Question

If sin x and cos x are the root of the equation ax^2-bx-1=0 then find the relation a and b

Answer 1

Solution

ax²-bx-1=0

=>ax²+(-b)X+(-1)=0

now the solution of the equation

ax²+bx+c=0

$x(1) = \frac{ - b + \sqrt{b {}^{2} - 4ac } }{2a} \\ x(2) = \frac{ - b - \sqrt{b {}^{2} - 4ac } }{2a} \\ for \: the \: above \: equation \\ x(1) = \frac{ - ( - b) + \sqrt{( - b) {}^{2} - 4 \times a \times ( - 1) } }{2a} = \\ \: \: \: \: \: \: \: \: \: = \frac{b + \sqrt{b {}^{2} + 4a } }{2a} = sin(a)\\ similarly....... \\ x(2) = \frac{b - \sqrt{b {}^{2} + 4a } }{2a} = \cos(a) \\ now.... \\ \sin {}^{2} (a) + \cos {}^{2} (a) =(\frac{b + \sqrt{b {}^{2} + 4a } }{2a} ) {}^{2} + (\frac{b - \sqrt{b {}^{2} + 4a } }{2a} ) {}^{2} \\ = > 1 = \frac{(b + \sqrt{b {}^{2} + 4a } ) {}^{2} }{4a {}^{2} } + \frac{(b - \sqrt{b {}^{2} + 4a } ) {}^{2} }{4a {}^{2} } \\ = > 1 = \frac{(b + \sqrt{b {}^{2} + 4a }) {}^{2} + (b - \sqrt{b {}^{2} + 4a} ) {}^{2} }{4a {}^{2} } \\ = > 4a {}^{2} = 2(b {}^{2} + b {}^{2} + 4a ) \\ = > 2a {}^{2} = 2b {}^{2} + 4a \\ = > 2a {}^{2} </u><u>-</u><u> </u><u>4a</u><u> </u><u>-</u><u> </u><u>2 b {}^{2} </u><u>=</u><u>0</u><u>\\ = > a {}^{2} </u><u>-2a</u><u> </u><u>-</u><u> b {}^{2} = 0$

here the algebraic formula which is used is,

(x+y)²+(x-y)²=2(x²+y²)

Answer 2

Answer:

a² - 2a - b² = 0

Step-by-step explanation:

Given---> Sinx and Cosx are root of the equation

ax² - bx - 1 = 0

To find ---> Relation between a and b

Solution----> ATQ,

ax² - bx - 1 = 0

We know that ,

Sum of roots = - coefficient of x / Coefficient of x²

Product of roots = Constat term / Coefficient of x²

ATQ, Roots of equation are Sinx and Cosx , so

Sum of roots = - ( -b / a )

=> Sinx + Cosx = b / a ........................(1)

Product of roots = - 1 / a

=> Sinx Cosx = - 1 / a .............................(2)

Now from equation (1)

Sinx + Cosx = b / a

Squaring both sides we get,

=> ( Sinx + Cosx )² = ( b / a )²

We have an identity as follows

( a + b )² = a² + b² + 2ab , applying it here we get

=> Sin²x + Cos²x + 2 Sinx Cosx = b²/a²

We have a formula as follows,

Sin²θ + Cos²θ = 1 , applying it we get,

=> 1 + 2 ( -1 / a ) = b² / a²

=> 1 - ( 2 / a )= b² / a²

Multiplying equation by a² , we get

=> a² - a² ( 2 / a ) = a² ( b² / a² )

=> a² - 2 a = b²

=> a² - 2a - b² = 0