If cos-7-cos3-7-cos5-7 are the roots of the equation 8x3-4x2-4x-1-0The value of -7-3-7-5-7-

The correct option is A 4 f(x)=8x^3 4x^2 4x+1 has roots =cosπ/7, cos3π/7, cos5π/7 Now, #160; #160; #160; #160; #160; #160; # ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VIII

Mathematics

Area of Polygons

If cosπ7,co...

Question

If $cos \frac{π}{7}, cos \frac{3 π}{7}, cos \frac{5 π}{7}$ are the roots of the equation $8 x^{3} - 4 x^{2} - 4 x + 1 = 0$
The value of $sec \frac{π}{7} + sec \frac{3 π}{7} + sec \frac{5 π}{7} =$

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $4$
$f (x) = 8 x^{3} - 4 x^{2} - 4 x + 1$

has roots $= cos \frac{π}{7}, cos \frac{3 π}{7}, cos \frac{5 π}{7}$

Now,
$sec \frac{π}{7} + sec \frac{3 π}{7} + sec \frac{5 π}{7}$

$= \frac{1}{cos \frac{π}{7}} + \frac{1}{cos \frac{3 π}{7}} + \frac{1}{cos \frac{5 π}{7}}$

$∴$ we need equation with roots $\frac{1}{x_{1}}, \frac{1}{x_{2}}, \frac{1}{x_{3}}$

Let $\frac{1}{x} = t$

$x = \frac{1}{t}$

$f (x) = \frac{8}{t^{3}} - \frac{4}{t^{2}} - \frac{4}{t} + 1 = 0$

$8 - 4 t - 4 t^{2} + t^{3} = 0$

$t^{3} - 4 t^{2} - 4 t + 8 = 0$

Sum of roots $= 4$

$∴ sec \frac{π}{7} + sec \frac{3 π}{7} + sec \frac{5 π}{7} = 4$

Suggest Corrections

Similar questions

$c o s \frac{π}{7}$ . $c o s \frac{3 π}{7}$ . $c o s \frac{5 π}{7}$ are the roots of the equation $8 x^{3} - 4 x^{2} - 4 x + 1 = 0$ . Then the value of $s e c \frac{π}{7} + s e c \frac{3 π}{7} + s e c \frac{5 π}{7}$ is

Q. If the roots of the equation

8 x^{3} - 4 x^{2} - 4 x + 1 = 0

are

cos \frac{π}{7}, cos \frac{3 π}{7}, cos \frac{5 π}{7}

, then show that

sec \frac{π}{7} + sec \frac{3 π}{7} + sec \frac{5 π}{7} = 4

$c o s \frac{π}{7}$ . $c o s \frac{3 π}{7}$ . $c o s \frac{5 π}{7}$ are the roots of the equation $8 x^{3} - 4 x^{2} - 4 x + 1 = 0$ . Then the value of $s i n \frac{π}{14} . s i n \frac{3 π}{14} . s i n \frac{5 π}{14}$ is

Q. If

cos \frac{2 π}{7}, cos \frac{4 π}{7}, cos \frac{6 π}{7}

are roots of the equation

8 x^{3} + 4 x^{2} - 4 x - 1 = 0

.The value of

cos \frac{π}{7} cos \frac{2 π}{7} cos \frac{3 π}{7}

is equal to

Q. The value of

cos 0 + cos \frac{π}{7} + cos \frac{2 π}{7} + cos \frac{3 π}{7} + cos \frac{4 π}{7} + cos \frac{5 π}{7} + cos \frac{6 π}{7} =

Join BYJU'S Learning Program

If cosπ7,cos3π7,cos5π7 are the roots of the equation 8x3−4x2−4x+1=0The value of secπ7+sec3π7+sec5π7=

The correct option is A 4f(x)=8x3−4x2−4x+1has roots =cosπ7,cos3π7,cos5π7Now, secπ7+sec3π7+sec5π7=1cosπ7+1cos3π7+1cos5π7∴ we need equation with roots 1x1,1x2,1x3Let 1x=tx=1tf(x)=8t3−4t2−4t+1=08−4t−4t2+t3=0t3−4t2−4t+8=0Sum of roots =4∴secπ7+sec3π7+sec5π7=4

If $cos \frac{π}{7}, cos \frac{3 π}{7}, cos \frac{5 π}{7}$ are the roots of the equation $8 x^{3} - 4 x^{2} - 4 x + 1 = 0$
The value of $sec \frac{π}{7} + sec \frac{3 π}{7} + sec \frac{5 π}{7} =$