Trigonometric Functions Class 11 Notes AP Inter 1st Year Maths Chapter 3

Students can go through AP Inter 1st Year Maths Notes 3rd Lesson Trigonometric Functions will help students in revising the entire concepts quickly.

Trigonometric Functions Class 11 Notes AP Inter 1st Year Maths 3rd Lesson

→ If in a circle of radius r, an arc of length l subtends an angle of θ radians, then l = rθ.

→ Radian measure = \(\frac{\pi}{180}\) × Degree measure

→ Degree measure = \(\frac{180}{\pi}\) × Radian measure

→ cos²x + sin²x = 1

→ 1 + tan²x = sec²x

→ 1 + cot²x = cosec²x

→ cos (2nπ + x) = cos x

→ sin (2nπ + x) = sin x

→ sin(-x) = -sin x

→ cos(-x) = cos x

→ cos(x + y) = cos x cos y – sin x sin y

→ cos(x – y) = cos x cos y + sin x sin y

→ cos(\(\frac{\pi}{2}\) – x) = sin x

→ sin(\(\frac{\pi}{2}\) – x) = cos x

→ sin(x + y) = sin x cos y + cos x sin y

→ sin(x – y) = sin x cos y – cos x sin y

→ cos(\(\frac{\pi}{2}\) + x) = -sin x

→ sin(\(\frac{\pi}{2}\) + x) = cos x

→ cos(π – x) = -cos x

→ sin(π – x) = sin x

→ cos(π + x) = -cos x

→ sin(π + x) = -sin x

→ cos(2π – x) = cos x

→ sin(2π – x) = -sin x

→ If none of the angles x, y and (x ± y) is an odd multiple of \(\frac{\pi}{2}\), then tan(x + y) = \(\frac{\tan x+\tan y}{1-\tan x \tan y}\)

→ tan(x – y) = \(\frac{\tan x-\tan y}{1+\tan x \tan y}\)

→ If none of the angles x, y and (x ± y) is a multiple of π, then cot(x + y) = \(\frac{\cot x \cot y-1}{\cot y+\cot x}\)

→ cot(x – y) = \(\frac{\cot x \cot y+1}{\cot y-\cot x}\)

→ cos 2x = cos²x – sin²x = 2 cos²x – 1 = 1 – 2 sin²x = \(\frac{1-\tan ^2 x}{1+\tan ^2 x}\)

→ sin 2x = 2 sin x cos x = \(\frac{2 \tan x}{1+\tan ^2 x}\)

→ tan 2x = \(\frac{2 \tan x}{1-\tan ^2 x}\)

→ sin 3x = 3 sin x – 4 sin³x

→ cos 3x = 4 cos³x – 3 cos x

→ tan 3x = \(\frac{3 \tan x-\tan ^3 x}{1-3 \tan ^2 x}\)

→ cos x + cos y = 2 cos \(\frac{x+y}{2}\) cos \(\frac{x-y}{2}\)

→ cos x – cos y = -2 sin \(\frac{x+y}{2}\) sin \(\frac{x-y}{2}\)

→ sin x + sin y = 2 sin \(\frac{x+y}{2}\) cos \(\frac{x-y}{2}\)

→ sin x – sin y = 2 cos \(\frac{x+y}{2}\) sin \(\frac{x-y}{2}\)

→ 2 cos x cos y = cos(x + y) + cos(x- y)

→ -2 sin x sin y = cos(x + y) – cos(x – y)

→ 2 sin x cos y = sin(x + y) + sin(x – y)

→ 2 cos x sin y = sin(x + y) – sin(x – y)

Example Problems

Question 1.
Convert 40°20′ into radian measure.
Solution:
We know that 180° = π radian
Hence, 40°20′ = 40\(\frac {1}{3}\) degree
= \(\frac{\pi}{180} \times \frac{121}{3}\) radian
= \(\frac{121 \pi}{540}\) radian
Therefore 40°20′ = \(\frac{121 \pi}{540}\) radian

Question 2.
Convert 6 radians into degree measure.
Solution:
We know that π radian = 180°.
Hence 6 radians = \(\frac{180}{\pi}\) × 6 degree
= \(\frac{1080 \times 7}{22}\) degree
= 343\(\frac {7}{11}\) degree
= 343 + \(\frac{7 \times 60}{11}\) minute [as 1° = 60′]
= 343° + 38′ + \(\frac {2}{11}\) minute [as 1′ = 60″]
= 343° + 38′ + 10.9″
= 343° 38′ 11″ approximately.
Hence, 6 radians = 343° 38′ 11″ approximately.

Question 3.
Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm (use π = \(\frac {22}{7}\)).
Solution:
Here l = 34.7 cm
and θ = 60° = \(\frac{60 \pi}{180}\) radian = \(\frac{\pi}{3}\)
Hence,, by r = \(\frac{1}{\theta}\), we have
r = \(\frac{37.4 \times 3}{\pi}\)
= \(\frac{37.4 \times 3 \times 7}{22}\)
= 35.7 cm

Question 4.
The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? (Use π = 3.14)
Solution:
In 60 minutes, the minute hand of a watch completes one revolution.
Therefore, in 40 minutes, the minute hand turns through \(\frac {2}{3}\) of a revolution.
Therefore, θ = \(\frac {2}{3}\) × 360° or \(\frac{4 \pi}{3}\) radian.
Hence, the required distance travelled is given by l = rθ
= 1.5 × \(\frac{4 \pi}{3}\) cm
= 2π cm
= 2 × 3.14 cm
= 6.28 cm

Question 5.
If the arcs of the same length in two circles subtend angles 65° and 110° at the centre, find the ratio of their radii.
Solution:
Let r₁ and r₂ be the radii of the two circles.
Given that θ₁ = 65°
= \(\frac{\pi}{180}\) × 65
= \(\frac{13 \pi}{36}\) radian
and θ₂ = 110°
= \(\frac{\pi}{180}\) × 110
= \(\frac{22 \pi}{36}\) radian
Let l be the length of each of the arcs.
Then l = r₁θ₁ = r₂θ₂, which gives
\(\frac{13 \pi}{36} \times \mathrm{r}_1=\frac{22 \pi}{36} \times \mathrm{r}_2\)
⇒ \(\frac{r_1}{r_2}=\frac{22}{13}\)
Hence r₁ : r₂ = 22 : 13

Multiple Choice Questions

Question 1.
The value of cos 5π is
(1) 0
(2) 1
(3) -1
(4) None of these
Answer:
(3) -1

Question 2.
The value of cos 1° cos 2° cos 3° ….. cos 179° is
(1) \(\frac{1}{\sqrt{2}}\)
(2) 0
(3) 1
(4) -1
Answer:
(2) 0

Question 3.
If sin θ + cosec θ = 2 then sin²θ + cosec²θ is equal to
(1) 1
(2) 4
(3) 2
(4) None of these
Answer:
(3) 2

Question 4.
If tan θ = \(\frac {1}{2}\) and tan φ = \(\frac {1}{2}\), then the value of θ + φ is
(1) \(\frac{\pi}{6}\)
(2) π
(3) 0
(4) \(\frac{\pi}{4}\)
Answer:
(4) \(\frac{\pi}{4}\)

Question 5.
The value of \(\frac{1-{tan}^2 15^{\circ}}{1+{tan}^2 15^{\circ}}\) is
(1) 1
(2) √3
(3) √3/2
(4) 2
Answer:
(3) √3/2

Question 6.
The value of sin(45° + θ) – cos (45° – θ) is
(1) 2 cos θ
(2) 2 sin θ
(3) 1
(4) 0
Answer:
(4) 0

Question 7.
The value of \(\cot \left(\frac{\pi}{4}+\theta\right) \cot \left(\frac{\pi}{4}-\theta\right)\) is
(1) -1
(2) 0
(3) 1
(4) None of these
Answer:
(3) 1

Question 8.
cos 2θ cos 2φ + sin²(θ – φ) sin²(θ + φ) is equal to
(1) sin 2(θ + φ)
(2) cos 2(θ + φ)
(3) sin 2(θ – φ)
(4) cos 2(θ – φ)
Answer:
(2) cos 2(θ + φ)

Question 9.
The value of sin 50° – sin 70° + sin 10° is equal to
(1) 1
(2) 0
(3) \(\frac {1}{2}\)
(4) 2
Answer:
(2) 0

Question 10.
If sin θ + cos θ = 1, then the value of sin 2θ is equal to
(1) 1
(2) \(\frac {1}{2}\)
(3) 0
(4) -1
Answer:
(3) 0