# Further Trig Questions (1 Viewer)

#### bujolover

##### Active Member
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.

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#### integral95

##### Well-Known Member
Welp I can't seem to use the "hence" method. But for otherwise I just did like

#### integral95

##### Well-Known Member
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.

• bujolover

#### bujolover

##### Active Member
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".

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! 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
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

• bujolover

#### bujolover

##### Active Member
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
Do you have any idea how to do 1(b) at all strictly using "hence"? #### pikachu975

##### I love trials
Moderator
Do you have any idea how to do 1(b) at all strictly using "hence"? I think the 'hence' means using the B value you found from -sin(5theta) = cosB so integral95 did it

#### integral95

##### Well-Known Member
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! Thank you for this. Alpha can't be zero, but it doesn't say that theta can't be zero.

• InteGrand

#### pikachu975

##### I love trials
Moderator
Alpha can't be zero, but it doesn't say that theta can't be zero.
Maybe by 'between 0 and 2pi' he means it's not included

• bujolover

#### bujolover

##### Active Member
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.

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#### bujolover

##### Active Member
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*

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#### integral95

##### Well-Known Member
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.