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Mathematics 18 Online
OpenStudy (anonymous):

prove that cos θ = 1/2 ( a + 1/a).

OpenStudy (lgbasallote):

i didnt know it was possible to turn a variable into another without a conversion thingy...

mathslover (mathslover):

is this only the question given? Please check it again

OpenStudy (athe):

What about "a"? What kind of variable?

OpenStudy (anonymous):

yea .. it is the way it is given in my note .. but what do you think is right ?

OpenStudy (anonymous):

@athe i dont know !

mathslover (mathslover):

well i dont think that it is right :

mathslover (mathslover):

Since we have a variable "a" in RHS .. the main question arises that what kind of variable "a" is right ?

OpenStudy (anonymous):

okay , then what do you think is the right question ?

mathslover (mathslover):

there are many possible questions .. hence i can not tell ..

OpenStudy (anonymous):

i dont know .. i just started with my trig lessons !

OpenStudy (anonymous):

:/

OpenStudy (athe):

if

OpenStudy (athe):

for example, if \[a=10 \] this equation senselessly.

hartnn (hartnn):

a might be something like tan(theta/2)...some half angle trigo function...

OpenStudy (anonymous):

i dont know .. i might very well close this question then !

OpenStudy (anonymous):

multiply first.

OpenStudy (anonymous):

thanks for the help anyway guys !

OpenStudy (anonymous):

I think it is given

OpenStudy (anonymous):

what is given ?

OpenStudy (anonymous):

and maybe problem is to find \[a^2+\frac{1}{a^2}\] or something like this

OpenStudy (anonymous):

@mukushla if the question is that how'd you do it ?

OpenStudy (anonymous):

square both sides

OpenStudy (anonymous):

by using 2cos^2 θ - 1 formula ?

OpenStudy (anonymous):

idk what is the original problem...but if \[a+\frac{1}{a}=2 \cos \theta\]then\[a^2+\frac{1}{a^2}+2=4 \cos^2 \theta\]

OpenStudy (anonymous):

so 4cos θ = 2 cos^2 2θ - 1 ?

OpenStudy (anonymous):

u dont need that formula

OpenStudy (anonymous):

then how do you do it ?

OpenStudy (anonymous):

its almost done....

OpenStudy (anonymous):

tnx for the help !

OpenStudy (anonymous):

np :)

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