Lesson Proving Trigonometry identities

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Proving Trigonometry identities


Problem 1

Prove an identity   sin%28x%29%2F%281-cos%28x%29%29 + %281-cos%28x%29%29%2Fsin%28x%29 = 2cosec(x).

Solution 

sin%28x%29%2F%281-cos%28x%29%29 +  %281-cos%28x%29%29%2Fsinx = %28sin%5E2%28x%29+%2B+%281-cos%28x%29%29%5E2%29%2F%28sin%28x%29%2A%281-cos%28x%29%29%29 =  = %282+-+2%2Acos%28x%29%29%2F%28sin%28x%29%2A%281-cos%28x%29%29%29 = 

%282%2A%281-cos%28x%29%29%29%2F%28sin%28x%29%2A%281-cos%28x%29%29%29 = 2%2Fsin%28x%29.


QED.

Problem 2

Prove an identity   1%2F%28cos%28x%29-1%29 + 1%2F%28cos%28x%29%2B1%29 = -2cot(x)*cosec(x).

Solution 

The left side is 

  1%2F%28cos%28x%29-1%29+%2B+1%2F%28cos%28x%29%2B1%29 =  = 

= %282cos%28x%29%29%2F%28%28cos%28x%29-1%29%2A%28cos%28x%29%2B1%29%29 = %282cos%28x%29%29%2F%28cos%5E2%28x%29-1%29 = -%282cos%28x%29%29%2F%281-cos%5E2%28x%29%29 = -%282cos%28x%29%29%2Fsin%5E2%28x%29%29 = -2cot%28x%29%2A1%2Fsin%28x%29.


Exactly as the right side is.  QED.

Problem 3

Prove an identity   %281+%2B+sin%28x%29%29%2Fcos%28x%29 + cos%28x%29%2F%281%2Bsin%28x%29%29 = 2sec(x).

Solution 

%281%2B+sin%28x%29%29%2Fcos%28x%29+%2Bcos%28x%29%2F%281+%2B+sin%28x%29%29 =  = %28%281%2Bsin%28x%29%29%5E2+%2B+cos%5E2%28x%29%29%2F%28cos%28x%29%2A%281%2Bsin%28x%29%29%29 = 

 =        ( <--- use sin%5E2%28x%29+%2B+cos%5E2%28x%29 = 1 )

= %282%2B2sin%28x%29%29%2F%28cos%28x%29%2A%281%2Bsin%28x%29%29%29 =  %282%2A%281%2Bsin%28x%29%29%29%2F%28cos%28x%29%2A%281%2Bsin%28x%29%29%29 =  = 2%2Fcos%28x%29 = 2*sec(x).


QED.

Problem 4

Prove an identity   sin%28a%29%2F%281%2Bsin%28a%29%29 - sin%28a%29%2F%281-sin%28a%29%29 = -2%2Atan%5E2%28a%29.

Solution 

sin%28a%29%2F%281%2Bsin%28a%29%29 - sin%28a%29%2F%281-sin%28a%29%29 = -2%2Atan%5E2%28a%29


Transform left side step by step

     sin%28a%29%2F%281%2Bsin%28a%29%29 - sin%28a%29%2F%281-sin%28a%29%29 = 

   = %28sin%28a%29%2F%281%2Bsin%28a%29%29%29%2A%28%281-sin%28a%29%29%2F%281-sin%28a%29%29%29 - %28sin%28a%29%2F%281-sin%28a%29%29%29%2A%28%281%2Bsin%28a%29%29%2F%281%2Bsin%28a%29%29%29 = 

   = %28sin%28a%29%2A%281-sin%28a%29%29%29%2F%281-sin%5E2%28a%29%29 - %28sin%28a%29%2A%281%2Bsin%28a%29%29%29%2F%281-sin%5E2%28a%29%29 =

   = %28sin%28a%29+-+sin%5E2%28a%29%29%2F%28cos%5E2%28a%29%29 - %28sin%28a%29+%2B+sin%5E2%28a%29%29%2F%28cos%5E2%28a%29%29 = 

   = %28-2%2Asin%5E2%28a%29%29%2Fcos%5E2%28a%29 = -2%2Atan%5E2%28a%29.

Problem 5

Prove that  cos(A) + cos(120°+A) + cos(120°-A) = 0.

Solution 

Use the addition formula for cosine

    cos(A + B) = cos(A)*cos(B) - sin(A)*sin(B)

(see the lesson Addition and subtraction formulas in this site). You will have 


cos(120°+A) = cos(120°)*cos(A) - sin(120°)*sin(A),

cos(120°-A) = cos(120°)*cos(A) + sin(120°)*sin(A).


Add these two equality (both sides). You will get

cos(120°+A) + cos(120°-A) = 2cos(120°)*cos(A).

Now use that cos(120°) = -1%2F2.  Hence, 2cos(120°) = -1.

Therefore,

cos(A) + cos(120°+A) + cos(120°-A) = cos(A) + 2cos(120°)*cos(A) = cos(A) - cos(A) = 0.

QED.


Problem 6

Prove an identity  sin%5E4%28x%29+-+cos%5E4%28x%29 = sin%5E2%28x%29+-+cos%5E2%28x%29.

Solution 
Let me start with this identity: a%5E2+-+b%5E2 = {a+b)*(a-b).

Are you familiar with it? 
If not, or if you are not sure, look into the lesson The difference of squares formula in this site.


OK. Now, apply this identity to the left side. You will get


sin%5E4%28x%29+-+cos%5E4%28x%29 = %28sin%5E2%28x%29+%2B+cos%5E2%28x%29%29%2A%28sin%5E2%28x%29+-+cos%5E2%28x%29%29.


But the first parentheses,  %28sin%5E2%28x%29+%2B+cos%5E2%28x%29%29 is equal to 1, as everybody knows.

Therefore, 

Sin%5E4%28x%29+-+Cos%5E4%28x%29 = sin%5E2%28x%29+-+cos%5E2%28x%29.

QED.


Problem 7

Prove that   = 3%2F2.

Solution 

   =        ( regroup the terms: 1 + 2 + 3 + 4 = (1+4) + (2+3) )

=  =    


     ( use the fact that  sin%28alpha%29 = sin%28pi-alpha%29 and apply it for each of the two parenthesed term )


= %282sin%5E4%28pi%2F8%29+%2B+2sin4%5E4%283pi%2F8%29%29 = 2%2A%28sin%5E4%28pi%2F8%29+%2B+sin%5E4%283pi%2F8%29%29

     ( Now use the identity sin%5E4%28alpha%29+%2B+cos%5E4%28alpha%29 =  = 1+-+2sin%5E2%28alpha%29%2Acos%5E2%28alpha%29 = 1+-+%281%2F2%29%2Asin%5E2%282%2Aalpha%29 ).

= 2%2A%281-%281%2F2%29%2Asin%5E2%282pi%2F4%29%29 = 2%2A%281-%281%2F2%29%2Asin%5E2%28pi%2F2%29%29 = 2%2A%281+-+%281%2F2%29%2A%28sqrt%282%29%2F2%29%5E2%29 = 2%2A%281-%281%2F2%29%5E2%29 = 2%2A%281-1%2F4%29 = 2%2A%283%2F4%29%29%29 = 3%2F2.

QED.


Problem 8

Prove an identity  cos%5E3%28A%29%2Acos%283A%29%2Bsin%5E3%28A%29%2Asin%283A%29 = cos%5E3%282A%29.

Solution 

The keys are these two formulas:

    cos%5E3%28A%29  = %281%2F4%29%2Acos%283A%29+%2B+%283%2F4%29%2Acos%28A%29,        (1)   and

    sin%5E3%28A%29  = %281%2F4%29%2Asin%283A%29+-+%283%2F4%29%2Asin%28A%29.        (2)

(see the lesson Powers of trigonometric functions in this site).  When applying them, you will get

cos^3(A)*cos(3A) =  %281%2F4%29%2Acos%283A%29+%2B+%283%2F4%29%2Acos%28A%29%29%2Acos%283A%29,   (3)  and

sin^3(A)*sin(3A) = %28-1%2F4%29%2Asin%283A%29+-+%283%2F4%29%2Asin%28A%29%29%2Asin%283A%29.   (4)


So, adding and expanding (3) and (4), you will get

  cos%5E3%28A%29%2Acos%283A%29%2Bsin%5E3%28A%29%2Asin%283A%29 = 

=  = 

= [%281%2F4%29%2Acos%5E2%283A%29+-+%281%2F4%29%2Asin%5E2%283A%29] + [%283%2F4%29%2Acos%28A%29%2Acos%283A%29+%2B+%283%2F4%29%2Asin%28A%29%2Acos%283A%29%29] = 

    For the first  bracket  [ . . ]  apply the formula cos(2x) = . . .   
    For the second bracket  [ . . ]  apply the formula cos(x-y) = . . .   You will get

= %281%2F4%29%2Acos%286A%29+%2B+%283%2F4%29%2Acos%28A-3A%29 = %281%2F4%29%2Acos%286A%29+%2B+%283%2F4%29%2Acos%282A%29 = 

    Now apply again the formula (1). You will get

= cos%5E3%282A%29.

Problem 9

Show that   2%2Asin%284pi%2F5%29%2Acos%28pi%2F5%29 = sin%282pi%2F5%29.

Solution 

First use the identity  


    sin%28pi-alpha%29 = sin%28alpha%29,


which is valid for any angle alpha.  From this identity, you get


    sin%284pi%2F5%29 = sin%28pi%2F5%29.


It gives you


    2%2Asin%284pi%2F5%29%2Acos%28pi%2F5%29 = 2%2Asin%28pi%2F5%29%2Acos%28pi%2F5%29.


Next use the identity  2%2Asin%28alpha%29%2Acos%28alpha%29 = sin%282%2Aalpha%29,  which is valid for any angle  alpha.  It gives you


    2%2Asin%284pi%2F5%29%2Acos%28pi%2F5%29 = 2%2Asin%28pi%2F5%29%2Acos%28pi%2F5%29 = sin%282pi%2F5%29.


It is what has to be proved.

Problem 10

Show that   cos%5E2%284pi%2F7%29 - sin%5E2%283pi%2F7%29 = cos%286pi%2F7+%29.

Solution 

First use the identity


    cos%284pi%2F7%29 = -cos%283pi%2F7%29,


which gives you


    cos%5E2%284pi%2F7%29 = cos%5E2%283pi%2F7%29.


Therefore, the left side of the given hypothetical identity becomes


    cos%5E2+%284pi%2F7%29 - sin%5E2+%283pi%2F7%29 = cos%5E2+%283pi%2F7%29 - sin%5E2+%283pi%2F7%29.    (1)


Next, use the trigonometric identity  


    cos(a)*cos(b) - sin(a)*sin(b) = cos(a+b).


It allows you to continue the line (1) in this way


    cos%5E2+%284pi%2F7%29 - sin%5E2+%283pi%2F7%29 = cos%5E2+%283pi%2F7%29 - sin%5E2+%283pi%2F7%29 = cos%283pi%2F7+%2B+3pi%2F7%29 = cos%286pi%2F7%29.


Thus 


    cos%5E2+%284pi%2F7%29 - sin%5E2+%283pi%2F7%29 = cos%286pi%2F7%29.


It is what has to be proved.

Problem 11

Prove that   cos%283%2Atheta%29 = 4%2Acos%5E3%28theta%29 - 3%2Acos%28theta%29.

Solution

This problem is to be solved in 2 easy steps.


Step 1 

cos(2a) = cos(a + a) = cos(a)*cos(a) - sin(a)*sin(a) = cos^2(a) - sin^2(a) = cos^2(a) - (1-cos^2(a)) = 2*cos^2(a) - 1.


Step 2 

cos(3a) = cos(2a + a) = cos(2a)*cos(a) - sin(2a)*sin(a) = (2cos^2(a)-1)*cos(a) - (2*sin(a)*cos(a))*sin(a) = 

                      = 2*cos^3(a) - cos(a) - 2*sin^2(a)*cos(a) = 2*cos^3(a) - cos(a) - 2*(1-cos^2(a))*cos(a) = 

                      = 2*cos^3(a) - cos(a) - 2*cos(a) + 2*cos^2(a) = 4*cos^3(a) - 3*cos(a).

The proof is completed.


Problem 12

Show that  cos%282pi%2F9%29  is a root of the equation  8x^3 - 6x + 1 = 0 .

Solution

            This problem has beautiful, nice, elegant and unexpected solution. 

Use the formula 


    cos%283%2Atheta%29 = 4%2Acos%5E3%28theta%29 - 3%2Acos%28theta%29.


This formula is valid for any angle theta.


    For its proof see the Problem 10 above. 


Let  theta = 2pi%2F9  and let  x = cos%282pi%2F9%29.


Notice that  3%2Atheta = %283%2A2%2Api%29%2F9 = 6pi%2F9 = %282%2F3%29%2Api = 120°.


Hence,  cos%283%2Atheta%29 = -1%2F2.


From the other side,  cos%283%2Atheta%29 = 4%2Acos%5E3%28theta%29 - 3%2Acos%28theta%29,  according to the formula above.


In other words, 


    4x%5E3 - 3x = -1%2F2.


Multiplying by 2 both sides and simplifying, you get


    8x%5E3 - 6x + 1 = 0.


It means that   x = cos%282pi%2F9%29  is the solution of the given equation.

The proof is completed.


Problem 13

Let  x  and  y  are acute angles,  tan(x).tan(y) = 1,  tan(x) - tan(y) = 2*sqrt(3).  Find  x  and  y  and prove that  x = 5y.

Solution 

Let x be the greater of the two given acute angles, and let y be the smaller.


Then, first, the equality  tan(x)*tan(y) = 1  implies that

    x + y = 90°.         (1)


Second, 

    tan(x-y) = %28tan%28x%29-tan%28y%29%29%2F%281+%2B+tan%28x%29%2Atan%28y%29%29 = 

         substitute given values for the numerator and denominator to get

              = %28%282%2Asqrt%283%29%29%29%2F%281%2B1%29 = %28%282%2Asqrt%283%29%29%29%2F2 = sqrt%283%29,

which implies  

    x - y = 60°.           (2)


From equations (1) and (2), by adding, you get

    2x = 90° + 60° = 150°;   hence,  x = 150%5Eo%2F2 = 75°.


Finally, substituting this value of x into (1), you get  y = 15°.


So, under the given conditions,  x = 75°  and  y = 15°.


In particular,  x = 5y.

Problem 14

Let  tan(A) = 1/3,  where  0 < A < pi%2F2.  Show that  4A = arctan%2824%2F7%29

Solution 

Use the formula for tan(2A)


    tan(2A) = %28tan%28A%29%2Btan%28A%29%29%2F%281-tan%5E2%28A%29%29 = %282%2Atan%28A%29%29%2F%281-tan%5E2%28A%29%29 = %28%282%2F3%29%29%2F%28%281-%281%2F3%29%5E2%29%29 = %28%282%2F3%29%29%2F%28%288%2F9%29%29 = 3%2F4.


Use the formula for tan(2A), again


    tan(4A) = %28tan%282A%29%2Btan%282A%29%29%2F%281-tan%5E2%282A%29%29 = %282%2Atan%282A%29%29%2F%281-tan%5E2%282A%29%29 = %28%286%2F4%29%29%2F%28%281-%283%2F4%29%5E2%29%29 = %28%286%2F4%29%29%2F%28%287%2F16%29%29 = 24%2F7.


Therefore,  4A = arctan%2824%2F7%29.

Problem 15

Without using a calculator,  find the value of  tan(6°) x tan(12°) x tan(18°) x . . . x tan(84°).

Solution 
The key to the solution is this identity  tan(x) = 1%2Ftan%28pi%2F2-x%29.

It gives


    tan(6°)*tan(84°) = 1

    tan(12°)*tan(78°) = 1

    tan(18)*tan(72°) = 1

    . . . and so on . . . 

    tan(42°)*tan(48°) = 1.



By multiplying all these identities, you will get finally


    tan(6°)*(tan(12°)*tan(18°)* . . . *tan(84°) = 1.


ANSWER.  This product is equal to 1.

Problem 16

Find an identity for   cos(4t)   in terms of   cos(t).

Solution 


Use the general trigonometric formula


    cos(2a) = 2cos^2(a) - 1.


twice.  It gives 


    cos(4t) = 2cos^2(2t) - 1 = 2*(2cos^2(t)-1)^2 - 1 = 2*(4cos^4(t) - 4cos^2(t) +1) - 1 = 

            = 8cos^4(t) - 8cos^2(t) + 1.     ANSWER


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Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.

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