Trigonometry (1 Viewer)

[tommykins]

Re: 回复: Re: Trigonometry

Isn't it f''(x)?

So what the hell is the point like if its f''(x) = 0 but the concavity does not change?

f''(x) = 0 inflextion point
f'(x) = 0 stationary point.

an inflextion point doesn't necessarily have to be a stationary point (that is, it doesn't have to have 0 gradient).

if it's f'(x) = f''(x) = 0 then it's a HORIZONTAL inflextion point.

also if concavity does no change then it is called a cusp, image here - http://curvebank.calstatela.edu/index/cusp.gif

 

[Aerath]

Re: 回复: Re: Trigonometry

For a point of inflexion, the gradient on both sides of the POI have to be the same sign, right? Or am I confusing it with something else?

 

[lyounamu]

Re: 回复: Re: Trigonometry

For a point of inflexion, the gradient on both sides of the POI have to be the same sign, right? Or am I confusing it with something else?

Yeah.

To tommykins: so what happens if you have a stationary point but there is no change in gradient? It is potentially a horizontal point of inflection, yeah?

 

[Wassup?]

That's what I was told by my teacher. Crossing out parts of a questions makes it appear as if it was a mistake and was never supposed to be part of the solution. He is a bit of a Nazi marker I guess.

 

[Aplus]

I got more problems for trig.

Given tanA= 1/2 and tanB=-2, where 180<270<360.
find the value of (secA-secB)/(cscA-cscB)

given 3sec² + 6tan² = 4 and 90<x><180.
Find the value of sinx+tanx.

given sinx = 2ab/ a²+b², where a<0
Express cosx and tanx in terms of a and b.

Given sin²x/(s+7cos²x = 1/9 where 270<x><360.
Find the value of tanx.

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[lyounamu]

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Good job, I totally forgot about that question.

 

[Aplus]

Apologies for the size of the image lol...

 

[tommykins]

That's what I was told by my teacher. Crossing out parts of a questions makes it appear as if it was a mistake and was never supposed to be part of the solution. He is a bit of a Nazi marker I guess.

Checked with a few sources.

You are correct.

Thanks for that insight.

 

[Aerath]

Apologies for the size of the image lol...

You're really anal about the neatness of your work eh?
You draw the bloody triangles with a ruler. (Except for the last one).

 

[Wassup?]

Checked with a few sources.

You are correct.

Thanks for that insight.

Yep, no probs. If you want to cross out an answer, do it really lightly.

 

[Aplus]

You're really anal about the neatness of your work eh?
You draw the bloody triangles with a ruler. (Except for the last one).

lol last one just rushed it.

 

[Fortian09]

lolz more trig stuff

simplify

cosxcos (x-(pi/4)) - sinxsin((pi/4)-x)

given tan A = 1/3 and tan B = 1/4, where A and B are acute angles. without using a calculator, find the value of sin(A-B)

given sin A + cosA = 2/5. find the values of:
a) sin2A
b)sin4A
c) sin⁴A + cos ⁴

 

[lolokay]

lolz more trig stuff

simplify

cosxcos (x-(pi/4)) - sinxsin((pi/4)-x)

given tan A = 1/3 and tan B = 1/4, where A and B are acute angles. without using a calculator, find the value of sin(A-B)

given sin A + cosA = 2/5. find the values of:
a) sin2A
b)sin4A
c) sin⁴A + cos ⁴

cosxcos (x-(pi/4)) - sinxsin((pi/4)-x)
= cosx(cosx/rt2 + sinx/rt2) - sinx(cosx/rt2 - sinx/rt2)
= ( cos²x + cosxsinx + sin²x - cosxsinx )/rt2
= 1/rt2

given tan A = 1/3 and tan B = 1/4, where A and B are acute angles. without using a calculator, find the value of sin(A-B)

tanA = 1/3
so sinA = 1/rt10, cosA = 3/rt10

tanB = 1/4
so sinB = 1/rt17, cosB = 4/rt17

sin(A-B)
= sinAcosB - sinBcosA
= 4/rt170 - 3/rt170
= 1/rt170 (calculator confirms)

given sin A + cosA = 2/5. find the values of:
a) sin2A
b)sin4A
c) sin⁴A + cos ⁴

(sin A + cosA)²
= s² + 2sc + c²
= 1 + 2sc = 4/25
2sc = sin2A = -21/25

4A = 2arcsin(-21/25)
sin4A = approx -0.91

(s+c)⁴ = s⁴ + 4s³c + 6s²c² + 4c³s + c⁴
= s⁴ + c⁴ + 2(2s³c + 2sc³ + 4s²c²) - 2s²c² = 16/625
2sc(s+c)² = -21/25(4/25)
= 2s³c + 2sc³ + 4s²c² = -84/625

s⁴ + c⁴ + 2(-84/625) - (441/625)/2 = 16/625
s⁴ + c⁴ = 809/1250

or just use A = 1/2 arcsin(-21/25) and plug into calculator

 

[Fortian09]

cambridge qn... lol
soo hard

tan 35 + tan 10 + tan35tan10=1 (looks easy, probs easy but i think not!)

 

[Mark576]

tan(35+10) = 1 = (tan 35 + tan 10)/(1 - tan35tan10)

&there4; LHS = 1 - tan35tan10 + tan35tan10 = 1 = RHS, as required.

 

[Fortian09]

is arcsin the same as sin^-1?

 

[tommykins]

yes

 

[Fortian09]

ahh icic thanks heaps tommy

hmm is anyone here good at gen equations?
coz i got some so if anyone's interested plz post

 

[Fortian09]

gee i hope people still post on this thread...
I have a few questions regarding trig. even though we're in yr 12 now...

umm ok...
I'm stupid so i dun get this qn...

1. There's the x and y axis, then in the second quad a line i drawn from origin to the coordinates P(-3,4) and the distance is 5, the angle from line to 1st quad x axis is Theta, find the values of sin, cos, tan, sec, cosec and cot (all in sin theta, cos theta, etc)

2. Determine the quads in which the angle Theta lies for each f the following cases.
a) cos Theta/ tan Theta > 0

 

[tommykins]

It is a triangle in the second quadrant.

In this quadrant, sinx and cosecx > 0 and cosx, tanx, cotx and secx < 0

Drawing the triangle, we can see that -
sin @ = 4/5
cosec@ = 5/4
cos@ = -3/5
sec@ = -5/3
tan@ = 4/3
cot@ = 3/4