TOPICS

Unit-3(Examples)

Trigonometric Functions

**Example-1 :-** Convert 40° 20′ into radian measure.

We know that 180° = π radian. Hence, 40° 20′ = 40 + 1/3 degree = π/180 x 121/3 radian = 121π/540 radian Therefore, 40° 20′ = 121π/540 radian

**Example-2 :-** Convert 6 radians into degree measure.

We know that π radian = 180°. Hence, 6 radians = 180/π x 6 degree = (1080 x 7)/22 degree = 343° + 7/11 degree = 343° + (7 x 60)/11 minute [1° = 60'] = 343° + 420/11 minute = 343° + 38' + 2/11 minute [1' = 60"] = 343° + 38' + 10.9" 6 radians = 343° 38' 11" approximately.

**Example-3 :-** Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm. (use π = 22/7)

Given that l = 37.4 cm and θ = 60° = 60π/180 radian = π/3 By using formula, r = l/θ = (37.4 x 3)/π = (112.2 x 7)/22 = 785.4/22 = 35.7 cm

**Example-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).

In 60 minutes, the minute hand of a watch completes one revolution. Therefore,in 40 minutes, the minute hand turns through 2/3 of a revolution. Therefore, θ = 2/3 x 360° = 4π/3 radian and r = 1.5 cm Hence, the required distance travelled is given By formula, l = r θ = 1.5 x 4π/3 = 2π = 2 x 3.14 = 6.28 cm

**Example-5 :-** If the arcs of the same lengths in two circles subtend angles 65°and 110° at the centre, find the ratio of their radii.

Let r₁ and r₂ be the radii of the two circles. Given that θ₁ = 65° = π/180 x 65 = 13π/36 radian θ₂ = 110° = π/180 x 110 = 22π/36 radian Let l be the length of each of the arc. Then l = r₁ θ₁ = r₂θ₂, which gives 13π/36 x r₁ = 22π/36 x r₂ i.e. r₁ /r₂ = 22/13 Hence, r₁ : r₂ = 22 : 13

**Example-6 :-** If cos x = –3/5 , x lies in the third quadrant, find the values of other five trigonometric functions.

Given that cos x = –3/5 (i) sec x = 1/cos x = -5/3 (ii) sin x = √1 - cos²x = √1 - (-3/5)²x = √1 - 9/25 = √16/25 = ±4/5 Since x lies in second quadrant, sin x will be negative. Therefore sin x = -4/5 (iii) cosec x = 1/sin x = -5/4 (iv) tan x = sin x/cos x = -4/5 ÷ (-3)/5 = -4/5 x (-5)/3 = 4/3 (v) cot x = 1/tan x = 3/4

**Example-7 :-** If cot x = –5/12, x lies in second quadrant, find the values of other five trigonometric functions.

Given that cot x = –5/12 (i) tan x = 1/cot x = -12/5 (ii) sec x = √1 + tan²x = √1 + (-12/5)²x = √1 + 144/25 = √169/25 = ±13/5 Since x lies in second quadrant, sec x will be negative. Therefore sec x = -13/5 (iii) cos x = 1/sec x = -5/13 (iv) sin x = √1 - cos²x = √1 - (-5/13)²x = √1 - 25/169 = √144/169 = ±12/13 Since x lies in second quadrant, sin x will be positive. Therefore sin x = 12/13 (v) cosec x = 1/sin x = 13/12

**Example-8 :-** Find the value of sin 31π/3.

We know that values of sin x repeats after an interval of 2π. Therefore, sin 31π/3 = sin(10π + π/3) = sin π/3 = √3/2

**Example-9 :-** Find the value of cos (–1710°).

We know that values of cos x repeats after an interval of 2π or 360°. Therefore, cos (–1710°) = cos (–1710° + 5 x 360°) = cos (–1710° + 1800°) = cos 90° = 0

**Example-10 :-** Prove that: 3sin π/6 . sec π/3 - 4sin 5π/6 . cot π/4 = 1

L.H.S. 3sin π/6 . sec π/3 - 4sin 5π/6 . cot π/4 = 3 x 1/2 x 2 - 4sin(π - π/6) x 1 = 3 - 4sin π/6 = 3 - 4 x 1/2 = 3 - 2 = 1 = R.H.S

**Example-11 :-** Find the value of sin 15°.

sin 15° = sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30° = 1/√2 x √3/2 - 1/√2 x 1/2 = √3/2√2 - 1/2√2 = (√3 - 1)/√2

**Example-12 :-** Find the value of tan 13π/12.

tan 13π/12 = tan(π + π/12) = tan π/12 = tan(π/4 - π/6) = (tan π/4 - tan π/6)/(1 + tan π/4 . tan π/6) = (1 - 1/√3)/(1 + 1/√3) = (√3 - 1)/(√3 + 1) By rationalising, = (√3 - 1)/(√3 + 1) x (√3 - 1)/(√3 - 1) = (√3 - 1)²/[(√3)² - (1)²] = 2 - √3

**Example-13 :-** Prove that:

**Example-14 :-** Show that: tan 3x . tan 2x . tan x = tan 3x – tan 2x – tan x

We know that 3x = 2x + x. Therefore, tan 3x = tan (2x + x) tan 3x = (tan 2x + tan x)/(1 - tan 2x . tan x) tan 3x – tan 3x tan 2x tan x = tan 2x + tan x tan 3x – tan 2x – tan x = tan 3x tan 2x tan x tan 3x tan 2x tan x = tan 3x – tan 2x – tan x

**Example-15 :-** Prove that: cos(π/4 + x) + cos(π/4 - x) = √2 cos x

**Example-16 :-** Prove that:

**Example-17 :-** Prove that:

**Example-18 :-** Find the principal solutions of the equation sin x = √3/2

We know that sin π/3 = √3/2 and sin 2π/3 = (π - π/3) = √3/2 Therefore, principal solutions are π/3 and 2π/3.

**Example-19 :-** Find the principal solutions of the equation tan x = -1/√3

We know that tan π/6 = 1/√3 tan(π - π/6) = -tan π/6 = -1/√3 and tan(2π - π/6) = -tan π/6 = -1/√3 Thus, tan 5π/6 = tan 11π/6 = -1/√3 Therefore, principal solutions are 5π/6 and 11π/6.

**Example-20 :-** Find the solution of sin x = -√3/2

Given: sin x = -√3/2 sin x = -sin π/3 sin x = sin(π + π/3) sin x = sin 4π/3 which gives, x = nπ + (-1)ⁿ 4π/6, where n ∈ Z.

**Example-21 :-** Solve cos x = 1/2

Given: cos x = 1/2 cos x = cos π/3 Therefore, x = 2nπ ± π/3, where n ∈ Z.

**Example-22 :-** Solve tan 2x = -cot(x + π/3).

Given: tan 2x = -cot(x + π/3) tan 2x = tan(π/2 + x + π/3) tan 2x = tan(x + 5π/6) Therefore, 2x = nπ + x + 5π/6 where n ∈ Z. or x = nπ + 5π/6 where n ∈ Z.

**Example-23 :-** Solve sin 2x – sin 4x + sin 6x = 0.

Given: sin 2x – sin 4x + sin 6x = 0 (sin 6x + sin 2x) – sin 4x = 0 2sin 4x cos 2x - sin 4x = 0 sin 4x (2cos 2x - 1) = 0 Therefore, sin 4x = 0 or 2cos 2x = 1 sin 4x = 0 or cos 2x = 1/2 4x = nπ or 2x = 2nπ ± π/3, where n ∈ Z. x = nπ/4 or x = nπ ± π/6, where n ∈ Z.

**Example-24 :-** Solve 2 cos² x + 3 sin x = 0

Given: 2 cos² x + 3 sin x = 0 2 (1 - sin²x) + 3 sin x = 0 2 - 2 sin²x + 3 sin x = 0 2 sin²x - 3 sin x - 2 = 0 (2sin x + 1)(sin x - 2) = 0 Hence, sin x = -1/2 or sin x = 2 But, sin x = 2 is not possible. Therefore, sin x = -1/2 = sin 7π/6 x = nπ + (-1)ⁿ 7π/6, where n ∈ Z.

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