Transcript MAT 1275: Introduction to Mathematical Analysis Dr. A. Rozenblyum

MAT 1275: Introduction to
Mathematical Analysis
Dr. A. Rozenblyum
VI. Trigonometric Identities and Equations
D. Multiple and Half-Angle Formulas
Multiple and Half-Angle Formulas
These formulas express trig functions of multiple (actually, double) angles and of halfangles through the trig functions of a single angle. As many formulas from section VI.C,
these formulas can be derived very easily from the basic formula for the cosine of the
difference or from some of its subsequent formulas.
Example 1. Derive the formula for the sine of a double angle:
sin 2 A  2 sin A  cos A .
Solution. Consider the sine of the sum formula (VI.C, example 6)
sin( A  B )  sin A  cos B  cos A  sin B
and put here B = A.
Example 2. Derive the following three formulas for the cosine of a double angle:
2
2
a) cos 2 A  cos A  sin A .
2
b) cos 2 A  1  2 sin A .
2
c) cos 2 A  2 cos A  1 .
Solution.
a) Consider the cosine of the sum formula (VI.C, example 5)
cos(A  B )  cos A  cos B  sin A  sin B
and put here B = A.
b) Substitute cos2 A  1  sin2 A into the formula cos 2 A  cos2 A  sin2 A .
2
2
2
2
c) Substitute sin A  1  cos A into the formula cos 2 A  cos A  sin A .
Multiple and Half-Angle Formulas
Example 3. Derive the formula for the tangent of a double angle:
tan 2 A 
2 tan A
.
1  tan 2 A
Solution. Consider the tangent of the sum formula (VI.C, example 8)
tan( A  B) 
tan A  tan B
1  tan A  tan B
and put here B = A.
Now, let’s derive half-angle formulas.
Example 4. Derive the formula for the sine of a half angle:
sin
A
1  cos A

.
2
2
Note. The “plus-minus” sign in this formula means thatsin
A
2
is not defined uniquely by
cos A . To get unique value, additional info is needed such as the quadrant where angle
A
is located.
2
Solution of the example 4. Consider the formula for the cosine of a double angle from
A
2 A
2
example 2 b): cos 2 A  1  2 sin A . Replace here A to
: cos A  1  2 sin
. Now,
2
2
A
solve this equation for sin :
2
A cos A  1
A
A 1  cos A
cos A  1  2 sin 2 ,
 sin 2 , sin2 
.
2
2
2
2
2
Finally, take square root from both sides.
Multiple and Half-Angle Formulas
Example 5. Derive the formula for the cosine of a half angle:
cos
A
1  cos A

.
2
2
Solution. It is similar to the example 4, just start with the formula from example 2 c). The
same note about “plus-minus” sign is true as for example 4.
Multiple and Half-Angle Formulas
Example 6. Derive the following two formulas for the tangent of a half angle:
A 1  cos A

.
2
sin A
A
sin A
b) tan 
.
2 1  cos A
a) tan
Solution.
a) Using results of examples 4 and 5, we have
tan
A
A
A
1  cos A
1  cos A
1  cos A
 sin  cos  


.
2
2
2
2
2
1  cos A
Multiply numerator and denominator by 1 cos A :
A
(1  cos A)  (1  cos A)
(1  cos A) 2
(1  cos A) 2 1  cos A
tan  



.
2
(1  cos A)  (1  cos A)
1  cos2 A
sin 2 A
sin A
We dropped here the “plus-minus” sign for the following reason. The numerator
1 cos A is never negative ( cos A  1 ). The denominator sin A has always the same sign
A
A




. Indeed, if 0  A  180 , then sin A  0 . In this case 0   90 , and also
2
2
A
A


tan  0. If 180  A  360 , then sin A  0 . In this case 90   180 , and also
2
2
A
tan  0.
2
b) Multiply top and bottom of the formula from part a) by 1 cos A :
as tan
A 1  cos A (1  cos A)  (1  cos A)
(1  cos2 A)
tan 


2
sin A
sin A  (1  cos A)
sin A  (1  cos A)
2
sin A
sin A


.
sin A  (1  cos A) 1  cos A
Multiple and Half-Angle Formulas
In conclusion, consider several specific problems.
Example 7. Find the exact values of sin 2 , cos 2 , and tan 2 , if sin 
3
and  is
5
in the second quadrant.
Solution. To use double angle formulas, calculate cos  . In the second quadrant cosine is
2
2
negative, therefore using the main identity sin   cos   1 , we have
2
4
 3
cos   1  sin    1      .
5
5
2
By the formula for the sine of a double angle (example 1),
3  4
24
sin 2  2 sin  cos  2       
.
5  5
25
To find cos 2 , any formula from the example 2 can be used. Let’s use the formula from
part b) (to work with original info):
2
7
 3
cos 2  1  2 sin   1  2    
.
25
5
2
To find tan 2 , we can use the formula for the tangent of a double angle (example 3),
however, it is easier to use the above results for sin 2 cos 2 :
tan 2 
Final answer: sin 2  
sin 2
24 7
24
   .
cos 2
25 25
7
24
7
24
, cos 2 
, tan 2  
25
25
7
Multiple and Half-Angle Formulas
Example 8. Find the exact values of sin
180    270 .



Solution. Because 180    270 , angle
so

2
2

2
, cos

2
, and tan

2
, if cos  

is located in the interval 90 
is in the second quadrant. In this quadrant, sin

2

 0 , cos  0 .
2
By the formula for the sine of a half angle (example 4) with the “plus” sign,
sin

2

1  cos
5
3
3 13

 1   / 2 

.
2
13
13
 13 
By the formula for the cosine of a half angle (example 5) with the “minus” sign,

1  cos
5
2
2 13

  1   / 2  

.
2
13
13
 13 
cos  
2
As in example 7, to find tan
tan

2
, it is easier to use the formula



3  2 
3
 sin  cos 

 .
2
2
2
2
13  13 
Final answer: sin

2


3
3 13

2 13
, cos  
, tan   .
13
2
13
2
2

2
5
and
13
 135 ,
Multiple and Half-Angle Formulas
Example 9. Solve the equation sin 2   cos  where 0    360 .
Solution. Using the sine of a double angle formula from example 1, we modify the
equation:
2 sin   cos    cos  , 2 sin   cos   cos   0 , cos  2 sin  1  0 .
The equation is split into two equations cos   0 and 2 sin   1  0 .
a) The equation cos  0 was solved in section VI.B, example 6. In the interval
0    360 it has two solutions90 and 270 .
b) The equation 2 sin   1  0 can be written as sin  
1
. This is a simplest equation
2
of the type sin x  a for –1< a < 0 that we considered in VI.C. In the interval
1 1
0    360 it has two solutions. To get them, consider the equation sin x   
2 2

for the reference (acute) angle x . Using a table from section V.C, x  30 . The two
solutions of the equation sin  
1
are
2
x  180  x  180  30  210 and x  360  x  360  30  330 .
Final answer: original equation has four solutions90 , 210 , 270 , and 330 .
Multiple and Half-Angle Formulas
x
2
Example 10. Solve the equation cos  cos x  0 where 0  x  2 .
Solution. We will use the formula for the cosine of a double angle from example 2 c):
cos 2 A  2 cos2 A  1 . Putting A 
x
2 x
, we have cos x  2 cos  1 . Substitute this
2
2
expression into the original equation:
x 
x 
x
x
x
x
cos   2 cos2  1  0 , cos  2 cos2  1  0 , 2 cos2  cos  1  0 .
2 
2 
2
2
2
2
x
We’ve got a quadratic equation with respect to cos . It can be factored:
2
x  
x 

 2 cos  1   cos  1  0 .
2  
2 

x
x
This equation is split into two equations 2 cos  1  0 and cos  1  0 . Each of them
2
2
x
1
x
can be written as simplest trig equations: cos  
and cos  1 .
2
2
2
Multiple and Half-Angle Formulas
Example 10 (continue)
x
2
1
in the interval 0  x  2 .
2
x
1
In we denote t  , the equation takes the form cost   . It is of the type cos t  a
2
2
0

t

2
 , this equation has two
that we considered in section VI.C. In the interval
1
solutions. One of them is obtuse angle (because a    0 ). To find it, we can consider
2
1 1
equation: cost     for the reference (acute) angle t  . Using a table from section
2 2

 2
V.C, t   . Therefore, t    t     
.
3
3
3
1
2
4
The second solution of the equation cost   is 2 
, sot 
. Now we back to
2
3
3
x
1
x
the equation cos   . Because  t , we have x = 2t. From here, the two solutions
2
2
2
x
1
2 4
4 8


of the equation cos   are x  2 
and x  2 
. The last value
2
2
3
3
3
3
8
 2 , so we reject it.
3
a) Let’s solve the equation cos  
x
2
Finally, the equation cos  
x
4
.
3
1
in the interval 0  x  2 has the only solution
2
Multiple and Half-Angle Formulas
Example 10 (continue)
b) Let’s solve the equation cos
x
 1.
2
In section VI.B, example 4, we solved the equation cos x  1 . The answer was the only
solution x = 0. The same solution is true for the equation cos
x
 1.
2
x
2
Final answer for example 10: in the interval 0  x  2 the equation cos  
two solutions: 0 and
4
.
3
Example 11. Prove the identity tan x  cot x  2 csc 2x .
Solution. Let’s modify the left side:
sin x cos x sin 2 x  cos2 x
1
tan x  cot x 



cos x sin x
cos x  sin x
cos x  sin x

2
2

 2 csc 2 x .
2 cos x  sin x sin 2 x
End of the Topic
1
2
has