Measurement of an Angle & Trigonometric Functions

Exercises

1)   Find the length of an arc of a circle of radius 5 cm subtending a central angle measuring 15ᵒ.

Solution:

Let s be the length of the arc subtending an angle θc at the centre of a circle of radius r.

Then,             θ = Here, r = 5 and θ = 15 = (15 x )c = ( )c

Therefore,   θ = ð ð             s = cm.

2)   Find the angle in radians between the hands of a clock at 7 : 20 p.m.

Solution:

We know that the hour hand completes one rotation in 12 hours while the minute hand completes one rotation in 60 minutes.

Therefore, Angle traced by the hour hand in 12 hours = 360

ð               Angle traced by the hour hand in 7 hours 20 minutes i.e. hours = ( ) = 220

Also, angle traced by the minute hand in 60 minutes = 360ᵒ

ð           The angle traced by the minute hand in 20 minutes = ( ) = 120ᵒ

Hence, the required angle between two hands = 220ᵒ - 120ᵒ = 100ᵒ.

3)  If sinA = and < A < π, then find cosA, tan2A.

Solution:

Since sin A > 0 and < A < π, then A is in II Quadrant.

Sin A = ð           sin² A = Therefore, cos² A = 1 - = ð           cos A = ( in II Quadrant)

Now, tan A = = = And tan 2A = = = = .

4)   Prove that cos510 cos330 + sin390 cos120 = -1.

Solution:

The values must be following:

cos 510 = cos(90 x 5 + 60) = - sin 60 = cos 330 = cos (90 x 3 + 60) = sin 60 = sin 390 = sin(90 x 4 + 30) = sin 30 = cos 120 = cos(90 x 1+ 30) = - sin 30 = Now, L.H.S.

= cos 510cos 330ᵒ + sin 390cos 120

= ( ) ( ) + ( ) ( )

= + ( )

= - ( ) = -1 = R.H.S.

Hence Proved.

5)  If A + B = , then prove that (1 + tanA)(1 + tanB) = 2.

Solution:

Given, A + B = ( )

Taking tangent both sides, we get, tan (A + B) = tan ( )

( ) = 1

Or

tan A + tan B = 1 – tan A tan B

or

tan A + tan B + tan A tan B = 1

Now, adding 1 both the sides, we get

tan A + tan B + tan A tan B + 1 = 2

(tan A + 1) + tan B (1 + tan A) = 2

Or

(tan A + 1) (1 + tan B) = 2

Or

(1 + tan A) (1 + tan B) = 2

Hence Proved.

6)   Draw the graph of cosx, sinx, tanx in [0, 2π).

Solution.   7)   Draw sinx, sin2x and sin3x on same graph and with same scale.

Solution. 8)  If tan(π cosθ) = cot(π sinθ), then prove that cos(θ - ) = ± Solution:

Given, tan (π cosθ) = cot (π sinθ)

Therefore, = ð    sin(π cosθ).sin(π sinθ) = cos(π cosθ).cos(π cosθ)

ð    cos(π sinθ).cos(π cosθ) - sin(π cosθ).sin(π sinθ) = 0

ð    cos(π sinθ + π cosθ) = 0    [since, cos(A – B) = cosA cosB – sinAsinB]

ð           cos(π sinθ + π cosθ) = cos ð           π sinθ + π cosθ = ð           sinθ + cosθ = Dividing and multiplying by √2, we get,

ð           2( sinθ + cosθ) = ð           2(sinθ sin45ᵒ + cosθ cos45ᵒ) = ð           (cosθ cos45ᵒ + sinθ sin45ᵒ) = ð           cos(θ - 45ᵒ) = [since, cos(A – B) = cosA cosB + sinAsinB]

ð           cos(θ - ) = .