If f x -cos x- cos x- cos x- sin x- sin x- sin x- sin 2- sin 2- sin 2-where - - - - - are real constant- then find the value of 2f - 4 -3f - 6 -f - 3

Expanding from R _ 3 f( x ) =[sin 2α #160; sin ((x+y) (x+β ))] sin 2β [sin[(x+y) (x+α )] +sin2y[sin ((x+β ) (x+α ))] f( x ) =sin2 α #160; s ...

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Byju's Answer

Standard XII

Mathematics

Definition of a Determinant

If f x =cos...

Question

If $f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos (x + α) & cos (x + β) & cos (x + γ) sin (x + α) & sin (x + β) & sin (x + γ) sin 2 α & sin 2 β & sin 2 γ \end{matrix} ∣ ∣ ∣ ∣ ∣$
where $α, β, γ$ are real constant, then find the value of $2 f (\frac{π}{4}) - 3 f (\frac{π}{6}) + f (\frac{π}{3})$

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Solution

Expanding from $R_{3}$
$f (x) = [sin 2 α sin ((x + y) - (x + β))] - sin 2 β [s i n [(x + y) - (x + α)] + sin 2 y [s i n ((x + β) - (x + α))]$
$f (x) = s i n 2 α sin (y - β) - sin 2 β sin (y - α) + sin 2 y sin (β - α)$
$f (x)$ doesn't depend on $x$
$2 f (\frac{π}{4}) - 3 f (\frac{π}{6}) + f (\frac{π}{3}) = 2 f - 3 f + f = 0$

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Similar questions

Q. Let

f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos (x + α) + i sin (x + α) & cos (x + β) + i sin (x + β) & cos (x + γ) + i sin (x + γ) sin (x + α) - i cos (x + α) & sin (x + β) - i cos (x + β) & sin (x + γ) - i cos (x + γ) sin (2 β - 2 γ) & sin (2 γ - 2 α) & sin (2 α - 2 β) \end{matrix} ∣ ∣ ∣ ∣ ∣

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Then the value of

f (α) + f (β) - 2 f (γ)

Q. If

f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos (x + α) & cos (x + β) & cos (x + γ) sin (x + α) & sin (x + β) & sin (x + γ) sin (β - γ) & sin (γ - α) & sin (α - β) \end{matrix} ∣ ∣ ∣ ∣ ∣

where

α, β, γ \in R

and

f (0) = - 2,

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30 \sum r = 1 | f (r) |

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If the value of determinant $∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & s i n (x + α) & c o s (x + α) 1 & s i n (x + β) & c o s (x + β) 1 & s i n (x + γ) & c o s (x + γ) \end{matrix} ∣ ∣ ∣ ∣ ∣$ is

Q. Let

f (x) = {sin}^{2} (x + α) + {sin}^{2} (x + β) - 2 cos (α - β) sin (x + α) sin (x + β)

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Q. If

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If f(x)=∣∣ ∣ ∣∣cos(x+α)cos(x+β)cos(x+γ)sin(x+α)sin(x+β)sin(x+γ)sin2αsin2βsin2γ∣∣ ∣ ∣∣where α, β, γ are real constant, then find the value of 2f(π4)−3f(π6)+f(π3)

Expanding from R3f(x)=[sin2αsin((x+y)−(x+β))]−sin2β[sin[(x+y)−(x+α)]+sin2y[sin((x+β)−(x+α))]f(x)=sin2αsin(y−β)−sin2βsin(y−α)+sin2ysin(β−α)f(x) doesn't depend on x2f(π4)−3f(π6)+f(π3)=2f−3f+f=0