Question

\sum _ { k = 1 } ^ { 3 } \cos ^ { 2 } ( ( 2 k - 1 ) \frac { \pi } { 12 } )​

Answer 1

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Asked on October 15, 2019 by

Mantasha Khattri

The value of

k=1

∑

13

sin(

4

π

+

6

(k−1)π

)sin(

4

π

+

6

kπ

)

1

is equal to

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VIDEO EXPLANATION

ANSWER

k=1

∑

13

sin

6

π

(sin(

4

π

+

6

kπ

)sin(

4

π

+(k−1)

6

π

))

sin[(

4

π

+

6

kπ

)−(

4

π

+(k−1)

6

π

)]

=2

k=1

∑

13

(cot(

4

π

+(k−1)

6

π

)−cot(

4

π

+

6

kπ

))

using compound angle formula.

=2(cot

4

π

−cot(

4

π

+

6

13π

))

=2(1−cot(

12

29π

))

=2(1−cot(

12

5π

))

=2(1−(2−

3

))

=2(−2+

3

)

=2(

3

−1)

Step-by-step explanation:

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