Summation of Series in Trigonometry

Trigonometric Series Formula

Below are the important trigonometric series formulas, including sine and cosine series, for quick reference and easy application in Class 11 problems.

Summation of Sine Series

The sum of a sine series with angles in arithmetic progression (A.P.) is:

$S = \sin \alpha + \sin (\alpha + \beta) + \sin (\alpha + 2\beta) + \cdots + \sin (\alpha + (n-1)\beta) = \frac{\sin \frac{n\beta}{2} \cdot \sin \left[\alpha + (n-1)\frac{\beta}{2}\right]}{\sin \frac{\beta}{2}}$

Proof

Let $S = \sin \alpha + \sin (\alpha + \beta) + \sin (\alpha + 2\beta) + \cdots + \sin (\alpha + (n-1)\beta)$

Multiply both sides by $2 \sin \frac{\beta}{2}$:

$2S \sin \frac{\beta}{2} = 2\sin\alpha \sin \frac{\beta}{2} + 2\sin(\alpha+\beta) \sin \frac{\beta}{2} + \cdots + 2\sin(\alpha+(n-1)\beta) \sin \frac{\beta}{2}$

Using the identity $2 \sin X \sin Y = \cos(X-Y) - \cos(X+Y)$, we get:

$2 \sin \alpha \sin \frac{\beta}{2} = \cos\left(\alpha - \frac{\beta}{2}\right) - \cos\left(\alpha + \frac{\beta}{2}\right)$

$2 \sin (\alpha + \beta) \sin \frac{\beta}{2} = \cos\left(\alpha + \frac{\beta}{2}\right) - \cos\left(\alpha + \frac{3\beta}{2}\right)$

$2 \sin (\alpha + 2\beta) \sin \frac{\beta}{2} = \cos\left(\alpha + \frac{3\beta}{2}\right) - \cos\left(\alpha + \frac{5\beta}{2}\right)$

$\cdots$

$2 \sin (\alpha + (n-1)\beta) \sin \frac{\beta}{2} = \cos\left[\alpha + (2n-3)\frac{\beta}{2}\right] - \cos\left[\alpha + (2n-1)\frac{\beta}{2}\right]$

Adding all terms:

$2 \sin \frac{\beta}{2} , S = \cos\left(\alpha - \frac{\beta}{2}\right) - \cos\left[\alpha + (2n-1)\frac{\beta}{2}\right]$

Using the sine formula for difference of cosines:

$2 \sin \frac{\beta}{2} , S = 2 \sin\left[\alpha + (n-1)\frac{\beta}{2}\right] \sin \frac{n\beta}{2}$

$\Rightarrow S = \frac{\sin \frac{n\beta}{2} \cdot \sin \left[\alpha + (n-1)\frac{\beta}{2}\right]}{\sin \frac{\beta}{2}}$

Summation of Cosine Series

Replacing $\alpha$ by $\frac{\pi}{2} + \alpha$ in the sine series formula, we get:

$\cos \alpha + \cos (\alpha + \beta) + \cos (\alpha + 2\beta) + \cdots + \cos (\alpha + (n-1)\beta) = \frac{\sin \frac{n\beta}{2}}{\sin \frac{\beta}{2}} \cdot \cos\left[\alpha + (n-1)\frac{\beta}{2}\right]$

Product and Transformation Formulas

Trigonometric series often involve products of sine and cosine functions. Using product-to-sum formulas simplifies these series and makes summation easier. These formulas are widely used in summation of series in trigonometry, cos series summation, and other advanced problems.

Product of Cosine Series Formula

The product of cosine functions can be converted into a sum using the formula:

$\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]$

Application:
This formula helps in summing series like $\cos \alpha \cos (\alpha+\beta) + \cos (\alpha+2\beta) \cos (\alpha+3\beta)$ efficiently. It is commonly used in summation of trigonometric series questions for Class 11 and competitive exams.

Product of Sine Series Formula

Similarly, sine products can be transformed using:

$\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$

$\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$

These formulas are extremely useful for trigonometric series examples where angles are in arithmetic progression, allowing conversion of complex products into simpler sums.

Cos Series Summation Formula

Specific cosine series formulas help in calculating sums like:

$\sum_{r=1}^{n} \cos r\theta = \frac{\sin \frac{n\theta}{2} \cdot \cos \frac{(n+1)\theta}{2}}{\sin \frac{\theta}{2}}$

$\sum_{r=1}^{n} \cos^2 r\theta = \frac{n}{2} + \frac{1}{2}\sum_{r=1}^{n} \cos 2r\theta$

These formulas are critical in summation of series in trigonometry class 11 and in solving problems for cos series summation formula applications.

Summation Techniques and Tricks

Efficient summation of trigonometric series requires strategic techniques. Using these methods simplifies complex algebraic and trigonometric series and saves time in exams.

Using Arithmetic Progression in Trigonometry

For series with angles in arithmetic progression (A.P.), the sum of sine or cosine series can be calculated using:

$\sum_{r=1}^{n} \sin (\alpha + (r-1)\beta) = \frac{\sin \frac{n\beta}{2} \cdot \sin \left[\alpha + (n-1)\frac{\beta}{2}\right]}{\sin \frac{\beta}{2}}$

$\sum_{r=1}^{n} \cos (\alpha + (r-1)\beta) = \frac{\sin \frac{n\beta}{2} \cdot \cos \left[\alpha + (n-1)\frac{\beta}{2}\right]}{\sin \frac{\beta}{2}}$

These formulas are widely used in summation of trigonometric series questions and trigonometric series examples.

Geometric Progression Approach

For geometric progression (G.P.) of trigonometric terms, the series can be summed using:

$\sum_{r=0}^{n-1} a r^k = a + ar + ar^2 + \cdots + ar^{n-1} = a \frac{1-r^n}{1-r}$,

where $r$ can be $\cos \theta$ or $\sin \theta$. This technique is useful in cos series summation formula and trigonometric series formula derivations.

Algebraic and Trigonometric Series Combination

Some problems mix algebraic coefficients with trigonometric terms:

$\sum_{r=1}^{n} r \sin r\theta, \quad \sum_{r=1}^{n} r \cos r\theta$

Common Mistakes and Tips

Mastering summation of series in trigonometry not only requires learning formulas but also understanding common pitfalls. Avoiding these mistakes ensures accurate solutions in Class 11 trigonometric series problems and competitive exams.

Formula Misapplication

One of the most frequent errors is applying the wrong formula to a series. To prevent this:

• Check if the series is arithmetic (A.P.) or geometric (G.P.).

• Identify whether it is a sine series, cosine series, or mixed series.

• Use the corresponding formula:

For example:

$\sum_{r=1}^{n} \sin (\alpha + (r-1)\beta) = \frac{\sin \frac{n\beta}{2} \cdot \sin \left[\alpha + (n-1)\frac{\beta}{2}\right]}{\sin \frac{\beta}{2}}$

$\sum_{r=1}^{n} \cos (\alpha + (r-1)\beta) = \frac{\sin \frac{n\beta}{2} \cdot \cos \left[\alpha + (n-1)\frac{\beta}{2}\right]}{\sin \frac{\beta}{2}}$

Applying these formulas incorrectly is a major source of errors in trigonometric series examples.

Incorrect Summation Limits

Another common mistake is misinterpreting the upper and lower limits of the series.

• Ensure the first term corresponds to $r=1$ or $r=0$ as given.

• Check if the last term is $n$ or $n-1$.

For example, the series:

$\sum_{r=0}^{n-1} \sin (\alpha + r\beta)$

differs from:

$\sum_{r=1}^{n} \sin (\alpha + (r-1)\beta)$

Even a small misplacement of the limits can lead to incorrect results in cos series summation formula applications.

Tricky Sign Errors

Handling negative angles and alternating signs is another source of mistakes:

• Remember that $\sin(-\theta) = -\sin \theta$ and $\cos(-\theta) = \cos \theta$.

• For alternating series like $\sum_{r=1}^{n} (-1)^{r-1} \sin r\theta$, carefully account for the sign of each term.

• When using product-to-sum formulas, track the positive and negative signs:

$\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$

Neglecting the sign can completely change the sum of a series, affecting Class 11 trigonometric series examples and competitive exam solutions.

Solved Examples Based on Trigonometric Series

Example 1: The value of $\sin \left(1^{\circ}\right)+\sin \left(3^{\circ}\right) \ldots+\sin \left(87^{\circ}\right)+\sin \left(89^{\circ}\right)$ is
1) $\frac{1}{\sin \left(1^{\circ}\right)}$
2) $\frac{1}{2 \sin \left(1^{\circ}\right)}$
3) $\frac{1}{2 \sin \left(2^{\circ}\right)}$
4) None of these

Solution:

$
\sin \left(1^{\circ}\right)+\sin \left(3^{\circ}\right) \ldots+\sin \left(87^{\circ}\right)+\sin \left(89^{\circ}\right)
$

Here $a=1^{\circ}, b=2^{\circ}$ and $n=45$

$
\begin{aligned}
& \text { Sum }=\frac{\sin \left(a+(n-1) \frac{b}{2}\right) \sin \frac{n b}{2}}{\sin \frac{b}{2}} \\
& =\frac{\sin \left(1+44 * \frac{2}{2}\right) \sin \frac{45 * 2}{2}}{\sin \frac{2}{2}} \\
& =\frac{1}{2 \sin \left(1^{\circ}\right)}
\end{aligned}
$
Hence, the answer is option 2.

Example 2: The sum of $\cos \left(1^{\circ}\right)+\cos \left(3^{\circ}\right) \ldots+\cos \left(87^{\circ}\right)+\cos \left(89^{\circ}\right)$ is
1) $\frac{1}{2 \cos \left(1^{\circ}\right)}$
2) $\frac{1}{\sin \left(1^{\circ}\right)}$
3) $\frac{1}{2 \sin \left(1^{\circ}\right)}$
4) None of these

Solution:

$
\cos \left(1^{\circ}\right)+\cos \left(3^{\circ}\right) \ldots+\cos \left(87^{\circ}\right)+\cos \left(89^{\circ}\right)
$
Here $a=1^{\circ}, b=2^{\circ}$ and $n=45$

$
\begin{aligned}
& \text { Sum }=\frac{\cos \left(a+(n-1) \frac{b}{2}\right) \sin \frac{n b}{2}}{\sin \frac{b}{2}} \\
& =\frac{\cos \left(1+44 * \frac{2}{2}\right) \sin \frac{45 * 2}{2}}{\sin \frac{2}{2}} \\
& =\frac{1}{2 \sin \left(1^{\circ}\right)}
\end{aligned}
$

Hence, the answer is option 3 .

Example 3: The value of $\cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdot \ldots \cdot \cos \frac{\pi}{2^{10}} \cdot \sin \frac{\pi}{2^{10}} \text { is: }$
Solution:

$\cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdot \cos \frac{\pi}{2^4} \cdot \ldots \ldots \ldots \ldots \cos \frac{\pi}{2^{10}} \cdot \sin \frac{\pi}{2^{10}}$

Here, $A=\frac{\pi}{2^{10}}, n=9$

So the given expression can be written as:

$\frac{\sin \left(2^9 \cdot \frac{\pi}{2^{10}}\right)}{2^9 \sin \left(\frac{\pi}{2^{10}}\right)} \times \sin \left(\frac{\pi}{2^{10}}\right)$

$=\frac{1}{2^9} \sin \frac{\pi}{2}$

$=\frac{1}{512}$

Example 4: The value of $\cos \frac{\pi}{19}+\cos \frac{3 \pi}{19}+\cos \frac{5 \pi}{19}+\ldots . .+\cos \frac{17 \pi}{19}$ is equal to (up to one decimal point):
1) $0.5$
2) $0$
3) $1$
4) $0.7$

Solution:

The trigonometric ratios for angles in all the four quadrants.

$
\begin{aligned}
& \cos \frac{\pi}{19}+\cos \frac{3 \pi}{19}+\cos \frac{5 \pi}{19}+\ldots .+\cos \frac{17 \pi}{19} \\
& \text { Here, } A=\frac{\pi}{19}, D=\frac{2 \pi}{19}, n=9 \\
& \because \cos A+\cos (A+D)+\cos (A+2 D)+\ldots \ldots . .+\cos (A+(n-1) D) \\
& =\frac{\sin \left(\frac{n D}{2}\right)}{\sin \frac{D}{2}} \cdot \cos \left(\frac{2 A+(n-1) D}{2}\right)=\frac{\sin 9 \times \frac{\pi}{19}}{\sin \frac{\pi}{19}} \times \cos \left(\frac{\frac{\pi}{19}+\frac{17 \pi}{19}}{2}\right) \\
& =\frac{\sin \frac{9 \pi}{19}}{\sin \frac{\pi}{19}} \times \cos \frac{9 \pi}{19} \\
& =\frac{1}{2} \cdot \frac{\sin \left(\frac{18 \pi}{19}\right)}{\sin \frac{\pi}{19}}=\frac{1}{2} \cdot \frac{\sin \frac{\pi}{19}}{\sin \frac{\pi}{19}}=\frac{1}{2}
\end{aligned}
$

Example 5: The value of the sum $\sum_{k=1}^n\left(\tan 2^{k-1} \cdot \sec 2^k\right)$ is
1) $\tan 2^n$
2) $\tan 2^n-1$
3) $\tan 2^n-\tan 1$
4) $\cos 2^n-\cos 2$

Solution:

$
\sum_{k=1}^n\left(\tan 2^{k-1} \cdot \sec 2^k\right)
$

$
\begin{aligned}
& \text { As } \sec \left(2.2^{k-1}\right)=\frac{1}{\cos \left(2.2^{k-1}\right)} \\
& =\sum_{k=1}^n \tan 2^{k-1} \times \frac{1+\tan ^2 2^{k-1}}{1-\tan ^2 2^{k-1}} \\
& =\sum_{k=1}^n\left(\frac{\tan (2)^{k-1}\left[2-1+\tan ^2 2^{k-1}\right]}{1-\tan ^2 2^{k-1}}\right) \\
& =\sum_{k=1}^n\left(\frac{2 \tan 2^{k-1}}{1-\tan ^2 2^{k-1}}-\tan (2)^{k-1}\right) \\
& =\sum_{k=1}^n\left(\tan 2^k-\tan 2^{k-1}\right) \\
& =\tan 2-\tan (1)+\tan 2^2-\tan 2+\ldots \tan 2^n-\tan 2^{n-1} \\
& =\tan 2^n-1
\end{aligned}
$

Hence, the answer is the option 3.

List of Topics Related to the Summation of Series

Below is the list of topics related to the summation of series in trigonometry for easy reference and practice.

NCERT Resources

NCERT resources for Class 11 Trigonometric Functions provide comprehensive study material, including Chapter 3 notes, solutions, and Exemplar answers. These resources help students practice summation of trigonometric series, strengthen concepts, and excel in exams.

NCERT Class 11 Chapter 3 - Trigonometric Functions Notes

NCERT Class 11 solutions for Chapter 3 - Trigonometric Functions

NCERT Exemplar solutions for Class 11 Chapter 3 - Trigonometric Functions

Practice Questions on Summation of Series

Practice questions on summation of series in trigonometry help students apply formulas, solve sine and cosine series problems, and master Class 11 trigonometric series examples efficiently.

Summation Of Series In Trigonometry - Practice Question

We have provided the list of practice questions based on the following topics: