Doubt in trigo

What if we aren't given the values? Because values of 0,30,45,60 and 90 degrees are known to us.

Then there's none other better option than to leave it at 2/sin18 - 2/sin54.

Waise, i have never been able to remember those values myself, nor have i ever bothered or tried to... :P, they're only 2b looked at when needed.

Because I saw this in an MCQ paper. Four options were given:

1 , 2, 3, 4

How do we know what the answer is, that too at class X level?

Chinta mat karo...u'll be asked to derive them in XIth :P

Smtimes the ppr setters presume tht u know the particular info, when u are actually not supposed to....so v can't do much about it, the best tht can be done is try n remember the values frm now onwards. :)

Okay.. Thank you so much! :)

Incase you don't remember the value of sin 18° then, calculate it as follows:

Let A = 18°
=> 5A = 90°
=> 3A + 2A = 90°
=> 3A = 90° - 2A
=> Sin 3A = Cos 2A [since , sin (90°-A) = cos A]
=> 3 sinA - 4sin³A = 1 - 2sin²A
=> -4sin³A + 2sin²A + 3 sinA - 1 = 0
=> 4sin³A - 2sin²A - 3 sinA + 1 =0

On putting sin A = x we get,

4x³ - 2x² -3x + 1 = 0

by trial 1 is a root of the above equation.. and so by facgtor theorem (x - 1) will be a factor
of the above eqn...

4x³ - 4x²+ 2x² - 2x - x + 1 = 0

=> 4x² (x - 1) + 2x (x - 1) - 1(x - 1) = 0
=> (x - 1)(4x² + 2x - 1) = 0

On solving you will get the value of x.

you'll get x = √5 - 1/4 or, 1 - √5 / 4
but since, x can't be negative....

so x = √5 - 1 / 4...

or sin 18° = √5 - 1 / 4...