IIT JEE , AIEEE Forum - Mathematics , Trignometry

spandana

1+cos²

+cos³

+cos⁴

+.........=4+

3

if 0<

<90. then

=

shakirshafi12

is cos Q missing between 1 and cos^2 Q if not then
it is an infinite Geometric progression from cos^2 Q
sum of inf G.P
1+cos^2 Q/(1-cos Q) = 4+sqroot(3)
cos^2 Q/ (1- cos Q) = 3+sqroot(3)
put 3+sqroot(3)=a;
cos^2 Q+a cos Q -a=0;
solve the eqn
cos Q=(sqroot(a^2+4a)-a)/2;
where a=3+sqroot(3)

prats7575

If 0<

<

/2 then every cos^2

, cos^3

, cos ^4

> 0.

Then at every step, the value will be increse. So the value will never decrese and for the infinite value, the ANS will be infinitive.

deepak_agarwal

the second post is absolutely rite.....the previous one is wrong

prats7575

sneha.bagri

I think first ans is correct because the value will decrease as the power increasesas cos

value is <1 as 0<

<90

sampathkaushik

if there's no cos

then we can add and subtract cos

to get the infinite GP as shakirshafi says.

since sum of infinite GP is a/(1-r), where r is the common ratio, which should be <1; and since cos in the given domain satisfies this criterion, the formula can be applied

(NOTE: I shall replace

with x for easier typing)

hence we get...

{ 1 + cosx + cos²x + cos³x + ......} - cosx = 4+

3

1/(1-cosx) - cosx = 4 +

3

cos²x + cosx (3 +

3) - (3 +

3) = 0
using the quadratic formula we get...

cosx ={ -3 -

3 +(12)^0.25 } /2 and { -3 -

3 -(12)^0.25 } /2

but since (12)^0.25 is less than 3+

3 for sure; hence both solutions obtained for cosx are negative;

but this isn't possible, as in the given domain, cosx is positive;

hence NO SOLUTION exists for the equation in the given domain...

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