Trig Ratios (1 Viewer)

[Lukybear]

How do you find tan105 as a exact value? Please some1 tell me.

 

[lychnobity]

Tan (75 + 30)

But, first find tan 75

 

[Lukybear]

Is there a rule for that?? I dont get it??

 

[Lukybear]

Also sin2x=cosx. Ive never dont that as well, how are we suppose to do it?

 

[hermand]

Tan (75 + 30)

But, first find tan 75

nah, tan(45+60)

Is there a rule for that?? I dont get it??

do you know double angle rules, etc? i'll do a worked solution for you in my next post.

 

[Lukybear]

I dont think so. Our prelim course has only done up to basic trig i.e. 1+tan^2x=sec^2x

Also; how do you do this?
sqrt3cotx+1=cosecx

 

[hermand]

 

[Lukybear]

thxs

 

[addikaye03]

Let all values be in terms of degrees, no denotion

tan(105)=tan(45+60)

=tan45+tan60/(1-tan45tan60

Now from our exact value triangles tan45=1 and tan60=rt 3

therefore (1+rt3)/(1-rt3) #

Im almost 100% sure that would get full marks but just incase i would consider rationalising it:

(1+rt3)(1+rt3)/(1+rt3)(1-rt3)

=4+2rt3/-2

=-2-rt3

Edit: Ah, beat me to it

 

[kurt.physics]

Also; how do you do this?
sqrt3cotx+1=cosecx

This is an example of a transcendental equation. The best method, and the method which you would be required to use in the HSC, is to graph the two functions, then estimate their values from the intersection of the graph.

 

[hermand]

I dont think so. Our prelim course has only done up to basic trig i.e. 1+tan^2x=sec^2x

Also; how do you do this?
sqrt3cotx+1=cosecx

root all on the lhs or what part?

 

[hermand]

Also sin2x=cosx. Ive never dont that as well, how are we suppose to do it?

[double angle rule]

you can't cancel the cos's out because they can equal zero.


&
when cosx=0, x=90, 270. [did that without calculator, might want to check it].
when 2sinx-1=0 --> 2sinx=1 --> sinx=1/2 then x=30, 150.

 

[addikaye03]

I dont think so. Our prelim course has only done up to basic trig i.e. 1+tan^2x=sec^2x

Also; how do you do this?
sqrt3cotx+1=cosecx

rt3(cosx/sinx)+1=1/sinx

(rt3cosx+sinx)/sinx=1/sinx

rt3cosx+sinx=1

Now we can use the tranformations form, where acosx+bsinx=Rcos(@-alpha)

R=rt(a^2+b^2)

tan (alpha)=b/a

Have a go, if you can't do it then i'll help you out

 

[addikaye03]

This is an example of a transcendental equation. The best method, and the method which you would be required to use in the HSC, is to graph the two functions, then estimate their values from the intersection of the graph.

lol riiiiigggghhhhttt

 

[Lukybear]

ahh thxs thxs. My school still isnt up to that yet.

 

[gurmies]

For tan105, I would probably call it -tan75 and then - tan (30 + 45)...for some reason I didn't see that 105 = 60 + 45 -.-

 

[addikaye03]

For tan105, I would probably call it -tan75 and then - tan (30 + 45)...for some reason I didn't see that 105 = 60 + 45 -.-

haha aw alex... you can think of some very efficent and clean methods...this wasn't one of them man lol

 

[hermand]

For tan105, I would probably call it -tan75 and then - tan (30 + 45)...for some reason I didn't see that 105 = 60 + 45 -.-

i'm obviously simple-minded =pp.

 

[gurmies]

haha aw alex... you can think of some very efficent and clean methods...this wasn't one of them man lol

Every dawg has it's day xD. I remember this was in one of our prelim exams, and I did it the 45 + 60 way...Didn't get any wiser -.-

 

[addikaye03]

Every dawg has it's day xD. I remember this was in one of our prelim exams, and I did it the 45 + 60 way...Didn't get any wiser -.-

hahaha you know that i know your great ability at maths man, im just joking.