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Introduction to Trigonometry Ex- 8.2

 Hi Friends and Champs

Before we proceed to next exercise , lets discuss about some common ratios for some specific angles of trigonometric ratios.

Angles

0°

30°

45°

60°

90°

Sin

0

 1/2

 1/√2

 √3/2

 1

Cos

1

 √3/2

 1/√2

 1/2

 0

Tan

0

 1/√3

 1

 √3

 ∞

Cot

∞

 √3

 1

 1/√3

 0

Cosec

∞

 2

 √2

 2/√3

 1

Sec

1

 2/√3

 √2

 2

 ∞


No need to remember this table. By some trick you can easily calculate all above given common ratios.

Remember only 3 ratios ,one for 30° or 60°,one for 45° and one for 0° or 90°. In my case ,I personally remember only three ratios cos60° = 1/2 , tan 45° = 1 and sin 90° =1, now start the game.

First draw two right angle triangle , in which one is right angle triangle ΔABC with angles 30°,60°and 90°, and second right angle triangle ΔPQR has angles 45°,45°and 90°, just like below figures.

As you already know that Cos60° = 1/2 
therefore Cos60° = 1/2 = Base / Hypotenuses  ,
From Pythagoras theorem, we can easily calculate third side .
For example, 

since cos 60° = 1/2 = B/H
and since H2 = B2 + P2
=> P2 = 4 - 1
=> P = √3 ,

now we can easily calculate other ratios for 30° and 60°. 
For example
If, I want to calculate tan30°,
now , since tan30° = Perpendicular / Base 
                               = BC / AB 
                               = 1 / √3 as simple as that !!

Similarly as we know that 
tan45° = 1 
           = 1/1 
           = P / B 
and since H2 = B2 + P2
=> H2 = 1 + 1
=> H = √2 , 
We can now calculate any ratio for 45° easily for other ratios also.

To remember ratios for 0° and 90°, it's easy to remember cos 0° = 1 so it will be easy to remember.
cos 0° = 1 = sin 90°.
sin 0° = 0 = cos 90° 
and since tan 0° = sin 0° / cos 0°
                           = 0/1 
                           = 0 
                          =cot 90°

and cot 0°         = 1/ tan 0°
                         = 1/0 
                         = ∞ 
                      = tan 90°

and cosec 0°     = 1/ sin 0° 
                         = 1/0 
                         = ∞ 
                      =sec 90°

and sec 0°         = 1/ cos 0°
                         = 1/1 
                         =1 
                        = cosec 90°

I hope you understand it, practice 1 or 2 times with draw triangle yourself to having this on your fingertip. 

Enjoy!
 
Exercise 8.2
1. Evaluate the following : 
(i) sin 60° cos 30° + sin 30° cos 60°
Solution:
    sin 60° cos 30° + sin 30° cos 60° = √3/2 x √3/2 + 1/2 x 1/2
                                                   = 3/4 + 1/4
                                                   =4/4 = 1 Ans.
(ii) 2 tan2 45° + cos2 30° – sin2 60°
 Solution:
    2 tan2 45° + cos2 30° – sin2 60°= 2 x (1)2 + (√3/2)2 - (√3/2)2
                                                   = 2 + 3/4 - 3/4
                                                   =2 Ans.
(iii) cos 45° / (sec 30° + cosec 30°)
Solution:
    cos 45°/( sec 30° + cosec 30°) = (1/√2) / ( 2/√3  + 2 )
                                                  =(1/√2) /[(2+ 2√3) / (√3 )]
                                                  =√3 / √2(2 + 2√3 )
                                                  =[√3 x (√3 -1)] / [2√2(1 + √3 )(√3 -1)] 
                                                                            multiply and divide by (√3 -1)
                                                  =(3 - √3) / [2√2 x (3-1)]
                                               =(3 - √3) / [4√2)
                                               =√2(3 - √3) / [4√2x√2) 
                                                multiply and divide by √2
                                              =(3√2 - √6) / 8 Ans.

(iv) [ sin 30° + tan 45°– cosec 60° ] / [ sec 30° + cos 60° + cot 45°]
Solution:
[ sin 30° + tan 45°– cosec 60° ] / [ sec 30° + cos 60° + cot 45°]

=[ 1/2 + 1 - 2/√3 ] / [ 2/√3 + 1/2 + 1]
=[ (√3 + 2√3 - 4) / 2√3 ] / [ (4 + √3 + 2√3) /2√3]
=(3√3 -4) / (4 + 3√3)
=(3√3 -4)(3√3 - 4) / (3√3 - 4)(4 + 3√3) multiple and divide by (3√3 - 4)
=(27 + 16 - 24√3) / (27 - 16)
=(43 - 24√3) / 11 Ans.

 (v) [ 5 cos2 60°  + 4sec2 30° -  tan2 45° ] / [sin2 30° + cos2 30° ]
Solution:
[ 5 cos2 60°  + 4sec2 30° -  tan2 45° ] / [sin2 30° + cos2 30° ]
=[ 5 x (1/2)2 + 4 (2/√3)2 - (1)2] / [ (1/2)2 + (√3/2)2]
=[ 5/4 + 16/3 -1 ] / [ 1/4 + 3/4 ]
=[(15 + 64 -12)/12 ] / 1
=67 / 12 Ans.

2. Choose the correct option and justify your choice 
 (i) 2tan30° / ( 1 + tan2 30° ) = 
 (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
Solution:
2tan30° / ( 1 + tan2 30° ) = (2 x 1 / √3) / [ 1 +(1 / √3)2)
= (2 / √3) / [ 1 +1 /3)  
=(6 / 4√3)
=(6 √3 / 12)
= √3 / 2
= sin 60° option (A) is correct. Ans.

(ii) [1 - tan245] / [1 +  tan245 ] = 
(A) tan 90° (B) 1(C) sin 45° (D) 0

Solution:
[1 - tan245] / [1+ tan245 ] = (1 - 1) / (1 + 1)
=0
= option (D) is correct. Ans.

(iii) sin 2A = 2 sin A is true when A = 
 (A) 0° (B) 30° (C) 45°(D) 60°
Solution:
option A is correct, since A =0 is the only condition when both vlaues in LHS and RHS will be equal i.e. equal to ZERO. 

(iv) 2tan30° /  ( 1- tan230 ) = 

 (A) cos 60°  (B) sin 60°  (C) tan 60°  (D) sin 30°

Solution:

2tan30° /  ( 1- tan230 ) = 2 x 1/√3 / ( 1 - 1/3 )

=2/√3 / (2/3) = √3 = tan 60° Ans.

3. If tan (A + B) = √3  and tan (A – B) = 1/√3 ; 0° < A + B <= 90°; A > B, find A and B.   

Solution : 

Since tan (A + B) = √3

                           = tan60°

=> A + B = 60 --------(1)

similarly

A - B = 30 ----------(2)

From 2A = 90 

=> A = 45°  Ans

From (1) put value of A

B = 60 - 45

=> B = 15°  Ans

4. State whether the following are true or false. Justify your answer. 

 (i) sin (A + B) = sin A + sin B. 

Solution: 

If sin (A + B) = sin A + sin B is true , then it should be true in all cases. 

Let < A =30 and <B = 60

then LHS = sin ( 30 +60 ) = sin90 =

and RHS = sinA + sinB = sin30 + sin60 = 1/2 + √3/2 =(1+√3) / 2

Hence LHS≠RHS   , so the statement is False

 (ii) The value of sin θ increases as θ increases. 

If the given statement is true , then it should be true for all angles

Sin 0° = 0

Sin 30° = 1/2

Sin 45° = 1/√2

Sin 60° = √3 / 2

Sin 90° = 1

and 0 < 1/2 < 1/√2 < √3 / 2 < 1

so the statement is True.

 (iii) The value of cos θ increases as θ increases. 

Since cos 0° = 1

cos 30° = √3 / 2

cos 60° = 1/2

and 1 > √3 / 2 > 1/2 

so the given statement is False.

 (iv) sin θ = cos θ for all values of θ.

False , since sin 0° ≠cos 0°

 (v) cot A is not defined for A = 0°

True


Kindly support me , with your comment .



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