Proving Trigonometric Identities 193a62

TRIGONOMETRY

Proving Trigonometric Identities

REVIEW Quotient Identities

Reciprocal Identities

Pythagorean Identities

sin 2   cos2   1 1  tan   sec  2

2

cot2   1  csc2 

sin xcot x  cos x tan x  cos x  sin 2 x

Let’s start by working on the left side of the equation….

sin xcot x  cos x tan x  cos x  sin 2 x  cos x  sin x  sin x    cos x    cos x   sin x 

Rewrite the inside the second parenthesis by using the quotient identities

sin xcot x  cos x tan x  cos x  sin 2 x  cos x  sin x  sin x    cos x    cos x   sin x 

Simplify

sin xcot x  cos x tan x  cos x  sin 2 x  cos x  sin x  sin x  sin x       sin x  1  sin x 

To add the fractions inside the parenthesis, you must multiply by one to get common denominators

sin xcot x  cos x tan x  cos x  sin 2 x  cos x sin 2 x  sin x     sin x sin x 

Now that you have the common denominators, add the numerators

sin xcot x  cos x tan x  cos x  sin 2 x  cos x  sin 2 x  sin x   sin x  

Simplify

sin xcot x  cos x tan x  cos x  sin 2 x cos x  sin 2 x  cos x  sin 2 x

Since the left side of the equation is the same as the right side, you’ve successfully proven the identity!

On to the next problem….

cos x  sin x  cos x  sin x 4

4

2

Let’s start by working on the left side of the equation….

2

cos x  sin x  cos x  sin x 4

cos

2

4



2

2



x  sin x cos x  sin x  2

2

We’ll factor the using the difference of two perfect squares technique

2

cos x  sin x  cos x  sin x 4

cos

2

4

2



x  sin x 1  2

Using the Pythagorean Identities the second set of parenthesis can be simplified

2

cos x  sin x  cos x  sin x 4

4

2

2

cos x  sin x  cos x  sin x 2

2

2

2

Since the left side of the equation is the same as the right side, you’ve successfully proven the identity!

On to the next problem….

cos x tan x  sec x  1  sin x

Let’s start by working on the right side of the equation….

cos x tan x  sec x  1  sin x cos x  1  sin x     1  sin x  1  sin x 

Multiply by 1 in the form of the conjugate of the denominator

cos x tan x  sec x  1  sin x cos x(1  sin x)  2 1  sin x

Now, let’s distribute in the numerator….

cos x tan x  sec x  1  sin x cos x  cos x sin x  2 cos x

… and simplify the denominator

cos x tan x  sec x  1  sin x cos x cos x sin x   2 2 cos x cos x

‘Split’ the fraction and

simplify

cos x tan x  sec x  1  sin x 1 sin x   cos x cos x

Use the Quotient and Reciprocal Identities to rewrite the fractions

cos x tan x  sec x  1  sin x

 sec x  tan x

And then by using the commutative property of addition…

cos x tan x  sec x  1  sin x

tan x  sec x  tan x  sec x

… you’ve successfully proven the identity!

One more….

1 1 2   2 csc x 1  cos x 1  cos x

Let’s work on the left side of the equation…

1 1 2   2 csc x 1  cos x 1  cos x

1 1  1  cos x   1  cos x        1  cos x  1  cos x 1  cos x  1  cos x 

Multiply each fraction by one to get the LCD

1 1 2   2 csc x 1  cos x 1  cos x

1  cos x 1  cos x   2 2 1  cos x 1  cos x

Now that the fractions have a common denominator, you can add the numerators

1 1 2   2 csc x 1  cos x 1  cos x

1  cos x  1  cos x  2 1  cos x

Simplify the numerator

1 1 2   2 csc x 1  cos x 1  cos x

2  2 1  cos x

Use the Pythagorean Identity to rewrite the denominator

1 1 2   2 csc x 1  cos x 1  cos x

1 2 2  sin x

Multiply the fraction by the constant

1 1 2   2 csc x 1  cos x 1  cos x

2 csc2 x  2 csc2 x

Use the Reciprocal Identity to rewrite the fraction to equal the expression on the right side of the equation