1. Trigonometry Identities and Equations
Pythagorean Identities

2. Pythagorean Identities
Let 𝑷 𝒄𝒐𝒔𝜽, 𝐬𝐢𝐧 𝜽 represent any point on the unit circle, 𝒙 𝟐 + 𝒚 𝟐 =𝟏. Substituting 𝒄𝒐𝒔𝜽 for 𝒙 and 𝒔𝒊𝒏𝜽 for 𝒚 produces a fundamental Pythagorean identity.
Fundamental Pythagorean and Identity
𝒔𝒊𝒏 𝟐 𝜽+ 𝒄𝒐𝒔 𝟐 𝜽=𝟏

3. Pythagorean Identities
Dividing each term of 𝒔𝒊𝒏 𝟐 𝜽+ 𝒄𝒐𝒔 𝟐 𝜽=𝟏 by either 𝒔𝒊𝒏 𝟐 𝜽 or 𝒄𝒐𝒔 𝟐 𝜽 leads to two additional Pythagorean identities that are listed below. Each of the three Pythagorean identities can be written in more than one way.
Three Pythagorean Trigonometry Identities
Some Equivalent Forms
𝒔𝒊𝒏 𝟐 𝜽+ 𝒄𝒐𝒔 𝟐 𝜽=𝟏
𝒔𝒊𝒏 𝟐 𝜽=𝟏− 𝒄𝒐𝒔 𝟐 𝜽 or
𝒄𝒐𝒔 𝟐 𝜽=𝟏− 𝒔𝒊𝒏 𝟐 𝜽
𝒕𝒂𝒏 𝟐 𝜽+𝟏= 𝒔𝒆𝒄 𝟐 𝜽
𝒕𝒂𝒏 𝟐 𝜽=𝟏− 𝒔𝒆𝒄 𝟐 𝜽 or
𝒔𝒆𝒄 𝟐 𝜽=𝟏− 𝒕𝒂𝒏 𝟐 𝜽
𝒄𝒐𝒕 𝟐 𝜽+𝟏= 𝒄𝒔𝒄 𝟐 𝜽
𝒄𝒐𝒕 𝟐 𝜽=𝟏− 𝒄𝒔𝒄 𝟐 𝜽 or
𝒄𝒔𝒄 𝟐 𝜽=𝟏− 𝒄𝒐𝒕 𝟐 𝜽

4. Pythagorean Identities
Example 1
Using a Pythagorean Identity to Simplify an expression
If 𝑦 = cos 𝐴 sec 𝐴− cos 𝐴 , then which choice is equivalent to 𝑦?
(1) 𝑐𝑜𝑠 2 𝐴
(2) cos 𝐴− sin 𝐴
(3) 𝑠𝑖𝑛 2 𝐴
(4) cot 𝐴−1
Replace sec 𝐴 with 𝟏 𝐜𝐨𝐬 𝑨 :
Solution:
𝒚 = 𝒄𝒐𝒔 𝑨 𝒔𝒆𝒄 𝑨− 𝒄𝒐𝒔 𝑨
= 𝒄𝒐𝒔 𝑨 𝟏 𝐜𝐨𝐬 𝑨 − 𝐜𝐨𝐬 𝑨
=𝟏− 𝒄𝒐𝒔 𝟐 𝑨
= 𝒔𝒊𝒏 𝟐 𝑨

5. Pythagorean Identities
Example 2
Using a Pythagorean Identity to Simplify an expression
If csc 𝑥≠1 which expression equivalent to 𝑐𝑜𝑡 2 𝑥 1 + csc 𝑥 ?
(1) 1− csc 𝑥
(2) csc 𝑥+1
(3) −csc 𝑥
(4) csc 𝑥− cot 𝑥
Solution:
Rewrite the numerator using the Pythagorean identity, 𝑐𝑜𝑡 2 𝑥= 𝑐𝑠𝑐 2 𝑥−1:
𝒄𝒐𝒕 𝟐 𝒙 𝟏+ 𝐜𝐬𝐜 𝒙 = 𝒄𝒔𝒄 𝟐 𝒙−𝟏 𝟏+ 𝐜𝐬𝐜 𝒙
= 𝐜𝐬𝐜 𝒙+𝟏 𝐜𝐬𝐜 𝒙−𝟏 𝟏+ 𝐜𝐬𝐜 𝒙
Factor:
= 𝐜𝐬𝐜 𝒙−𝟏

6. Proving Trigonometric Identities
Proving that a trigonometry equation is an identity involves showing that the two sides of the equation can be made to look exactly alike. Start with the more complicated side of the equation, and express it using only sines and cosines. Continue to work on the same side of the equation by doing one or more of the following:
Factoring
Combining and simplifying fractional terms.
Making a substitution using a known trigonometric identity such as a quotient, reciprocal, or Pythagorean identity.

7. Proving Trigonometric Identities
Example 3
Proving a Trigonometric Identity
Prove that the following equation is an identity for all values of 𝜃 for which the expression are defined:
𝐬𝐞𝐜 𝜽− 𝐜𝐨𝐬 𝜽= 𝐬𝐢𝐧 𝜽 𝐭𝐚𝐧 𝜽
Solution:
Draw a vertical boundary line separating the left and right sides of the equation. Working independently on each side of the equation, change secant and tangent to sine and cosine. Then simplify

8. Proving Trigonometric Identities
Example 3
Proving a Trigonometric Identity
Prove that the following equation is an identity for all values of 𝜃 for which the expression are defined:
𝐬𝐞𝐜 𝜽− 𝐜𝐨𝐬 𝜽= 𝐬𝐢𝐧 𝜽 𝐭𝐚𝐧 𝜽
𝟏 𝐜𝐨𝐬 𝜽 − 𝐜𝐨𝐬 𝜽
Use reciprocal identity:
Rewrite the second term cos 𝜽 , as a fraction having cos 𝜽 as its denominator:
𝟏 𝐜𝐨𝐬 𝜽 − 𝒄𝒐𝒔 𝟐 𝜽 𝐜𝐨𝐬 𝜽
𝟏−𝒄𝒐𝒔 𝟐 𝜽 𝐜𝐨𝐬 𝜽
Combine fractions:

9. Proving Trigonometric Identities
Example 3
Proving a Trigonometric Identity
Prove that the following equation is an identity for all values of 𝜃 for which the expression are defined:
𝐬𝐞𝐜 𝜽− 𝐜𝐨𝐬 𝜽= 𝐬𝐢𝐧 𝜽 𝐭𝐚𝐧 𝜽
𝒔𝒊𝒏 𝟐 𝜽 𝐜𝐨𝐬 𝜽
Use the Pythagorean identity in the numerator:
𝒔𝒊𝒏 𝟐 𝜽 𝐜𝐨𝐬 𝜽
𝒔𝒊𝒏 𝜽 𝒔𝒊𝒏 𝜽 𝐜𝐨𝐬 𝜽
Change the right side to sines and cosines:
Multiply on the right side:
𝒔𝒊𝒏 𝟐 𝜽 𝐜𝐨𝐬 𝜽 =
𝒔𝒊𝒏 𝟐 𝜽 𝐜𝐨𝐬 𝜽

10. Solving Trigonometric Equations
Trigonometry equations that are not identities are solved in much the same ways as algebraic equations, but with two important exceptions:
A trigonometric equation is generally solved in two stages.
First Solve for the trigonometric function.
Then find the angle.
There may be more than one solution since a trigonometric function is positive in two quadrants and negative in two quadrants.
A substitution using Pythagorean, quotient, or reciprocal identity may be needed.

11. Solving Trigonometric Equations
Example 1
Solving a Linear Trigonometric Equation
Solve 2 𝑠𝑖𝑛 𝑥+1=0 for 𝑥, where 0≤𝑥<2𝜋:
Solution:
Solve for 𝒔𝒊𝒏 𝒙 :
If 2 𝑠𝑖𝑛 𝑥+1=0 , then 𝑠𝑖𝑛 𝑥=− 1 2
Because 𝑥= 𝑠𝑖𝑛 2 − 1 2 , the reference angle is 𝜋 6 .
Solve for ∠𝒙:
Sine is negative in Quadrant III and IV, so there are two possible solutions
𝑸 𝑰𝑰𝑰 : 𝒙 𝟏 =𝝅+ 𝝅 𝟔 = 𝟕𝝅 𝟔
or 𝑸 𝑰𝑽 : 𝒙 𝟐 =𝟐𝝅− 𝝅 𝟔 = 𝟏𝟏𝝅 𝟔
So the solution are 𝟕𝝅 𝟔 and 𝟏𝟏𝝅 𝟔

12. Solving Trigonometric Equations
Example 2
Solving a Quadratic Trigonometric Equation
Solve 𝒕𝒂𝒏 𝟐 𝒙= 𝒕𝒂𝒏 𝒙+𝟐 , where 𝟎°≤𝒙<𝟑𝟔𝟎°, for 𝑥 to the nearest tenth of a degree.
Solution:
Rewrite 𝒕𝒂𝒏 𝟐 𝒙= 𝒕𝒂𝒏 𝒙+𝟐 in standard form. Then solve by factoring.
Solve for 𝒕𝒂𝒏 𝒙 :
𝒕𝒂𝒏 𝟐 𝒙− 𝒕𝒂𝒏 𝒙−𝟐=𝟎
𝒕𝒂𝒏 𝒙+𝟏 𝒕𝒂𝒏 𝒙−𝟐 =𝟎
𝒕𝒂𝒏 𝒙+𝟏 =0
or 𝒕𝒂𝒏 𝒙−𝟐 =0

13. Solving Trigonometric Equations
Example 2
Solving a Quadratic Trigonometric Equation
Solve 𝒕𝒂𝒏 𝟐 𝒙= 𝒕𝒂𝒏 𝒙+𝟐 , where 𝟎°≤𝒙<𝟑𝟔𝟎°, for 𝑥 to the nearest tenth of a degree.
Solution:
Rewrite 𝒕𝒂𝒏 𝟐 𝒙= 𝒕𝒂𝒏 𝒙+𝟐 in standard form. Then solve by factoring.
Solve for ∠𝒙:
If 𝑡𝑎𝑛 𝑥=−1 , the reference angle is 45°.
Tangent is negative in Quadrant II and IV:
Thus:
𝑸 𝑰𝑰 : 𝒙 𝟏 =𝟏𝟖𝟎°−𝟒𝟓=𝟏𝟑𝟓°
or 𝑸 𝑰𝑽 : 𝒙 𝟐 =𝟑𝟔𝟎°−𝟒𝟓°=𝟑𝟏𝟓°
Use the calculator to find the reference angle
If 𝑡𝑎𝑛 𝑥=2 , 𝑥= 𝑡𝑎𝑛 −1 2.
∠𝒙=𝟔𝟑.𝟒°
Because tangent is positive in Quadrant I and III:
𝑸 𝑰 : 𝒙 𝟑 ≈𝟔𝟑.𝟒°
or 𝑸 𝑰𝑰𝑰 : 𝒙 𝟒 =𝟏𝟖𝟎°+𝟔𝟑.𝟒°=𝟐𝟒𝟑.𝟒°

14. Solving Trigonometric Equations
Example 2
Solving a Quadratic Trigonometric Equation
Solve 𝒕𝒂𝒏 𝟐 𝒙= 𝒕𝒂𝒏 𝒙+𝟐 , where 𝟎°≤𝒙<𝟑𝟔𝟎°, for 𝑥 to the nearest tenth of a degree.
Solution:
Rewrite 𝒕𝒂𝒏 𝟐 𝒙= 𝒕𝒂𝒏 𝒙+𝟐 in standard form. Then solve by factoring.
Use the calculator to find the reference angle
Solve for ∠𝒙:
If 𝑡𝑎𝑛 𝑥=2 , 𝑥= 𝑡𝑎𝑛 −1 2.
∠𝒙=𝟔𝟑.𝟒°
Tangent is positive in Quadrant I and II:
or 𝑸 𝑰𝑽 : 𝒙 𝟐 =𝟐𝝅− 𝝅 𝟔 = 𝟏𝟏𝝅 𝟔
𝑸 𝑰 : 𝒙 𝟏 =𝟏𝟖𝟎°−𝟒𝟓
Hence there are four solutions: 𝟔𝟑.𝟒°, 𝟏𝟑𝟓°, 𝟐𝟒𝟑.𝟒° and 𝟑𝟏𝟓°

15. Solving Trigonometric Equations Requiring Substitution
If a trigonometry equation contains two different trigonometric functions, use a trigonometric identity to transform the equation into an equivalent equation that contains the same trigonometric function.

16. Solving Trigonometric Equations Requiring Substitution
Example 3
Solving a Trigonometric Equation Using Substitution
Solve for x to the nearest of a degree: 3 𝑐𝑜𝑠 2 +5 𝑠𝑖𝑛 𝑥=4 0°≤𝑥≤360°
Solution:
Because the equation contains two different trigonometric functions, one should be eliminated. Transform the original equation into an equivalent equation that contains only the sine function by replacing with:
𝟑 𝒄𝒐𝒔 𝟐 𝒙+𝟓 𝒔𝒊𝒏 𝒙=𝟒
𝟑 𝟏− 𝒔𝒊𝒏 𝟐 𝒙 +𝟓 𝒔𝒊𝒏 𝒙=𝟒
𝟑 −𝟑𝒔𝒊𝒏 𝟐 𝒙+𝟓 𝒔𝒊𝒏 𝒙=𝟒
𝟑𝒔𝒊𝒏 𝟐 𝒙−𝟓 𝒔𝒊𝒏 𝒙+𝟏=𝟎
Solve for 𝒔𝒊𝒏𝒙 using the quadratic formula:

17. Solving Trigonometric Equations Requiring Substitution
Example 3
Solving a Trigonometric Equation Using Substitution
Solve for x to the nearest of a degree: 3 𝑐𝑜𝑠 2 +5 𝑠𝑖𝑛 𝑥=4 0°≤𝑥≤360°
Solution:
𝟑𝒔𝒊𝒏 𝟐 𝒙−𝟓 𝒔𝒊𝒏 𝒙+𝟏=𝟎
Solve for 𝒔𝒊𝒏𝒙 using the quadratic formula:
𝒔𝒊𝒏 𝒙 = −𝒃± 𝒃 𝟐 −𝟒𝒂𝒄 𝟐𝒂
= −(−𝟓)± (−𝟓) 𝟐 −𝟒(𝟑)(𝟏) 𝟐(𝟑)
Let 𝑎= 3, 𝑏 = −5 and 𝑐 = 1
= 𝟓± 𝟏𝟑 𝟔
= 𝟓±𝟑.𝟔𝟎𝟓𝟓𝟓 𝟔

18. Solving Trigonometric Equations Requiring Substitution
Example 3
Solving a Trigonometric Equation Using Substitution
Solve for x to the nearest of a degree: 3 𝑐𝑜𝑠 2 +5 𝑠𝑖𝑛 𝑥=4 0°≤𝑥≤360°
Solution:
𝟑𝒔𝒊𝒏 𝟐 𝒙−𝟓 𝒔𝒊𝒏 𝒙+𝟏=𝟎
= 𝟓±𝟑.𝟔𝟎𝟓𝟓𝟓 𝟔
Solve for ∠𝒙:
𝒔𝒊𝒏 𝒙 = 𝟓−𝟑.𝟔𝟎𝟓𝟓𝟓 𝟔
𝒔𝒊𝒏 𝒙 = 𝟓+𝟑.𝟔𝟎𝟓𝟓𝟓 𝟔
≈𝟎.𝟐𝟑𝟐𝟒
≈𝟏.𝟑𝟗𝟑𝟒𝟑
𝒙= 𝒔𝒊𝒏 −𝟏 𝟎.𝟐𝟑𝟐𝟒
Reject since the maximum value of sin⁡𝑥 is 1.
𝒙 𝑹𝒆𝒇 =𝟏𝟑.𝟓°
𝑸 𝑰 : 𝒙 𝟏 =𝟏𝟑.𝟓°
The equation has two solution: 𝟏𝟑.𝟓° and 𝟏𝟔𝟔.𝟓°
𝑸 𝑰𝑰 : 𝒙 𝟐 =𝟏𝟖𝟎°−𝟏𝟑.𝟓°=𝟏𝟔𝟔.𝟓°