IIT JEE , AIEEE Forum - Mathematics , Trignometry -

BALGANESH

challenge to all experts

these sums will prove whether you are good in maths or not

do any one

x ^2 +5 /2 =x - 2cos( x - ) to prove ( + ) is an odd muttiple of 5

or

sin^4 /a + cos ^4/b = a = b

to prove sin ²ⁿ / a ^n-1 + cos ²ⁿ / = 1 /( a+b )^n-1

BALGANESH

waterdemon ,iitbipin where are you

rooney

Cant make head or tail of what you are asking. Ask clearly with proper typing.

rohith291991

ok second problem....rewrite as a quadratic in sin²A ...then we get (a+b)sin⁴A-2asin²A+a²/(a+b)=0 solving we get sin²A=a/(a+b) similarly solving by quadratic we get cos²A=b/(a+b) substituting in the second relation we get aⁿ⁺¹/(a+b)²ⁿ + bⁿ⁺¹/(a+b)²ⁿ =1/(a+b)^n-1 but since sin²A=a/(a+b) we get cos²A=1-sin²A=root(b²+2ab)/(a+b)=b/(a+b) hence 2ab=0

a=0 or b=0 now in the relation if we suppose a=0 then we get 1/(b)^n-1= 1/(b)^n-1which is always true....or if b=0 then we get 1/(a)^n-1= 1/(a)^n-1which is again always true.. hence the given relation is always true... hence proved....but im having a difficulty understanding if it is mathematically correct to multiply initially by ab on both sides as the value of ab is zero....can anyone explain this part?? btw balganesh plz be careful while typing...i had to guess the question from the symmetry of the problem...lots of typing errors...

rohith291991

anyone??

rohith291991

reply guys...,no one seems to be taking interest...

feynmann

Guessing the qn 1

complet the square . Apply| cos ( x )| <= 1

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