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2 sin ^(2) "" (3pi)/(4) + 2 cos ^(2) "" ...

`2 sin ^(2) "" (3pi)/(4) + 2 cos ^(2) "" (pi)/( 4) + 2 sec ^(2) "" (pi)/(3) =10`

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2 sin ^(2) "" (pi)/(6) + cosec ^(2) "" (7pi)/(6) cos ^(2) "" (pi)/(3) = 3/2

sin ^(2) " (pi)/(6) + cos ^(2) "" (pi)/(3) - tan ^(2) " (pi)/(4) =- 1/2

sin ^2 ""(2pi )/(7) + sin ^2"" (3pi )/(14) + sin ^2 ""(11 pi)/(14) + sin ^2 ""(5 pi)/(7) is

To find the sum sin^(2) ""(2pi)/(7) + sin^(2)""(4pi)/(7) +sin^(2)""(8pi)/(7) , we follow the following method. Put 7theta = 2npi , where n is any integer. Then " " sin 4 theta = sin( 2npi - 3theta) = - sin 3theta This means that sin theta takes the values 0, pm sin (2pi//7), pmsin(2pi//7), pm sin(4pi//7), and pm sin (8pi//7) . From Eq. (i), we now get " " 2 sin 2 theta cos 2theta = 4 sin^(3) theta - 3 sin theta or 4 sin theta cos theta (1-2 sin^(2) theta)= sin theta ( 4sin ^(2) theta -3) Rejecting the value sin theta =0 , we get " " 4 cos theta (1-2 sin^(2) theta ) = 4 sin ^(2) theta - 3 or 16 cos^(2) theta (1-2 sin^(2) theta)^(2) = ( 4sin ^(2) theta -3)^(2) or 16(1-sin^(2) theta) (1-4 sin^(2) theta + 4 sin ^(4) theta) " " = 16 sin ^(4) theta - 24 sin ^(2) theta +9 or " " 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta -7 =0 This is cubic in sin^(2) theta with the roots sin^(2)( 2pi//7), sin^(2) (4pi//7), and sin^(2)(8pi//7) . The sum of these roots is " " sin^(2)""(2pi)/(7) + sin^(2)""(4pi)/(7) + sin ^(2)""(8pi)/(7) = (112)/(64) = (7)/(4) . The value of (tan^(2)""(pi)/(7) + tan^(2)""(2pi)/(7) + tan^(2)""(3pi)/(7))xx (cot^(2)""(pi)/(7) + cot^(2)""(2pi)/(7) + cot^(2)""(3pi)/(7)) is

To find the sum sin^(2) ""(2pi)/(7) + sin^(2)""(4pi)/(7) +sin^(2)""(8pi)/(7) , we follow the following method. Put 7theta = 2npi , where n is any integer. Then " " sin 4 theta = sin( 2npi - 3theta) = - sin 3theta This means that sin theta takes the values 0, pm sin (2pi//7), pmsin(2pi//7), pm sin(4pi//7), and pm sin (8pi//7) . From Eq. (i), we now get " " 2 sin 2 theta cos 2theta = 4 sin^(3) theta - 3 sin theta or 4 sin theta cos theta (1-2 sin^(2) theta)= sin theta ( 4sin ^(2) theta -3) Rejecting the value sin theta =0 , we get " " 4 cos theta (1-2 sin^(2) theta ) = 4 sin ^(2) theta - 3 or 16 cos^(2) theta (1-2 sin^(2) theta)^(2) = ( 4sin ^(2) theta -3)^(2) or 16(1-sin^(2) theta) (1-4 sin^(2) theta + 4 sin ^(4) theta) " " = 16 sin ^(4) theta - 24 sin ^(2) theta +9 or " " 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta -7 =0 This is cubic in sin^(2) theta with the roots sin^(2)( 2pi//7), sin^(2) (4pi//7), and sin^(2)(8pi//7) . The sum of these roots is " " sin^(2)""(2pi)/(7) + sin^(2)""(4pi)/(7) + sin ^(2)""(8pi)/(7) = (112)/(64) = (7)/(4) . The value of (tan^(2)""(pi)/(7) + tan^(2)""(2pi)/(7) + tan^(2)""(3pi)/(7))/(cot^(2)""(pi)/(7) + cot^(2)""(2pi)/(7) + cot^(2)""(3pi)/(7)) is

To find the sum sin^(2) ""(2pi)/(7) + sin^(2)""(4pi)/(7) +sin^(2)""(8pi)/(7) , we follow the following method. Put 7theta = 2npi , where n is any integer. Then " " sin 4 theta = sin( 2npi - 3theta) = - sin 3theta This means that sin theta takes the values 0, pm sin (2pi//7), pmsin(2pi//7), pm sin(4pi//7), and pm sin (8pi//7) . From Eq. (i), we now get " " 2 sin 2 theta cos 2theta = 4 sin^(3) theta - 3 sin theta or 4 sin theta cos theta (1-2 sin^(2) theta)= sin theta ( 4sin ^(2) theta -3) Rejecting the value sin theta =0 , we get " " 4 cos theta (1-2 sin^(2) theta ) = 4 sin ^(2) theta - 3 or 16 cos^(2) theta (1-2 sin^(2) theta)^(2) = ( 4sin ^(2) theta -3)^(2) or 16(1-sin^(2) theta) (1-4 sin^(2) theta + 4 sin ^(4) theta) " " = 16 sin ^(4) theta - 24 sin ^(2) theta +9 or " " 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta -7 =0 This is cubic in sin^(2) theta with the roots sin^(2)( 2pi//7), sin^(2) (4pi//7), and sin^(2)(8pi//7) . The sum of these roots is " " sin^(2)""(2pi)/(7) + sin^(2)""(4pi)/(7) + sin ^(2)""(8pi)/(7) = (112)/(64) = (7)/(4) . The value of tan^(2)""(pi)/(7)tan ^(2)""(2pi)/(7) tan ^(2)""(3pi)/(7) is

Show that sin^(2)""(pi)/(18) + sin^(2)""(pi)/(9) + sin^(2)""(7pi)/(18) + sin^(2)"" (4pi)/(9) = 2 .

cos^2 ""(pi)/(8) + cos ^2 ""(3 pi)/(8) + cos ^2 ""(5 pi)/(8) + cos ^2 ""(7 pi)/(8) is

Prove that cos^(2)""pi/8+cos^(2) ""(3pi)/(8)+cos^(2)""(5pi)/(8)+cos^(2)""(7pi)/(8)=2

If z_(1)= sqrt(2)(cos ""(pi)/(4)+ isin""(pi)/(4))" and "z_(2)= sqrt(3)(cos ""(pi)/(3)+ isin"" (pi)/(3)) , then |z_(1)z_(2)| is

NCERT TAMIL-TRIGONOMETRIC FUNCTIONS -EXERCISE 3.3
  1. 2 sin ^(2) "" (pi)/(6) + cosec ^(2) "" (7pi)/(6) cos ^(2) "" (pi)/(3) ...

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  2. cot ^(2) "" (pi)/(6) + cosec ""(5pi)/(6) + 3 tan ^(2) ""(pi)/(6) =6

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  3. 2 sin ^(2) "" (3pi)/(4) + 2 cos ^(2) "" (pi)/( 4) + 2 sec ^(2) "" (pi)...

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  4. Find the value of: (i) sin 75^(@) (ii) tan 15 ^(@)

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  5. Prove that cos (pi/4-x) cos (pi/4-y)- sin (pi/4-x) sin(pi/4-y) =si...

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  6. (tan ((pi)/(4) +4))/( tan ((pi)/(4) -x ))= ((1 + tan x )/( 1- tan x ))...

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  7. (cos (pi +x) cos (-x))/( sin (pi -x) cos ((pi )/(2) + x))= cot ^(2) x

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  8. cos ((3pi )/( 2 ) + x) cos (2pi + x) [cot ((3pi)/( 2)- x ) + cot (2pi ...

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  9. sin (n +1) x sin (n +2) x + cos (n +1) x cos (n +2) x = cos x

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  10. Prove that cos ((3pi)/(4)+x)-cos ((3pi)/(4)-x)=-sqrt2 sin x

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  11. sin ^(2) 6x - sin ^(2) 4x = sin 2x sin 10x

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  12. cos ^(2) 2x -cos ^(2) 6x = sin 4x sin 8x

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  13. sin 3 x + 2 csin 4x + sin 6x = 4 cos ^(2) x sin 4x

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  14. cot 4x (sin 5x + sin 3x) = cot x (sin 5x - sin 3x)

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  15. (cos 9x - cos 5x )/( sin 17 x- sin 3x )=- (sin 2x)/( cos 10x )

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  16. (sin 5x + sin 3x )/( cos 5x + cos 3x ) = tan 4x

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  17. (sin x - sin y)/( cos x + cos y) = tan ""(x-y)/(2)

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  18. (sin x + sin 3x )/( cos x + cos 3x )= tan 2x

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  19. (sin x - sin 3x)/( sin ^(2) x - cos ^(2) x) = 2 sin x

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  20. (cos 4x +cos 3x + cos 2x )/( sin 4x + sin 3x + sin 2x)= cot 3x

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