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Need help with Core Mathematics 3

Just a couple of questions I'm struggling with, can anyone lend a hand? :redface:

-----------------
3) Find the function f'(x) where f(x) is:
f) (1+sinx) / (cosx)

I've used the quotient rule and got to (cos²x - sinx - sin²x) / cos²x

I can't figure out where to go from here though. The answer is apparently (1 + sinx) / cos²x but I don't know how to get to that... I'm probably just missing something really obvious or maybe made a stupid mistake...

-----------------
13) Solve, in the interval 0° (< or equal to) x < 360° , the following equations. Give answers to the nearest 0.1° .

a) 3cosx = 2sin(x +60°)
b) sin (x+30°) + 2sinx = 0
c) cos (x+25°) + sin (x+65°) = 1
d) cos x = cos (?+60°)
e) tan (x-45°) = 6tanx
f) sinx + cosx = 1

(replaced theta symbols with x's, answer will be in degrees)

I missed the lesson where things like this were tought but apparently even with the lesson it's still incredibly difficult to do. I'd like to give it a try though but right now it makes no sense to me. :frown:

Thanks in advance. :colondollar:

(edited 13 years ago)

Reply 1

Chemhistorian

4

Original post by Wolfiexe

Just a couple of questions I'm struggling with, can anyone lend a hand? :redface:

-----------------
3) Find the function f'(x) where f(x) is:
f) (1+sinx) / (cosx)

I've used the quotient rule and got to (cos²x - sinx - sin²x) / cos²x

I can't figure out where to go from here though. The answer is apparently (1 + sinx) / cos²x but I don't know how to get to that... I'm probably just missing something really obvious or maybe made a stupid mistake...

-----------------
13) Solve, in the interval 0° ? ? < 360° , the following equations. Give answers to the nearest 0.1° .

a) 3cos? = 2sin(? +60°)
b) sin (?+30°) + 2sin? = 0
c) cos (?+25°) + sin (?+65°) = 1
d) cos ? = cos (?+60°)
e) tan (?-45°) = 6tan?
f) sin? + cos? = 1

I missed the lesson where things like this were tought but apparently even with the lesson it's still incredibly difficult to do. I'd like to give it a try though but right now it makes no sense to me. :frown:

Thanks in advance. :colondollar:

For the trig, you want to get them all in to single trig functions, eg for divide through by cos to get 3 = 2tan(x+60)

Divide by 2: 3/2 = tan(x+60), solve as normal using a cast diagram and make sure you take 60 from your answers to get x :smile:

HOpe that helped!

Reply 2

Chemhistorian

4

And for f dont fall into the trap of sin^2x + cos^2x = 1, to square it youd have to square everything which isnt the same as sin^2x + cos^2x :smile:

Reply 3

Craiky1506

9

Original post by Wolfiexe

Just a couple of questions I'm struggling with, can anyone lend a hand? :redface:

-----------------
3) Find the function f'(x) where f(x) is:
f) (1+sinx) / (cosx)

I've used the quotient rule and got to (cos²x - sinx - sin²x) / cos²x

I can't figure out where to go from here though. The answer is apparently (1 + sinx) / cos²x but I don't know how to get to that... I'm probably just missing something really obvious or maybe made a stupid mistake...

-----------------
13) Solve, in the interval 0° ? ? < 360° , the following equations. Give answers to the nearest 0.1° .

a) 3cos? = 2sin(? +60°)
b) sin (?+30°) + 2sin? = 0
c) cos (?+25°) + sin (?+65°) = 1
d) cos ? = cos (?+60°)
e) tan (?-45°) = 6tan?
f) sin? + cos? = 1

I missed the lesson where things like this were tought but apparently even with the lesson it's still incredibly difficult to do. I'd like to give it a try though but right now it makes no sense to me. :frown:

Thanks in advance. :colondollar:

For question 3, I think it's maybe best to split the fraction up - so you get secx + tanx, then differentiate this (I'm assuming you've learnt how to differentiate these of course).

For question 13 - you need to learn the double angle formulae - once you do, it's just a case of putting what you have into the equations, simplifying and solving as normal.

Reply 4

ScottishShortiex

18

Original post by Wolfiexe

Just a couple of questions I'm struggling with, can anyone lend a hand? :redface:

-----------------
3) Find the function f'(x) where f(x) is:
f) (1+sinx) / (cosx)

I've used the quotient rule and got to (cos²x - sinx - sin²x) / cos²x

I can't figure out where to go from here though. The answer is apparently (1 + sinx) / cos²x but I don't know how to get to that... I'm probably just missing something really obvious or maybe made a stupid mistake...

[

3) f(x) = (1 + sinx) / (cos x)

f'(x) = (cos x)(cos x) - (1 + sin x)(-sinx) / cos^2 x (USING THE QUOTIENT RULE)

= cos^2 x - [-sin x - sin^2 x] / cos^2 x

= cos^2 x + sinx + sin^2 x / cos^2 x

= (1 + sinx)/ cos^2 x (since cos^2 x + sin^2 x = 1)

Reply 5

Craiky1506

9

Original post by Chemhistorian

For the trig, you want to get them all in to single trig functions, eg for divide through by cos to get 3 = 2tan(x+60)

Divide by 2: 3/2 = tan(x+60), solve as normal using a cast diagram and make sure you take 60 from your answers to get x :smile:

HOpe that helped!

This is wrong... sin(x+60)/cos(x) does not equal tan(x+60), it has to be cos(x+60) on the bottom as well for that to be true. The best way that I can see is to use double angle formulae and simplify.

Reply 6

Wolfiexe

OP

2

Ah, I messed up with the add and minus signs using the quotient rule I think, I'll also double check my differentation. Thanks =P

I'll look up the double angle formula and see what it is, I remember how to do the CAST diagram as well so I'll give that a go too.

Thanks alot guys, very fast responses! :biggrin:

Edit: Can anyone tell me what the double angle formula is? I googled it and got a few different formulas so I'm not sure on which one I should be paying attention to. :P

(edited 13 years ago)

Reply 7

Craiky1506

9

Original post by Wolfiexe

Ah, I messed up with the add and minus signs using the quotient rule I think, I'll also double check my differentation. Thanks =P

I'll look up the double angle formula and see what it is, I remember how to do the CAST diagram as well so I'll give that a go too.

Thanks alot guys, very fast responses! :biggrin:

Edit: Can anyone tell me what the double angle formula is? I googled it and got a few different formulas so I'm not sure on which one I should be paying attention to. :P

It depends on which question you're doing, for example in 13a you need the double angle formula for sin. But just to let you know, the double angle formulae are:

Sin(A+B) = SinACosB + CosASinB
Cos(A+B) = CosACosB - SinASinB (NOTE THE CHANGE OF SIGN!)
Tan(A+B) = (TanA + TanB)/(1-TanATanB)

Reply 8

ztibor

10

Original post by Wolfiexe

Just a couple of questions I'm struggling with, can anyone lend a hand? :redface:

-----------------
3) Find the function f'(x) where f(x) is:
f) (1+sinx) / (cosx)

I've used the quotient rule and got to (cos²x - sinx - sin²x) / cos²x

I can't figure out where to go from here though. The answer is apparently (1 + sinx) / cos²x but I don't know how to get to that... I'm probably just missing something really obvious or maybe made a stupid mistake...

-----------------
13) Solve, in the interval 0° ? ? < 360° , the following equations. Give answers to the nearest 0.1° .

a) 3cos? = 2sin(? +60°)
b) sin (?+30°) + 2sin? = 0
c) cos (?+25°) + sin (?+65°) = 1
d) cos ? = cos (?+60°)
e) tan (?-45°) = 6tan?
f) sin? + cos? = 1

I missed the lesson where things like this were tought but apparently even with the lesson it's still incredibly difficult to do. I'd like to give it a try though but right now it makes no sense to me. :frown:

Thanks in advance. :colondollar:

For a) You are almost right
THe derivative of denominator is (-sinx) and multiplying with the numerator and substracting them the sign will be posiitve

for B)
a. USe the addition formula
3cosx=2(sinxcos60+cosxsin60) (cos60=1/2 sin60=V3/2
(3-V3)cosx=sinx -> tanx
b.sin(x+30)+2sinx=0 similarly
V3/2sinx+1/2cosx=-2sinx
1/2cosx=(-2-V3/2)sinx ->tanx
c. cosx cos25 -sinx sin 25 +sinx cos65 +cosx sin65=1 considering that
sin65=cos25 and cos65=sin25
2cosx cos25 =1 ->cosx
d cosx=cos(x+60) it means
1st : x= (x+60) +k360 ->x
2nd: x=-(x+60) +k360 ->x
e. tan(x-45)=6tanx tan45=1 and tan(A-B)=(tanA-tanB)/(1+tanAtanB)
(tanx-1)/(1+tanx)=6tanx ->tanx quadratic
f.
sinx+cosx=1 so
(sinx+cosx)^2=1
sin^2x+2sinxcosx+cos^2x=1 as sin^2x+cos^2x=1
sinxcosx=0 ->sinx ->cosx

(edited 13 years ago)

Reply 9

anshul95

11

Well if you've learnt about sec x and its derivative yet the best thing thing to is to rewrite the function as sec x +tan x