- #1

prove that psquared +qsquared +rsquared +2pqr =1

i need a comprehensive solution

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- Thread starter mathelord
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- #1

prove that psquared +qsquared +rsquared +2pqr =1

i need a comprehensive solution

- #2

The Bob

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Let us get the ball rolling. What have you done towards it?

The Bob (2004 ©)

The Bob (2004 ©)

- #3

HallsofIvy

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Looks like the cosine law to me.

- #4

Zurtex

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To me it quite clearly reads as:VietDao29 said:Do you mean:

if [tex]\arccos p + \arccos q + \arccos r = 180[/tex] then [tex]\sqrt{p} + \sqrt{q} + \sqrt{r} + 2pqr = 1[/tex]?

Maybe I am wrong, but if you mean that, it's a wrong problem.

p = 0.573576

q = 0.422618

r = 0.5

Am I missing something?

Viet Dao,

[tex]p^2 + q^2 + r^2 + 2pqr = 1[/tex]

- #5

VietDao29

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Dear me, how can I made such a stupid mistake?

Viet Dao,

Viet Dao,

- #6

HallsofIvy

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saying that "arc cos[p] +arc cos[q] +arc cos[r] =180" is simply saying that θ+φ+ψ= 180 or that θ, φ, ψ are angles in a triangle.

- #7

bao_ho

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acos(p radians)+acos(q radians)+acos(r radians) = 180 degrees = π radians?

- #8

bao_ho

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ϖ radians

- #9

bao_ho

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Π radians

- #10

bao_ho

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3.1415... radians

- #11

HallsofIvy

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- #12

bao_ho

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if the angles are given in degrees then

p*p + q*q + r*r +2*p*q*r != 1

p*p + q*q + r*r +2*p*q*r != 1

- #13

Gokul43201

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And why is that ? p, q, r are NOT angles !bao_ho said:if the angles are given in degrees then

p*p + q*q + r*r +2*p*q*r != 1

- #14

Gokul43201

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"arc cos" is simply the inverse cosine or "cos ^{-1}"

- #15

- #16

VietDao29

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No-one's going to do the homework for you. Here's a hint:

[tex]\alpha = \arccos p \Rightarrow \cos \alpha = p[/tex]

[tex]\beta = \arccos q \Rightarrow \cos \beta = q[/tex]

[tex]\zeta = \arccos r \Rightarrow \cos \zeta = r[/tex]

You have [tex]\alpha + \beta + \zeta = 180[/tex]

[tex]p^2 + q^2 + r^2 + 2pqr= \cos ^2 \alpha + \cos ^2 \beta + \cos ^2 \zeta + 2 \cos \alpha \cos \beta \cos \zeta= ?[/tex]

Viet Dao,

[tex]\alpha = \arccos p \Rightarrow \cos \alpha = p[/tex]

[tex]\beta = \arccos q \Rightarrow \cos \beta = q[/tex]

[tex]\zeta = \arccos r \Rightarrow \cos \zeta = r[/tex]

You have [tex]\alpha + \beta + \zeta = 180[/tex]

[tex]p^2 + q^2 + r^2 + 2pqr= \cos ^2 \alpha + \cos ^2 \beta + \cos ^2 \zeta + 2 \cos \alpha \cos \beta \cos \zeta= ?[/tex]

Viet Dao,

Last edited:

- #17

Gaz031

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You could start by doing:mathelord said:

prove that psquared +qsquared +rsquared +2pqr =1

i need a comprehensive solution

[tex]\cos ((arc cos p + arc cos q) + arc cos r)=-1[/tex]

[tex]\cos (arc cosp + arc cosq)cos(arc cosr) - \sin (arc cosp+arc cosq) \sin (arc cosr)=-1[/tex]

[tex]rpq-r\sin (arc cosp) \sin (arc cosq))-q\sin (arc cosp) \sin (arc cosr) - p\sin (arc cos q)\sin (arc cosr) = -1[/tex]

You can then use identities like [tex]\sin x =\pm \sqrt{1-cos^{2} x}[/tex] to give [tex]\sin (arc cosp) =\pm \sqrt{1-p^{2}}[/tex] and so tidy your expression to give the desired one.

- #18

geniusprahar_21

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=> cos(A+B) = cos(180 - c) => - cosC = cosa.cosb - sina.sinb. squaring both sides,

change all sin square to cos square. take things common, you'll get the answer. don't forget to use... cosA = p, cosB = q, cosC = r

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