For α = 13 2 π , cos 2 α + cos 2 2 α + cos 2 3 α + cos 2 4 α + cos 2 5 α + cos 2 6 α =

∑ cos 2 k α = 2 6 + ( c o s 2 α + c o s 4 α + c o s 6 α + c o s 8 α + c o s 10 α + c o s 12 α ) . Since 13 α = 2 π : 12 α = 2 π − α , 10 α = 2 π − 3 α , 8 α = 2 π − 5 α , so the bracket equals cos α + cos 2 α + … + cos 6 α = − 2 1 . The sum is 2 6 − 2 1 = 4 11 .

Start Free Mock LoginAlready registered? Login here

Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

MathematicsTrigonometric Ratios & IdentitiesSummation of SeriesMedium2 minQB

MathematicsMediumsingle choice

Passage / Comprehension

To evaluate an expression of the form $sin a + sin (a + d) + \dots + sin (a + (n - 1) d)$ or $cos a + cos (a + d) + \dots + cos (a + (n - 1) d)$ we can multiply each term by $2 sin \frac{d}{2}$ and apply the transformation formula to the resulting products. Using this information, answer the following.

For $α = \frac{2 π}{13}$ , $cos^{2} α + cos^{2} 2 α + cos^{2} 3 α + cos^{2} 4 α + cos^{2} 5 α + cos^{2} 6 α =$

Options:

Answer:

Solution:

$\sum cos^{2} k α = \frac{6 + ( c o s 2 α + c o s 4 α + c o s 6 α + c o s 8 α + c o s 10 α + c o s 12 α )}{2}$ . Since $13 α = 2 π : 12 α = 2 π - α, 10 α = 2 π - 3 α, 8 α = 2 π - 5 α$ , so the bracket equals $cos α + cos 2 α + \dots + cos 6 α = - \frac{1}{2}$ . The sum is $\frac{6 - \frac{1}{2}}{2} = \frac{11}{4}$ .

Stream:JEESubject:MathematicsTopic:Trigonometric Ratios & IdentitiesSubtopic:Summation of Series

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...