Formulas on CBSE Class 11 Maths Chapter 3 - Trigonometric Functions

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In the above-given picture of a right-angled triangle, the longest side is known as hypotenuse. The other two opposite sides of the hypotenuse are called adjacent and opposite. The angle between the adjacent and opposite is 90°.

The important trigonometric formula for class 11 are as follows:

1.  sin A = opposite/hypotenuse = a/c

2. cos A = adjacent/hypotenuse  = b/c      

3. tan A = opposite/adjacent  =  a/b = (a/c)/(b/c) = sin A/cos A

4. cosec A = 1/sin A= hypotenuse/opposite= c/a

5. sec A = 1/cos A= hypotenuse/adjacent= c/b

6. cot A = 1/tan A= adjacent/opposite= cos A/sin A= b/a

Below given is the list of special trigonometry formulas for class 11. They are:

1. Pythagorean Identities

The trigonometric identities that are related to Pythagorean Theorem are:

1. \[\sin^{2}A + cos^{2}A = 1\] 

2. \[1 + tan^{2}A = sec^{2} \text{or } secA - tanA = \frac{1}{secA + tanA}\]

3. \[1 + cot^{2}A = cosec^{2} \text{or } cosecA - cotA = \frac{1}{cosecA + cotA}\]

4. \[sin2A = 2 sin A. cosA\]

5. \[cos2A = cos^{2}A - sin^{2}A\]

6. \[tan2A = \frac{2 tanA}{(1 - tan^{2}A)}\]

7. \[cot 2A = \frac{(cot^{2}A - 1)}{2 cotA}\]

2. (i) sin (-A) = -sin A

(ii) cosec (-A) = -cosec A

(iii) cos (-A) =cos A

(iv) sec (-A) =sec A

(v) tan (- A) = -tan A

(vi) cot (-A)= -cot A

3. (i) sin(π/2 - A) = cos A

(ii) cos(π/2 - A) = Sin A

(iii) tan(π/2 - A) = Cot A

(iv) sec(π/2 - A) = cosec A

(v) cosec(π/2 - A) = sec A

(vi) cot(π/2 - A) = tan A

4. (i) sin(π/2 + A) = cos A

(ii) cos(π/2 + A) = -sin A

(iii) tan(π/2+ A) = -cot A

(iv) cot(π/2 + A) = -tan A

(v) sec(π/2 + A) = -cosec A

5. (i) sin(π - A) = sin A

(ii) cos(π - A) = -cos A

(iii) tan(π- A) = -tan A

(iv) cosec(π - A) = -cosec A

(v) sec(π - A) = -sec A

(vi) cot(π - A) = -cot A

6. Sum and Difference identities

For angles A and B, the following relationships are given below:

(i) sin(A+B) = sin (A)cos(A)+cos(A)sin(A)

(ii) cos(A+B) = cos(A)cos(B)+sin(A)sin(B)

(iii) tan (A+B) = {tan(A)+ tan(B)}/1-tan(A)tan(B)

(iv) sin(A-B) = sin(A)cos(B)-cos(A)sin(B)

(v) cos(A-B) = cos(A)cos(B)-sin(A)sin(B)

(vi) tan(A-B) = {tan(A)-tan(B)}/1+tan(A)tan(B)

7. Law of Sines

\[\frac{a}{sin A} = \frac{b}{sinB} = \frac{c}{sinC}\]

8. Law of Cosines

(i) \[c^{2} = a^{2} + b^{2} - 2ab cos C\]

(ii) \[a^{2} = b^{2} + c^{2} - 2bc cosA\]

(iii) \[b^{2} = a^{2} + c^{2} - 2ac cos B\]

9. Law of Tangents

\[\frac{a - b}{a + b} = \frac{tan[\frac{1}{2} (A - B)]}{tan[\frac{1}{2} (A + B)]}\]

10. (i) sin(3π/2 - A) = cos A

(ii) cos(3π/2 - A) = -sin A

(iii) tan(3π/2 - A) = cot A

(iv) cot(3π/2 - A) = tan A

(v) cosec(3π/2 - A) = -sec A

(vi) sec(3π/2 - A) = -cosec A

11. (i) sin(3π/2 + A) = -cos A

(ii) cos(3π/2 + A) = sin A

(iii) tan(3π/2 + A) = -cot A

(iv) cot(3π/2 + A) = -tan A

(v) cosec(3π/2 + A) = -sec A

(vi) sec(3π/2 + A) = cosec A

Solved Examples

1. Prove that \[cos^{2}a + cos^{2} (a + \frac{\pi}{3}) + cos^{2} (a - \frac{\pi}{3}) = \frac{3}{2}\].

Answer. We have,

\[L.H.S = \frac{1 + cos2a}{2} + \frac{1 + cos(2a + \frac{2 \pi}{3})}{2} + \frac{1 + cos (2a - \frac{2 \pi}{3})}{2}\]

\[= \frac{1}{2} [ 3 + cos 2a + cos (2a + \frac{2 \pi}{3}) + cos (2a - \frac{2 \pi}{3})]\]

\[= \frac{1}{2} [3 + cos2a + 2 cos 2a cos \frac{ 2 \pi}{3}\]

\[= \frac{1}{2} [ 3 + cos 2a + 2 cos 2a cos (\pi - \frac{\pi}{3})]\]

\[= \frac{1}{2} [3 + cos 2a - cos 2a cos \frac{\pi}{3}]\]

\[= \frac{1}{2} [3 + cos 2a - cos 2a\]

\[= \frac{3}{2} = R.H.S\]

2. Find the value of \[tan \frac{\pi}{8}\].

Answer. Let \[a = \frac{\pi}{8}\]. Then, \[2a = \frac{\pi}{4}\].

Now, \[tan 2a = \frac{2 tan a}{1 - tan^{2}a}\]

or \[tan \frac{\pi}{4} = \frac{2 tan \frac{\pi}{8}}{1 - tan^{2} \frac{\pi}{8}}\]

Let \[b = tan \frac{\pi}{8}\]. Then \[1 = \frac{2b}{1 - b^{2}}\]

or, \[b^{2} + 2b - 1 = 0\]

Therefore, \[b = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}\]

Since \[\frac{\pi}{8}\] lies in the first quadrant, \[b = tan \frac{\pi}{8}\] is positive. Hence,

\[tan \frac{\pi}{8} = \sqrt{2} - 1\].

Conclusion 

CBSE Class 11 Maths Chapter 3 is Trigonometric Functions. This chapter is mostly based on equations and formulas along with some other concepts. Therefore, we, at Vedantu, have provided a list of formulas that are easy to learn and memorize, in a simple PDF format that can be printed and kept as study material for all students. We recommend that students make full use of this concise piece of information and prepare themselves for the upcoming exams.