# core 3 questions!!!!Watch

This discussion is closed.
#1
Hey. Have core 3 exam on Thursday so was just wondering if anyone could help me with these few questions. (I will rate if someone tells me how!!!!)

1. Found in part (i) that 5sinx + 12cosx = 13sin(x+63.78)
part (ii) asks to find the max and min points of f(x) where f(x)=30/(5sinx+12cosx). And to state the values of x (0<x<360) for which they occur.

2. The angle A is such that secA+tanA=2. Show that secA-tanA = 0.5 and hence find the exact value of cosA.

3. By using the substitution t=tan0.5x prove that cosecx-cotx =tan0.5x

4. The angle A(in radians) satifies the equation tan2A=sinA
(i) Show that either sinA=0 or 2cosA=cos2A

Thankyou
0
13 years ago
#2
(Original post by Looontag)
Hey. Have core 3 exam on Thursday so was just wondering if anyone could help me with these few questions. (I will rate if someone tells me how!!!!)

1. Found in part (i) that 5sinx + 12cosx = 13sin(x+63.78)
part (ii) asks to find the max and min points of f(x) where f(x)=30/(5sinx+12cosx). And to state the values of x (0<x<360) for which they occur.

2. The angle A is such that secA+tanA=2. Show that secA-tanA = 0.5 and hence find the exact value of cosA.

3. By using the substitution t=tan0.5x prove that cosecx-cotx =tan0.5x

4. The angle A(in radians) satifies the equation tan2A=sinA
(i) Show that either sinA=0 or 2cosA=cos2A

Thankyou
Is C3 like proper Hard?!?
0
#3
(Original post by Vijay1)
Is C3 like proper Hard?!?
yep
0
13 years ago
#4
(Original post by Looontag)
yep

Question 4(i) Just solve for A to get pi, or 180 degrees.
Then just substituite in SinA to get 0.
0
13 years ago
#5
For the first part max and min points occur when the sin() bit = 1 and -1 so just stick the numbers in for the answer.
0
13 years ago
#6
I'll try number 3:

3. By using the substitution t=tan0.5x prove that cosecx-cotx =tan0.5x
LHS = 1/sinx - 1/tanx = 1/sinx - cosx/sinx = (1 - cosx)/sinx

=> (1 - cosx)/2sin(x/2)cos(x/2)

=> (1 - [cos²(x/2) - sin²(x/2)])/2sin(x/2)cos(x/2)

=> [2sin²(x/2)]/2sin(x/2)cos(x/2)

=> sin(x/2)/cos(x/2) => tan(x/2) = RHS

I havent used the substitution, but I seem to have done it anyway.

Phew.
0
13 years ago
#7
I did number 3 entirely different! (still quite tough though) and I did use the substitution!

3. By using the substitution t=tan0.5x prove that cosecx-cotx =tan0.5x

Take it as 2 parts. i.e. call cosec x - A and cot x – B

Part B is easier to start with (as you can use it to simplify part A later on.
cot x = 1 / tan x = 1 / tan(2 x 0.5x)
cot x = 1 / (2 tan 0.5x / 1 – (tan 0.5x)2)
Replace by t
cot x = 1/ (2t / 1- t2)
cot x = (1- t2)/2t

A - cosec x. This is the toughest. Using 1 + cot2 x = cosec2 x and root all:
cosec x = (1 + cot2 x) ^ 0.5
cosec x = (1 + [(1- t2)/2t]2) ^ 0.5 (see how part B is handy!)
cosec x = (1 + (t4 – 2t2 + 1)/4t2) ^ 0.5
cosec x = (4t2/4t2 + (t4 – 2t2 + 1)/4t2) ^ 0.5
cosec x = ( (t4 + 2t2 + 1)/4t2 ) ^ 0.5
cosec x = ( [(t2 + 1)^2] / [2t ^ 2] ) ^ 0.5
cosec x = (t2 + 1) / 2t

Now do A – B
cosec x – cot x = (t2 + 1) / 2t - (1- t2)/2t = 2t2 / 2t = t = tan0.5x

Hope this helps.
I have done questions 2 and 4. will post in a minute.
0
13 years ago
#8
1. Found in part (i) that 5sinx + 12cosx = 13sin(x+63.78)
part (ii) asks to find the max and min points of f(x) where f(x)=30/(5sinx+12cosx). And to state the values of x (0<x<360) for which they occur.

once you have part (i),
(ii) f(x) = 30/(5sinx+12cosx) = 30/ 13sin(x+63.78)

since sin of anything cannot occur greater than 1, 30/13(1) is the maximum
since sin of anything cannot occur less than -1, 30/13(-1) is the minimum

if you picture the sine curve between 0 and 360, 1 occurs at sin90 and -1 at sin270

putting these values into the formula to find out the corresponding x's, x-63.78=90 and x-63.78=270

you should be able to go from there
0
13 years ago
#9
2. The angle A is such that secA+tanA=2. Show that secA-tanA = 0.5 and hence find the exact value of cosA.

sec A + tan A = 2
(sec A + tan A) (sec A - tan A) = 2(sec A – tan A)
sec2A – tan2A = 2(sec A – tan A)
1 = 2(sec A – tan A)
0.5 = (sec A – tan A)

sec A + tan A = 2
sec A – tan A = 0.5

using simultaneous methods (not sure if your allowed to do it!) add the 2 equations:

2sec A = 5/2
sec A = 5/4
cos A = 4/5

4. The angle A(in radians) satifies the equation tan2A=sinA
(i) Show that either sinA=0 or 2cosA=cos2A

This was surprisingly easy:

tan2A = sinA

sin2A / cos2A = sinA
sin2A = sinA x cos2A

sin 2A = 2sinAcosA

2sinAcosA = sinA x cos2A (divide through by sin A)
2cosA = cos2A

Also, since we divided through by sinA, it means that one solution is sinA = 0.

I will do no.1 tomorrow (I am very tired now)
Goodnight all
1
13 years ago
#10
4. The angle A(in radians) satifies the equation tan2A=sinA
(i) Show that either sinA=0 or 2cosA=cos2A

tan2A = sin2A/cos2A
= 2(sinA)cosA/cos²A+sin²A
=(sinA)

sinA=0 (ie 0=0) or:

.2cosA/cos²A-sin²A = 1
2cosA=2cos²A-1
2cosA=cos2A

2cosA/
0
X
new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Cranfield University
Cranfield Forensic MSc Programme Open Day Postgraduate
Thu, 25 Apr '19
• University of the Arts London
Open day: MA Footwear and MA Fashion Artefact Postgraduate
Thu, 25 Apr '19
• Cardiff Metropolitan University
Sat, 27 Apr '19

### Poll

Join the discussion

#### Have you registered to vote?

Yes! (24)
50%
No - but I will (1)
2.08%
No - I don't want to (1)
2.08%
No - I can't vote (<18, not in UK, etc) (22)
45.83%