Can anyone help me with this trig question plz? (1 Viewer)

[CJ251]

Hi everyone. Hope ur all goin gud. Not long for the HSC.
Well i got this question for trig. show that if 0 < x < pie on four
then cos2x>2sin squared x minus one?

I cant write maths things here. thanks heaps if u cud help me.

 

[SoulSearcher]

You can either use 2sin^2 x - 1 or 2sin²x - 1 for notation like that.

As for your question, well I don't get "show that if 0 then".

 

[CJ251]

Hi everyone. Hope ur all goin gud. Not long for the HSC.
Well i got this question for trig. show that if 0 < x<pie on four
then cos2x>2sin squared x minus one?

I cant write maths things here. thanks heaps if u cud help me.[/QUOTE]

 

[CJ251]

You can either use 2sin^2 x - 1 or 2sin²x - 1 for notation like that.

As for your question, well I don't get "show that if 0 then".

Thanks the question came out wrong.
heres the question hope i get it right here.

Show that if 0 < x < pie on 4, then cosx > 2sin^2 x - 1

 

[SoulSearcher]

Well you can draw both graphs on the same set of axes and that should show that cosx > 2sin²x -1 for 0 < x < pi/4.

An alternate way is to covert 2sin²x -1 into -cos2x, and since cos2x is positive or 0 for the domain 0 < x < pi/4, -cos2x would be negative in that domain, and since cosx is positive for the domain 0 < x < pi/4, then cosx > -cos2x and hence that cosx > 2sin²x -1 for 0 < x < pi/4.

 

[BIRUNI]

instead showing cosx > 2sin^2 x - 1 you can show 2sin^x-1-cosx<0
so -2cos(3x/2)cos(x/2)<0

cos(3x/2)cos(x/2)<0 obviously when 0<x<pi/4 cos(3x/2) and cos(x/2) are bigger than zero so their addition namely cos(3x/2)cos(x/2) is bigger than zero and it is proven
there are so many other alternative ways let me see if i can type them later

 

[SoulSearcher]

That works, but I don't know if he knows sums and products of sine and cosine functions.

 

[BIRUNI]

method 2: x is between o and pi/4 then cosx is between 0.7071... and 1

for x between o and pi/4 sin x is between o and 0.7971... so 2sin^2x-1 is between o and -1
from these two results it is obvious that cosx>2sin^x-1

 

[BIRUNI]

method3: 2sin^x-1=-cos2x
for x between o and pi/4 cosx is between 1 and 0.7071..
for x between o and pi/4 cos2x is betwwen 1 and 0.3826.. so -cos2x is between -1 and -0.3826.. hence from these two cosx>-cos2x
or you could alternatively say that cosx+cos2x is obviously biiger than zero so the inequality holds

 

[BIRUNI]

are these enough?

but as soulsearcher mentioned, the simplest and quickes way is graphing.

 

[CJ251]

Sweet. I get it. Thanks heaps. I appreciate it.
Im not very good at 3 unit so i'll definitely ask more questions. lol. But i'll always try.
Thanks everyone, good luc studyin and hope u could help me l8r.

 

[CJ251]

Hey im back. I got another question. I HATE TRIG.
Okay.
Find all the angles x with 0 < or equal x < or equal to 2pie for which sinx + cosx = 1
Once again thank you for helping me.

 

[SoulSearcher]

Well, again, draw the graphs of sinx and cosx on the same set of axes, and you should notice that this sum only works when x = 0, x = pi/2 and x = 2pi.

 

[CJ251]

Wow that was quick. Thanks heaps. Your the best soulsearcher.

 

[BIRUNI]

there are so many ways such as t formula, subsidary angles, squaring boths sides and etc but I advise auxilary method
you can memorize this
asinx+bcosx=rsin(x+alpha)
asinx-bcosx=rsin(x-alpha)
acosx-bsinx=rcos(x+alpha)
acosx+bsinx=rcos(x-alpha)

r^2=a^2+b^2 but r>0
alpha=tan^(-1)(b/a)

 

[pLuvia]

You can use the auxiliary method, you can change sinx+cosx=1 into sqrt{2}sin(x+pi/4)=1, which then you solve for x, a much quicker method

 

[onebytwo]

Hey im back. I got another question. I HATE TRIG.
Okay.
Find all the angles x with 0 < or equal x < or equal to 2pie for which sinx + cosx = 1
Once again thank you for helping me.

- use t-formulae sinx=2t(1+t²) and cosx=(1-t²)/(1+t²)
remeber to test pi as a soln because if pi is an answer it wont show in your results when using the t formulae

- or transformations, let sinx+cosx= Rsin(x+@)
= Rsinxcos@+Rsin@cosx
then Rsin@=1 (eqtn 1)
then Rcos@=1 (eqtn 2)
from 1, we get sin@=1/R and from 2, we get cos@=1/R
from these we get tan@=1, so @=pi/4
squaring 1 and 2, then adding we get R²=2, so R=rt2
then sinx+cosx=rt2sin(x+pi/4)=1
then solving the underlined
sin(x+pi/4)=1/rt2
then x+pi/4=pi/4 or x+pi/4=3pi/4
so x=0 or x=2pi or x=pi/2

EDIT: way too slow

 

[Riviet]

there are so many ways such as t formula, subsidary angles, squaring boths sides and etc but I advise auxilary method
you can memorize this
asinx+bcosx=rsin(x+alpha)
asinx-bcosx=rsin(x-alpha)
acosx-bsinx=rcos(x+alpha)
acosx+bsinx=rcos(x-alpha)

r^2=a^2+b^2 but r>0
alpha=tan^(-1)(b/a)

You also need to know how to derive these results, as shown by onebytwo in the previous post.

 

[CJ251]

Sweet thanks. Okay jus a stupid question. For sin (theta - alfa) = 1/rt2
it equlas pi/4, 3pie/4, 9pi/4, 11pi/4,.......
Um how do we actually come to these results. I thought it was jus the first two but i cudnt get the rest.

 

[followme]

The general formula

([FONT=宋体]θ-[/FONT]α)= πn + (-1)ⁿsin^-1(1/√2)
where n is an integer
n=0 ([FONT=宋体]θ-[/FONT]α)= π/4
n=1 ([FONT=宋体]θ-[/FONT]α)=3π/4...

ususally the question will indicate a range eg 0≤x≤2pi, or it will ask u to find teh general solution.