Need Help from 3U Cambrige Book. (1 Viewer)

[currysauce]

Suppose that X is an acute angle and cosX=4/5. Using the results sin²x=1/2(1-cos2x) and cos²x=1/2(1+cos2x), find the exact value of:

i) cos(X/2)



New questions too, nothing to do with above,

If cosX=31/50 and is in the 4th quad. , find

cos X/2

 

[Slidey]

"Suppose that X is an acute angle and cosX=4/5. Using the results sin²x=1/2(1-cos2x) and cos²x=1/2(1+cos2x), find the exact value of:

i) cos(X/2)"

cos^2(X/2)=(1+cosX)/2
cos(X/2)=sqrt([1+cosX]/2)
cos(X/2)=sqrt([1+4/5]/)2
cos(X/2)=3/sqrt(10)

"If cosX=31/50 and is in the 4th quad. , find

cos X/2"

X/2 must be in the second quadrant, so the answer shall be negative.
cos(X/2)=-sqrt([1+cosX]/2)

cos(X/2)=-sqrt([1+31/50]/2)
cos(X/2)=-9/10

EDIT: Fixed an angle - I had cosA for some reason.

 

[currysauce]

thankyou good sir, more may follow

 

[currysauce]

new questions

Eliminate A from these parametrics.


(i) x= 2tan(A/2) y = cosA



(ii) x=3sinA y = 6sin2A


2. Write down exact. value of cos 45 = 1/root2

b) show that :cos 22.5 = 1/2 root (2+root(2))
c) show that cos 11.25 = 1/2 root(2+root(2 + root2)))

3. Show that root (8-4root(3)) = root 6-root2

 

[Slidey]

(i) x= 2tan(A/2) y = cosA

y=cosA=(1-t^2)/(1+t^2)
x=2t

Go from there.

(ii) x=3sinA y = 6sin2A

x^2=9sin^2(A)
y^2=144sin^2(A)cos^2(A)
y^2=144sin^2(A).(1-sin^2(A))
y^2=144sin^2(A)-144sin^4(A)
sin^2(A)=x^2/9
y^2=16x^2-16x^4/9
y^2=16(9x^2-x^4)/9
y=+4sqrt(9x^2-x^4)/3

EDIT: Thanks, m_isk

 

[m_isk]

(ii) x=3sinA y = 6sin2A


y^2=36sin^2(A)cos^2(A)
y^2=36sin^2(A).(1-sin^2(A))
y^2=36sin^2(A)-36sin^4(A)
sin^2(A)=x^2/9
y^2=4x^2-4x^4/9
y=2sqrt(x^2-x^4)/3

Slide, correct me if i'm wrong, but if y=6sin2A, then y=12sinAcosA
and y^2 =144sin^2(A)cos^2(A)

 

[Slidey]

Oh... more. OK.

"2. Write down exact. value of cos 45 = 1/root2

b) show that cos 22.5 = 1/2 root (2+root(2))
c) show that cos 11.25 = 1/2 root(2+root(2 + root2)))"

b)
22.5=45/2
cos(45/2)=sqrt([1+cos45]/2) (that identity again)
cos(22.5)=sqrt(2+sqrt2)/2 #

c)

11.25=22.5/2 Repeat above.

"3. Show that root (8-4root(3)) = root 6-root2"

(sqrt6-sqrt2)^2=(6+2)-2sqrt12=8-4sqrt3
.'. sqrt6-sqrt2=sqrt(8-4sqrt3) #

 

[currysauce]

Last question for 2day

Show that tan 165 = root3 - 2 [DONE THIS]


then hence, show that tan 82.5 = root(6) + root(3) + root(2) +2


THANKS FOR ALL YOUR HELP GUYS

 

[currysauce]

^^ anyone? thanks

 

[haboozin]

Last question for 2day

Show that tan 165 = root3 - 2 done


then hence, show that tan 82.5 = root(6) + root(3) + root(2) +2


THANKS FOR ALL YOUR HELP GUYS

tan165 = - tan(45 - 30)
=-((tan45 - tan30)/(1+tan45tan30))

then expand...

= sqrt(3) - 2


tan 82.5 = tan (45 + 30 + 12.5) and it is half the angle of 165..but i dont know from there.... im not too sure...?

 

[jumb]

tan165 = - tan(45 - 30)
=-((tan45 - tan30)/(1+tan45tan30))

then expand...

= sqrt(3) - 2


tan 82.5 = tan (45 + 30 + 12.5) and it is half the angle of 165..but i dont know from there.... im not too sure...?

You should be thinking of tan(82.5) = tan(165/2)

 

[FinalFantasy]

let 165=x
82.5=x\2
tanx=tan (2* (x\2))=2tan (x\2)\1-tan²(x\2)=sqrt(3)-2
2tan(x\2)=(1-tan²(x\2))(sqrt(3)-2)
2tan(x\2)=sqrt3-2-tan²(x\2)(sqrt3-2)
tan²(x\2)(sqrt3-2)+2tan(x\2)-(sqrt3-2)=0
tan(x\2)=-2+-sqrt[4+4(sqrt3-2)(sqrt3-2)]\2(sqrt3-2)
=-2+-sqrt[4+4(3-4sqrt3+4)]\2(sqrt3-2)
=-2+-sqrt(4+28-16sqrt3)\2(sqrt3-2)
=-2+-4sqrt(2-sqrt3)\2(sqrt3-2)
=-1+-2sqrt(2-sqrt3)\(sqrt3-2)
=1+-2sqrt(2-sqrt3)\(2-sqrt3)
=2+root3+2\sqrt(2-root3)
den by calculator 2\sqrt(2-root3)=root2+root 6
therefore
tan 82.5 = root(6) + root(3) + root(2) +2

i cheated a bit unless u can show algebraically 2\sqrt(2-root3)=root2+root 6

 

[Slidey]

"i cheated a bit unless u can show algebraically 2\sqrt(2-root3)=root2+root 6"

LHS=2sqrt(2-sqrt3)/(2-sqrt3)=sqrt(8-4sqrt3)/(2-sqrt3)=(sqrt6-sqrt2)(2+sqrt3)
=(2sqrt6+3sqrt3-2sqrt2-sqrt6)=sqrt6+sqrt2=RHS

 

[FinalFantasy]

"i cheated a bit unless u can show algebraically 2\sqrt(2-root3)=root2+root 6"

LHS=2sqrt(2-sqrt3)/(2-sqrt3)=sqrt(8-4sqrt3)/(2-sqrt3)=(sqrt6-sqrt2)(2+sqrt3)
=(2sqrt6+3sqrt3-2sqrt2-sqrt6)=sqrt6+sqrt2=RHS

hahaha
go captain!
now it's a complete proof=P

 

[haboozin]

damn, im so stupid, i should've read Slide_Rule's proofs from the previous questions, they solved it as tan(165/2)..

 

[bunnykun]

BLESS THIS POST OMG
[edit: needed caps to show how this post was a beacon of light and hope in a dark abyss otherwise known as 3u maths]