trig, thx (2 Viewers)

[chewy123]

find the value b if
b=cos(2b - 30)

 

[shaon0]

find the value b if
b=cos(2b - 30)

b=cos(2b - 30)
b=cos2b.cos30-sin2b.sin30
b=(sqrt(3)(2cos^2(b)-1)-2cosb.sinb)/2
2b + 2cosb.sinb=(sqrt(3)(2cos^2(b)-1)
2b + 2cosb.sinb=(sqrt(3)(cos^2(b)-sin^2(b))

 

[bored of sc]

cos(2b-30)
= cos2b.cos30 + sin2b.sin30
= (cos²b - sin²b).[root 3]/2 + 2.sinb.cosb.1/2
= [root3]/2.(cosb-sinb)(cosb+sinb) + sinb.cosb
= b

No idea how to find the value of b though.

 

[shaon0]

cos (2b - 30)
= cos2b.cos30 + sin2b.sin30
= (cos²b - sin²b).[root 3]/2 + 2.sinb.cosb.1/2
= [root3]/2.(cos²b-(1-cos² b)) + sinb.cosb
= [root3]/2.(2cos²b-1) + sinb.cosb
= [root3]/2.((root2.cosb-1)(root2.cosb+1) + sinbcosb

No idea how to find the value of b though .

Yeah this is a weird question...i don't know how to do it either.

 

[bored of sc]

Yeah this is a weird question...i don't know how to do it either.

Maybe the value of b has b's in it.

If that's the case:
b = [root3]/2.(cosb-sinb)(cosb+sinb) + sinb.cosb

 

[Aerath]

Maybe the value of b has b's in it.

If that's the case:
b = [root3]/2.(cosb-sinb)(cosb+sinb) + sinb.cosb

Don't think you can do that.
Is this even 2U? =\

 

[bored of sc]

Don't think you can do that.
Is this even 2U? =\

Good point. It's 3U.

 

[lolokay]

the answer is 0.8809896803

(guess and check. I have no idea how to solve this.)

 

[chewy123]

lol, sry, sinb=cos(2b - 30)

 

[lolokay]

lol, sry, sinb=cos(2b - 30)

sinb = sin(90 - 2b + 30) = sin(180-b)
b = 120 - 2b or 180 - b = 120 - 2b
3b = 120
b = 40

or 60 = -b
b = -60

actually I think it's 40 +-120k, -60 +-360k

 

[Aerath]

lol, sry, sinb=cos(2b - 30)

LOL - you had me stumped for ages there.

 

[shaon0]

lol, sry, sinb=cos(2b - 30)

far out! now it makes sense, easy question lol.

 

[kingnothing999]

hey

does any1 know how the calculation of angles of any magnitude works.

thanks

 

[lyounamu]

^How do you want this explained?

Post a question that you don't know. I shall respond to that.

 

[lolokay]

hey

does any1 know how the calculation of angles of any magnitude works.

thanks

any magnitude? like, using taylor/maclaurin series
well, if that's what you mean there's a fairly simple explanation of it here: http://www.intmath.com/Series-expansion/Infinite-series-expansion.php

 

[kingnothing999]

well like these question:

find all quadrants where:

a) sin x > 0
b) tan x < 0 and cos x > 0

its got to do with the ASTC (All Stations To Central) rule.

quadrant 1= ALL
quadrant 2= SIN+
quadrant 3= TAN+
quadrant 4= COS+

I hope you know what i mean coz i dont know how to draw the graph on here.

 

[lyounamu]

well like these question:

find all quadrants where:

a) sin x > 0
b) tan x < 0 and cos x > 0

its got to do with the ASTC (All Stations To Central) rule.

quadrant 1= ALL
quadrant 2= SIN+
quadrant 3= TAN+
quadrant 4= COS+

I hope you know what i mean coz i dont know how to draw the graph on here.

a) sin x > 0 in the first and the second quadrants.

b) tan x < 0 in 2nd and 4th quadrants and cos x > 0 in the first and the fourth quadrants. Therefore, the quadrant that satisfies both is the fourth quadrant.

Yeah, memorise the All Stations To Chinese (olympic) rule.

 

[kingnothing999]

yeh, but im not really sure how it works
like for eg:

tan 150 degrees will be negative (quad 2)
cos 225 degrees will be negative (quad 3)
sin 120 degrees will be positive (quad 2)
cot 300 degrees will be negative (quad 4)

this may sound stupid, but like how do you figure out which quadrants they are in

hope i dont bother you too much

 

[tacogym27101990]

the first quadrant is 0 to 90
the 2nd quad is 90 to 180
the third is 180 to 270
and the fourth is 270 to 360

 

[kingnothing999]

ok... thanks that sorta helps

um can you help me with this question

question:
tan 50degrees

answer:
tan 150degrees= tan (180-30)
acute angle=30degrees
tan is negative in quadrant 2

(I dont get how it is in quad 2, thats the only prob)

can someone plz help out here?