The correct option is C 32 #160; #160; cos ^2π12+cos ^23π12+cos ^25π12 =cos ^2π12+cos ^2π4+cos ^25π12=cos ^2π12+cos ^2(π2 π12)+(1√(2))^2= cos ...

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Standard XII

Mathematics

Complex Numbers

∑k=13cos2 2k-...

Question

$\sum_{k = 1}^{3} {cos}^{2} ((2 k - 1) \frac{π}{12}) =$

0

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\frac{1}{2}

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- \frac{1}{2}

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\frac{3}{2}

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Solution

The correct option is C $\frac{3}{2}$
${cos}^{2} \frac{π}{12} + {cos}^{2} \frac{3 π}{12} + {cos}^{2} \frac{5 π}{12} = {cos}^{2} \frac{π}{12} + {cos}^{2} \frac{π}{4} + {cos}^{2} \frac{5 π}{12} = {cos}^{2} \frac{π}{12} + {cos}^{2} (\frac{π}{2} - \frac{π}{12}) + (\frac{1}{\sqrt{2}})^{2} = {cos}^{2} \frac{π}{12} + {sin}^{2} \frac{π}{12} + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2}$

Therefore, Answer is $\frac{3}{2}$

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Similar questions

Q. If

{tan}^{- 1} [20 \sum k = 0 sec (\frac{5 π}{12} + \frac{k π}{2}) sec (\frac{5 π}{12} + \frac{(k + 1) π}{2})] = - {tan}^{- 1} a

, then the value of

a

c o s^{- 1} (\frac{π}{3} + s e c^{- 1} (- 2))

Q. Find $\sum_{k=0}^{\infty}12(\dfrac{2}{3})^k - \sum_{\infty}^{k=0}18(\dfrac{-1}{2})^k =$

Q. Find the values of k for which the roots are real and equal in each of the following equations:

(i)

k x^{2} + 4 x + 1 = 0

(ii)

k x^{2} - 2 \sqrt{5} x + 4 = 0

(iii)

3 x^{2} - 5 x + 2 k = 0

(iv)

4 x^{2} + k x + 9 = 0

(v)

2 k x^{2} - 40 x + 25 = 0

(vi)

9 x^{2} - 24 x + k = 0

(vii)

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

(viii)

x^{2} - 2 (5 + 2 k) x + 3 (7 + 10 k) = 0

(ix)

(3 k + 1) x^{2} + 2 (k + 1) x + k = 0

(x)

k x^{2} + k x + 1 = - 4 x^{2} - x

(xi)

(k + 1) x^{2} + 2 (k + 3) x + (k + 8) = 0

(xii)

x^{2} - 2 k x + 7 k - 12 = 0

(xiii)

(k + 1) x^{2} - 2 (3 k + 1) x + 8 k + 1 = 0

(xiv)

(2 k + 1) x^{2} + 2 (k + 3) x + (k + 5) = 0

(xvii)

4 x^{2} - 2 (k + 1) x + (k + 4) = 0

Find the values of k for which the roots are real and equal in each of the following equations.
$(i) k x^{2} + 4 x + 1 = 0 (i i) k x^{2} - 2 \sqrt{5} x + 4 = 0 (i i i) 3 x^{2} - 5 x + 2 k = 0 (i v) 4 x^{2} + k x + 9 = 0 (v) 2 k x^{2} - 40 x + 25 = 0 (v i) 9 x^{2} - 24 x + k = 0 (v i i) 4 x^{2} - 2 k x + 1 = 0 (v i i i) x^{2} - 2 (5 + 2 k) x + 3 (7 + 10 k) = 0 (i x) (3 k + 1) x^{2} + 2 (k + 1) x + k = 0 (x) k x^{2} + k x + 1 = - 4 x^{2} - x (x i) (k + 1) x^{2} + 2 (k + 3) x (k + 8) = 0 (x i i) x^{2} - 2 k x + 7 k - 12 = 0 (x i i i) (k + 1) x^{2} - 2 (3 k + 1) x + 8 k + 1 = 0 (x i v) 5 x^{2} - 4 x + 2 + k (4 x^{2} 1 - 2 x - 1) = 0 (x v) (4 - k) x^{2} + (2 k + 4) x + (8 k + 1) = 0 (x v i) (2 k + 1) x^{2} + 2 (k + 3) x + (k + 5) = 0 (x v i i) 4 x^{2} - 2 (k + 1) x + (k + 4) = 0 (x v i i i) 4 x^{2} - 2 (k + 1) x + (k + 1) = 0$

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∑3k=1cos2((2k−1)π12)=

The correct option is C 32 cos2π12+cos23π12+cos25π12=cos2π12+cos2π4+cos25π12=cos2π12+cos2(π2−π12)+(1√2)2=cos2π12+sin2π12+12=1+12=32Therefore, Answer is 32

$\sum_{k = 1}^{3} {cos}^{2} ((2 k - 1) \frac{π}{12}) =$