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 = 15ᵒ = (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 = 360ᵒ

 

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

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 cos510ᵒ cos330ᵒ + sin390ᵒ cos120ᵒ = -1.

 

Solution:

 

      The values must be following:

      cos 510ᵒ = cos(90ᵒ x 5 + 60ᵒ) = - sin 60ᵒ =

      cos 330ᵒ = cos (90ᵒ x 3 + 60ᵒ) = sin 60ᵒ =

      sin 390ᵒ = sin(90ᵒ x 4 + 30ᵒ) = sin 30ᵒ =

      cos 120ᵒ = cos(90ᵒ x 1+ 30ᵒ) = - sin 30ᵒ =

Now, L.H.S.

                       = cos 510ᵒ cos 330ᵒ + sin 390ᵒ cos 120ᵒ

                       = () () + () ()

                       =  + ()

                       = - () = -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.

Graph 1.PNG
graph 2.PNG
graph 3.PNG
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


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

Solution.

graph 4.PNG
 

 

 

 

 

 

 

 

 

 

 


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 -  ) = .