The answer was 2√3cos(θ + (π/6)).) I keep getting 5π/3, and I don't know where I'm going wrong (I'm sure I'm wrong).

Thanks in advance! :)

P.S. π is actually pi, just before anyone asks, to prevent people wasting time asking this.
Your answer is right, along with theta = 0
 

bujolover

Active Member
Joined
Jan 5, 2017
Messages
154
Gender
Undisclosed
HSC
2018
Welp I can't seem to use the "hence" method. But for otherwise I just did like


Does anyone know how to solve this question using "hence"? I'm guessing I would lose marks if I did otherwise in an exam situation, since it only says "hence".

Your answer is right, along with theta = 0
I forgot to mention that alpha cannot be 0 or 90 deg., so only 5pi/3 would be correct in that case. So I was right after all! :D Thank you for this. :)

EDIT: I accidentally talked about alpha, but I meant to say that theta is between 0 and 90 (not including them). My bad. :(
 
Last edited:

pikachu975

I love trials
Moderator
Joined
May 31, 2015
Messages
2,540
Location
NSW
Gender
Male
HSC
2017
Hi guys,

1.

(a) ***I do not need help with this part*** Consider the statement cos[(π/2) + A] = +/- sinA. For which sign is this statement true for all A? For which A is the statement true for both signs?

Answer: Negative, A = 0, nπ (n is an integer)

(b) Taking A = 5θ in (a), write down the value of B such that -sin5θ = cosB. Hence find the least value of θ between 0 and 2π such that sin5θ + cos8θ = 0.
I actually haven't gotten any answers for this ^^, as I have little idea of how to do the question.

2. Hence, or otherwise, solve 2cosθ + 2cos(θ + (π/3)) = 3, for 0 < θ < 2π. (This is the second part of the question. The first part was to express the expression on the LHS in the form Rcos(θ + α), where R>0 and α is between 0 and 90 degrees (not including them). The answer was 2√3cos(θ + (π/6)).) I keep getting 5π/3, and I don't know where I'm going wrong (I'm sure I'm wrong).

Thanks in advance! :)

P.S. π is actually pi, just before anyone asks, to prevent people wasting time asking this.
2root3 cos(x+pi/6) = 3
cos(x+pi/6) = root3 / 2
x+pi/6 = pi/6, 11pi/6
x = 0, 10pi/6
x = 0, 5pi/3 but x>0
x = 5pi/3
 

integral95

Well-Known Member
Joined
Dec 16, 2012
Messages
780
Gender
Male
HSC
2013
Does anyone know how to solve this question using "hence"? I'm guessing I would lose marks if I did otherwise in an exam situation, since it only says "hence".



I forgot to mention that alpha cannot be 0 or 90 deg., so only 5pi/3 would be correct in that case. So I was right after all! :D Thank you for this. :)
Alpha can't be zero, but it doesn't say that theta can't be zero.
 

bujolover

Active Member
Joined
Jan 5, 2017
Messages
154
Gender
Undisclosed
HSC
2018
Alpha can't be zero, but it doesn't say that theta can't be zero.
In Q2, which is what you're talking about, it clearly says in the question "0 < θ < 2π". Check. :) And it was my bad that I talked about alpha.
 
Last edited:

bujolover

Active Member
Joined
Jan 5, 2017
Messages
154
Gender
Undisclosed
HSC
2018
Welp I can't seem to use the "hence" method. But for otherwise I just did like


Sorry I got back to you on this one so late. The answer given is actually 3pi/26, which is nowhere near what you got... :/ I have no idea how to get that; it's a very obscure number. But what you've done looks right. I don't understand. *scratching my head*
 
Last edited:

integral95

Well-Known Member
Joined
Dec 16, 2012
Messages
780
Gender
Male
HSC
2013
Sorry I got back to you on this one so late. The answer given is actually 3pi/26, which is nowhere near what you got... :/ I have no idea how to get that; it's a very obscure number. But what you've done looks right. I don't understand. *scratching my head*
I guess we have to step it up a notch with your knowledge of general solution.
So the equation could be re-written as the cos functions.



Sub in k = 1 in the 2nd equation to get your solution.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top