Question

if abc are non zero real numbers and alfa beta are solutions of the equation a cos teta +b sin teta =c then sin alfa +sin beta = 2bc/a^2+ b^2

​

Answer 1

Answer:

We have acosθ=c−bsinθ

Squaring both sides

⇒a

2

(1−sin

2

θ)=c

2

−2bcsinθ+b

2

sin

2

θ

⇒(a

2

+b

2

)sin

2

θ−2bcsinθ+(c

2

−a

2

)=0

Since α and β are the values of θ as given,

∴ roots of the above equation are sinα and sinβ.

∴sinα+sinβ= Sum of roots =

a

2

+b

2

2bc

and sinα.sinβ= Product of roots =

a

2

+b

2

c

2

−a

2

Answer 2

Answer:

We are given that a, b, and c are non-zero real numbers and $\alpha$ and $\beta$ are the solution of the equation.

We are asked to prove that  $sin\,\alpha +sin\,\beta =\dfrac{2bc}{a^2+b^2}$

Also, from the given relation, we have

$a\,cos\,\theta + b\,sin\,\theta=c$

Now by squaring and change in terms $sin\,\theta$, we get

$a^{2} cos^2{\theta} +b^2 sin^2\theta =c^2$

Now we know that $\bold{cos^{2}\theta=1-sin^2\theta}$

Therefore, our equation becomes:-

$a^2(1-sin^2\theta)+b^2sin^2\theta=c^2\\a^2-a^2\,sin^2\,\theta+b^2\,sin^2\,\theta=c^2\\a^2+sin^2\,\theta(b^2-a^2)=c^2\\a^2+sin^2\,\theta(a^2+b^2-2bc)=c^2$
Further simplifying, we get

$a^2+a^2sin^2\theta+b^2sin^2\theta-2bc\,sin^2\theta=c^2\\a^2+(a^2+b^2)sin^2\,\theta-2bc\,sin^2\theta=c^2\\(a^2+b^2)sin^2\,\theta-2bc\,sin^2\theta+(c^2-a^2)=0$

Now we are given that Its roots are $sin\alpha$ and $sin\beta$ as $\alpha$  and $\beta$  are the values of $\theta$

Therefore, the sum of roots are

$sin\,\alpha +sin\,\beta =\dfrac{2bc}{a^2+b^2}$

Hence it has been proved that if a, b, and c are non-zero real numbers then $sin\,\alpha +sin\,\beta =\dfrac{2bc}{a^2+b^2}$ .

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