NAME: _____________________________
PLEASE WRITE YOUR NAME ON ALL PAGES.
TEACHER: I MADARANG
Note: You can find more examples and discussion
SUBJECT: ADVANCED MATH WEEK 2
DUE: May 8th
in section 5.1 of your book.
PERIOD: __________________
WEEK 2: Trigonometric Identities
Use the identities below to:
Part 1. Evaluate trigonometric functions.
Part 2. Simplify trigonometric expressions.
Part 3. Use reciprocal, quotient, Pythagorean, cofunction and even/odd identities to simplify expressions.
(Proving identities)
Recall that in the Third Quarter, you received a master sheet on the trigonometric identities. See list
below:
PART I. Using Identities to Evaluate Functions.
EXAMPLE 1: Use the values to evaluate the missing
trig functions.
𝟕
𝟐𝟓
Given: 𝐭𝐚𝐧 𝒙 = 𝟐𝟒 , 𝐬𝐞𝐜 𝒙 = − 𝟐𝟒
Looking at the ratios above, we can say that y = 7, x = 24 and r = 25.
To find the other trig ratios, we will use these values.
Our answers will therefore be:
𝟕
𝟐𝟒
𝟕
𝐬𝐢𝐧 𝒙 =
,
𝐜𝐨𝐬 𝒙 = −
,
𝐭𝐚𝐧 𝒙 =
𝟐𝟓
𝟐𝟓
𝟐𝟒
𝟐𝟓
𝐜𝐬𝐜 𝒙 =
,
𝟕
𝟐𝟓
𝐬𝐞𝐜 𝒙 = −
,
𝟐𝟒
𝟐𝟒
𝐜𝐨𝐭 𝒙 =
𝟕
EXAMPLE 2:
𝟒
Given: 𝐜𝐨𝐬 𝜽 = 𝟕 𝒂𝒏𝒅 𝐭𝐚𝐧 𝜽 < 𝟎
Looking at the ratios above, we can say that x = 4 and
r = 7. Also, if x is positive and 𝐭𝐚𝐧 𝜽 is negative, we can
infer that y is also negative. To find y, we can use the
Pythagorean theorem
𝒙𝟐 + 𝒚 𝟐 = 𝒓 𝟐
𝟒𝟐 + 𝒚𝟐 = 𝟕𝟐
𝒚𝟐 = 𝟕𝟐 − 𝟒𝟐
𝒚 = ±√𝟒𝟗 − 𝟏𝟔 = ± √𝟑𝟑
Taking only the negative answer, we have 𝒚 = −√𝟑𝟑.
Our answers will be:
𝟒
√𝟑𝟑
√𝟑𝟑
𝐬𝐢𝐧 𝜽 = −
,
𝐜𝐨𝐬 𝜽 = ,
𝒕𝒂𝒏 𝜽 = −
𝟕
𝟕
𝟒
𝐜𝐬𝐜 𝜽 = −
𝟕√𝟑𝟑
,
𝟑𝟑
𝐬𝐞𝐜 𝜽 =
𝟕
,
𝟒
𝐜𝐨𝐭 𝜽 = −
** Don’t forget to rationalize radicals and
simplify fractions!
𝟒√𝟑𝟑
𝟑𝟑
NAME ______________________________________ PERIOD _______________
EXAMPLE 3:
𝝅
𝟑
Given: 𝐜𝐨𝐬 ( − 𝒙) = 𝒂𝒏𝒅 𝒔𝒊𝒏 𝒙 > 𝟎
EXAMPLE 4:
From the list of co-function identities in the first page,
𝝅
𝟑
𝐜𝐨𝐬 ( 𝟐 − 𝒙) = 𝐬𝐢𝐧 𝒙. This means that 𝐬𝐢𝐧 𝒙 = 𝟓.
So, y = 3, and r = 5. Using the Pythagorean theorem,
we can solve for x = 4.
From the list of even/odd identities, 𝒔𝒊𝒏 (−𝜽) = −𝒔𝒊𝒏 𝒙.
𝟐
𝟐
Since −𝒔𝒊𝒏 𝜽 = − , therefore, 𝒔𝒊𝒏 𝜽 = . This means
𝟑
𝟑
that y = 2 and r = 3. To find y, we again use the
Pythagorean Theorem:
𝒙𝟐 + 𝒚𝟐 = 𝒓𝟐
𝒙𝟐 + 𝟐𝟐 = 𝟑𝟐
𝒙𝟐 = 𝟑𝟐 − 𝟐𝟐
x= ±√𝟗 − 𝟒 = ± √𝟓
From the given value above, tan x is negative. Since y
is positive, then x must be negative, hence, 𝒙 = −√𝟓 .
Our six trig ratios will then be:
𝟐
𝟐√𝟓
√𝟓
𝐬𝐢𝐧 𝜽 = ,
𝐜𝐨𝐬 𝜽 = −
,
𝐭𝐚𝐧 𝜽 = −
𝟑
𝟑
𝟓
𝟐
Our six trig ratios will then be:
𝟑
𝟒
𝐬𝐢𝐧 𝒙 = ,
𝐜𝐨𝐬 𝒙 = ,
𝟓
𝟓
𝐜𝐬𝐜 𝒙 =
Given: 𝒔𝒊𝒏 (−𝜽) = −
𝟓
𝟓
,
𝟑
𝐬𝐞𝐜 𝒙 =
𝟓
,
𝟒
𝐭𝐚𝐧 𝒙 =
𝟑
𝟒
𝐜𝐨𝐭 𝒙 =
𝟒
𝟑
𝐜𝐬𝐜 𝜽 =
𝟑
,
𝟐
𝟐
𝟑
𝒂𝒏𝒅 𝐭𝐚𝐧 𝜽 = −
𝐬𝐞𝐜 𝜽 = −
𝟑√𝟓
,
𝟓
𝟐√𝟓
𝟓
𝐜𝐨𝐭 𝜽 = −
√𝟓
𝟐
**Again, don’t forget to rationalize radicals and
simplify fractions!
Use the given data to find the values of all six trig identities. Show your work at all times. Use binder
paper if needed.
NAME ______________________________________ PERIOD _______________
PART II. Simplifying trigonometric expressions. Use the identities to simplify the given expressions.
EXAMPLE 2: Simplify √𝟓 − 𝒙𝟐 where 𝒙 = √𝟓 𝐜𝐨𝐬 𝜽
EXAMPLE 1: Simplify √𝟐𝟓 − 𝒙𝟐 where 𝒙 = 𝟓 𝐬𝐢𝐧 𝜽
Discussion: Substitute the value of x into the
expression, and simplify.
Using the same steps as example 1,
𝟐
√𝟐𝟓 −
𝒙𝟐
= √𝟐𝟓 −
→ substitute x
= √𝟐𝟓 − 𝟐𝟓 𝒔𝒊𝒏𝟐 𝜽 → squaring 𝟓 𝐬𝐢𝐧 𝜽
= √𝟐𝟓(𝟏 − 𝒔𝒊𝒏𝟐 𝜽) → factoring out 25
= √𝟐𝟓 𝒄𝒐𝒔𝟐 𝜽
→ using the
(𝟓 𝐬𝐢𝐧 𝜽)𝟐
= √𝟐𝟓 ⋅ √𝒄𝒐𝒔𝟐 𝜽
= 𝟓 𝐜𝐨𝐬 𝜽
Pythagorean identity
→ separating the radicals
→ simplifying the radicals
√𝟓 − 𝒙𝟐 = √𝟓 − (√𝟓 𝐜𝐨𝐬 𝜽)
→ substitute x
𝟐
= √𝟓 − (√𝟓) 𝒄𝒐𝒔𝟐 𝜽 → squaring √𝟓 𝐜𝐨𝐬 𝜽
= √𝟓 − 𝟓 𝒄𝒐𝒔𝟐 𝜽 → simplifying the radical
= √𝟓(𝟏 − 𝒄𝒐𝒔𝟐 𝜽) → factoring out 5
= √𝟓𝒔𝒊𝒏𝟐 𝜽
→ using the
Pythagorean identity
𝟐
= √𝟓 ⋅ √𝒔𝒊𝒏 𝜽
→ separating the radicals
= √𝟓𝐬𝐢𝐧 𝜽
→ simplifying the radical
Use trigonometric substitution to simplify the given expressions. Show your work!
NAME ______________________________________ PERIOD _______________
PART III. Proving identities. You have been exposed to proving identities in the third quarter. The
exercises below are review to prepare you for next week’s identities.
𝟏
𝐜𝐨𝐬 𝒙 𝐬𝐢𝐧 𝒙
𝐜𝐨𝐬 𝒙
+ 𝐬𝐢𝐧 𝒙 (𝐜𝐨𝐬 𝒙) =
𝟏
𝟏+𝐜𝐨𝐬 𝒙
𝐜𝐨𝐬 𝒙
+ 𝐜𝐨𝐬 𝒙 =
𝐜𝐨𝐬 𝒙
𝟏+𝐜𝐨𝐬 𝒙
𝐜𝐨𝐬 𝒙
=
𝐜𝐨𝐬 𝒙
𝟏+𝐜𝐨𝐬 𝒙
𝐜𝐨𝐬 𝒙
𝟏+𝐜𝐨𝐬 𝒙
𝐜𝐨𝐬 𝒙
→ Subst reciprocal & quotient identities
→ Simplifying the fraction to get LCD
→Writing the expression as 1 fraction ∎
(𝒕𝒂𝒏𝟐 𝒙 + 𝟏) + 𝒄𝒐𝒕𝟐 𝒙 = 𝒄𝒔𝒄𝟐 𝒙 + 𝒕𝒂𝒏𝟐 𝒙 → Substitute 𝒔𝒆𝒄𝟐 𝒙
𝒕𝒂𝒏𝟐 𝒙 + 𝟏 + (𝒄𝒔𝒄𝟐 𝒙 − 𝟏) = 𝒄𝒔𝒄𝟐 𝒙 + 𝒕𝒂𝒏𝟐 𝒙
→ Substitute 𝒄𝒐𝒕𝟐 𝒙
𝒕𝒂𝒏𝟐 𝒙 + 𝟏 + 𝒄𝒔𝒄𝟐 𝒙 − 𝟏 = 𝒄𝒔𝒄𝟐 𝒙 + 𝒕𝒂𝒏𝟐 𝒙
→ taking out the ( )
𝒄𝒔𝒄𝟐 𝒙 + 𝒕𝒂𝒏𝟐 𝒙 = 𝒄𝒔𝒄𝟐 𝒙 + 𝒕𝒂𝒏𝟐 𝒙
→ +1 – 1 = 0 ∎