IIT JEE , AIEEE Forum - Mathematics , Trignometry

rahulraj1

Find A/2 if it satisfy 2cosA/2=root under (1+sinA)+ root under (1-sinA)

LAMPARD

I think this is true for all A.
See,square on both sides,
you get 4cos²A/2=2+2cosA
=>2(1+cosA)=2(1+cosA)
which is true for all A and hence the given expression is true for all A.

sandeepramesh

no it isnt. u forgot to take mod when out of the sqrt :)

prateekdhariwal

ur method is wrong.....modulus should have been taken on squaring both the sides...

spideyunlimited

Find A/2 if it satisfy 2cosA/2=root under (1+sinA)+ root under (1-sinA)

2. cosA/2 = sqrt(1+sinA) + sqrt (1-sinA)

2. cosA/2 = sqrt(sin^2 A/2 + cos^2 A/2 + 2.sinA/2.cosA/2) + sqrt (sin^2 A/2 + cos^2 A/2 - 2.sinA/2.cosA/2)

2. cosA/2 = (sinA/2 + cos A/2) + (sinA/2 - cosA/2)
2.cosA/2 = 2.sinA/2
tan A/2 = 1

tan A/2 = tan pi/4

Now use general solution :)

A/2 = 2n

+-

/4

So A/2 = 2n

+

/4 , 2n

-

/4

:)

FAKE IT FAKE IT :P

Decoder

here is another way..

look..1 + sinA is nothing but [cosA/2 + sinA/2 ]² ..

so.

1+ sinA = / cosA/2 + sinA/2 / .... ( / is modulus)..

1 - sinA = / cosA/2 - sinA/2 /..

u open these modulus by applying simple logic ..

that when and how...for wat values of A/2.. inside mod values r positive...

ur main focus must be on "-" wala mod..

u have to get 2cosA/2..

thus u want minus wala mod to be opened as -- cosA/2 - sinA/2...

this can happen only wen cosA/2 > sinA/2.. (as mod values we want to be positive)..

so this happens wen A/2 [-/4,/4]..

so so A [-/2,/2]..

Decoder

hellooo!! is my ans. correct

rahulraj1

No the correct ans is2npi-pi/4,2npi+pi/4

spideyunlimited

EDITED MY ERROR

I FORGOT TO TAKE GENERAL SOLUTION OF TAN X = TAN @ in the last step :)

REST IS CORRECT :)

Ask & Discuss Questions with Community & Experts