Transcript 10.3 Verify Trigonometric Identities

10.3 Verify Trigonometric
Identities
Trigonometric Identities
• Trigonometric Identity: a trigonometric
equation that is true for all values of the
variable for which both sides of the equation
are defined.
• There are 5 fundamental Trigonometric
Identities. (See page 628 in your book.)
1+ tan 2 θ = sec2θ
sin 
1
1

2
2
cos  cos 
2
1
sin 2 
1


2
2
cos  cos 
1  sin   cos 
2
2
cos2 
cos2 
1  sin   cos 
2
2
1  sin   cos 
2
2
1+ cot2 θ= csc2θ
1
1
1

2
tan  sin 2 
1
1
1

2
sin  sin 2 
cos 2 
cos2 
1
1

2
sin  sin 2 
sin 2  cos2 
1


2
2
sin  sin  sin 2 
sin 2   cos2   1
Section 9.3
Page 572
4
π
Given that sin q = 5 and 2 < q < π, find the values of
the other five trigonometric functions of q .
SOLUTION
STEP 1
Find cos q .
2
2
4
( ) + cos q =
5
2
cos q =
2
cos q =
cosq =
cosq =
Write Pythagorean identity.
1
Substitute 4 for sin q.
5
2
2
4
4
1 – ( ) Subtract ( ) from each side.
5
5
9
Simplify.
25
+– 3
Take square roots of each side.
5
– 3
Because q is in Quadrant
5
II, cos q is negative.
STEP 2
Find the values of the other four
trigonometric functions of q using the
known values of sin q and cos q.
4
5
4
sin q
tan q = cos q =
=– 3
–3
5
–3
5
3
cos q
–
cot q = sin q =
= 4
4
5
1
5
csc q = sin1q =
= 4
4
5
sec q = cos1 q =
5
1
–
= 3
3
–
5
Find the values of the other five trigonometric
functions of q.
1. cos q = 1 , 0 < q < π
6
2
SOLUTION
Find sin q .
STEP 1
sin 2q + cos 2q = 1
1 )2
1
6 =
sin2 q = 1 – ( 1 )2
6
sin2 q = 35
36
sin 2q + (
sinq =
sinq =

Write Pythagorean identity.
Substitute 1 for cos q .
6
2
1
Subtract ( ) from each side.
6
Simplify.
35
6
Take square roots of each side
35
6
Because q is in Quadrant
I, sin q is positive.
STEP 2
35
6
sin q
tan q = cos q =
= 35
1
6
1
6
6 35
csc q = sin1q =
= 35 =
35
35
6
1
6
cos q
35
1
cot q = sin q =
=
=
35
35
35
6
sec q = cos1 q = 1 = 6
1
6
Find the values of the other five trigonometric functions of q.
2. sin q = –3 , π < q < 3π
7
2
SOLUTION
STEP 1
Find cos q .
sin 2q + cos 2q = 1
Write Pythagorean identity.
( – 3)2 + cos2 q = 1
Substitute – 3 for sin q .
7
7
2
2
–
3
)
cos q = 1 – (
–3
7
Subtract 7 from each side.
2
cos q = 40
Simplify.
49
cos q = +– 2 10 Take square roots of each side.
7
cosq = – 2 10 Because q is in Quadrant lII,
7
cos q 18 is negative.
–2 10
2 10
cos q = 7
cot q = sin q
= 3
3
–
7
1
–7
1
csc q = sin q =
= 3
3
–
7
sec q = cos1 q = 1
– 2 10
7
7 10
= – 20
10.3 Assignment Day 1
Page 631, 3-9 all
10.3 Verify Trigonometric
Identities, day 2
1
Simplify the expression csc q cot q + sin q .
2
2
1
csc q cot q + sin q = csc q cot q + csc q
Reciprocal identity
2
= csc q (csc q – 1) + csc q Pythagorean identity
2
= csc q – csc q + csc q
3
3
= csc q
Distributive property
Simplify.
Simplify the expression tan ( π – q ) sin q.
2
tan ( π – q ) sin q = cot q sin q
Cofunction identity
2
cos q
= ( sin q ) ( sin q ) Cotangent identity
= cos q
Simplify.
Simplify the expression.
3. sin x cot x sec x
cos x
1
sin x ·
sin x · cos x
ANSWER
1
Substitute identity functions
Simplify
Simplify the expression.
4.
tan x csc x
sec x
sin x
1
cos x · sin x
Substitute identity functions
Simplify
1
cos x
ANSWER
1
Simplify the expression.
5.
cos π – q –1
2
1 + sin (– q )
sin ( θ ) – 1
1 – sin ( θ )
Substitute Cofunction identity;
Substitute Negative Angle identity
sin ( θ ) – 1
– 1(sin ( θ ) – 1)
ANSWER
–1
Factor
Simplify
10.3 Assignment Day 2
Page 631, 10-19 all
10.3 Verify Trigonometric
Identities, day 3
Verifying Trigonometric Identities
• When verifying an identity, begin with the expression
on one side.
• Use algebra and trigonometric properties to
manipulate the expression until it is identical to the
other side.
2
Verify the identity sec q2– 1 = sin q .
sec q
2
sec q – 1
sec q – 1
=
2
2
2
sec q
sec q
sec q
2
2
Write as separate fractions.
2
1
=1–(
)
sec q
Simplify.
= 1 – cos q
Reciprocal identity
= sin q
Pythagorean identity
2
2
cos x
Verify the identity sec x + tan x =
.
1 – sin x
1
sec x + tan x = cos x + tan x
Reciprocal identity
sin x
1
= cos x + cos x
Tangent identity
1 + sin x
= cos x
Add fractions.
1 + sin x 1 – sin x Multiply by 1 – sin x
= cos x
1 – sin x
1 – sin x
1 – sin 2x
= cos x (1 – sin x)
Simplify numerator.
cos2x
= cos x (1 – sin x)
Pythagorean identity
cos x
= 1 – sin x
Simplify.
Shadow Length
A vertical gnomon (the part of
a sundial that projects a
shadow) has height h. The
length s of the shadow cast by
the gnomon when the angle
of the sun above the horizon
is q can be modeled by the
equation below. Show that the
equation is equivalent to
s = h cot q .
h sin (90° – q )
s=
sin q
SOLUTION
Simplify the equation.
h sin (90° – q )
s=
sin q
h sin ( π – q )
2
=
sin q
Write original equation.
Convert 90° to radians.
cos q
= hsin
q
Cofunction identity
= h cot q
Cotangent identity
Verify the identity.
6. cot (– q ) = – cot q
SOLUTION
1
cot (– q ) = tan (– θ )
Reciprocal identity
1
= –tan ( θ )
Negative angle identity
= – cot θ
Reciprocal identity
Verify the identity.
7. csc2 x (1 – sin2 x) = cot2 x
SOLUTION
csc2 x (1 – sin2 x ) = csc2 x cos2x
1
= sin2 x cos2 x
= cot2 x
Pythagorean identity
Reciprocal identity
Tangent identity and
cotangent identities
Verify the identity.
8. cos x csc x tan x = 1
SOLUTION
sin x
cos x csc x tan x = cos x csc x cos
x
= cos x
=1
1 sin x
sin x cos x
Tangent identity and
cotangent identities
Reciprocal identity
Simplify
Verify the Identity.
9. (tan2 x + 1)(cos2 x – 1) = – tan2 x
SOLUTION
(tan2 x + 1)(cos2 x – 1) = – sec2 x (–sin2x)
1 (–sin2x)
= cos2 x
= –tan2 x
Pythagorean identity
Reciprocal identity
Tangent identity
and cotangent
identities
10.3 Assignment, day 3
Page 631, 25-32 all