Graphs of Trigonometric Functions
Graph of y = sin X:
Graph of y = cos X:
Graph of y = tan X
If in any triangle the angles be to one another as 1 : 2 : 3, prove that the corresponding sides are 1 :  : 2.
Solution: Let the angles be x , 2 x and 3 x .
Then, x + 2 x + 3 x = 180 ° Þ 6 x = 180 ° Þ x = 30 °
So, the angle are 30 ° , 60 ° and 90 °
Let a , b , c denote the sides apposite to these angles.
Þ a : b : c = sin30 ° : sin60 ° : sin90 °
Þ a : b : c =  Þ a : b : c = 1 :  : 2
Example: The angle of a triangle ABC are in A.P. and it is being given that b : c =  , find Ð A .
Solution: Since Ð A, Ð B, Ð C are in A.P.
\ 2 Ð B = Ð A + Ð C
Þ 3 Ð B = Ð A + Ð B + Ð C
Þ 3 Ð B = 180 ° Þ Ð B = 60 °
Now, 
Þ  Þ  [  b : c =  ]
Þ 
Þ sin C =  Þ Ð C = 45 °
\ Ð A = 180 ° – ( Ð B + Ð C) = 180 ° – (60 ° + 45 ° ) = 75 °
Law of cosine: This law is used to find the unknown parts of a triangle.
In  ABC , 
Example: The lengths of the sides of a triangle are 5, 10, and 12. Solve the triangle.
Solution:
1. 
a » 101.0 °
2. 
b » 24.1 °
3. q » 180 ° – (101.0 ° + 24.1 ° ) = 54.9 °
Trigonometric Identities :
• 
• 
• 
Example: Prove the following trigonometric identity.
Solution:
LHS = 
=  Taking LCM
= 
=  
=  
Example: Prove the following trigonometric identity
Solution:
LHS= 
=  Rationalizing denominator
= 
=  
= 
= 
=   and 
Addition Formulae:
• sin (A + B) = sin A cos B + cos A sin B
sin (A - B) = sin A cos B - cos A sin B
• cos (A + B) = cos A cos B - sin A sin B
cos (A - B) = cos A cos B + sin A sin B
• 
Example: Suppose that  and  , where  and  . Find cos ( a + b ).
Solution: Sketch right–triangle diagrams to help find cos a and cos b .
If  and  , then  .
 
If  and  , then 
Thus,
Example: Find the exact value of sin42 ° cos12 ° – cos 42 ° sin 12 ° .
Solution: Recognizing that this expression fits the formula for sin( u – v ), you can write
sin42 ° cos12 ° – cos 42 ° sin 12 ° = sin (42 ° – 12 ° )
= sin 30 °
= 
Example: Find all solutions of  in the interval [0, 2 p ).
Solution: Using sum and difference formulas, rewrite the equation as
sin x = 
sin x = – 
Therefore, the only solutions in the interval [0, 2 p ) are
x =  and x = 
Double Angle Formulae:
sin2A = 2 sinA cosA
cos2A = cos 2 A - sin 2 A
= 2 cos 2 A - 1
= 1 - 2 sin 2 A
Example: Find all solutions of 2 cos x + sin 2 x = 0
Solution: Begin by rewriting the equation so that it involves functions of x (rather than 2x). Then factor and solve as usual.
2 cos x + sin 2 x = 0 Write original equation
2 cos x + 2 sin x cos x = 0 Double–angle formula
2 cos x (1 + sin x ) = 0 Factor.
cos x = 0 and 1 + sin x = 0 Set factors equal to zero
x =  x =  Solutions in [0, 2 p )
Therefore, the general solution is
x =  + 2n p and x =  + 2n p
where n is an integer.
Example: Express sin 3 x in terms of sin x .
Solution: sin 3 x = sin(2 x + x)
= sin 2 x cos x + cos 2 x sin x
= 2 sin x cos x cos x + (1 – 2 sin 2 x ) sin x
= 2 sin x cos 2 x + sin x – 2 sin 3 x
= 2 sin x (1 – sin 2 x ) + sin x – 2 sin 3 x
= 2 sin x – 2 sin 3 x + sin x – 2 sin 3 x
= 3 sin x – 4 sin 3 x
The graph of trigonometrical functions :
Graph of y = sin X:
Graph of y = cos X:
Graph of y = tan X
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