10.3 verify trigonometric identities. trigonometric identities trigonometric identity: a...
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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
22
2
cos
1
cos
sin1
2
2
2 cos
sin
cos
11
2
2
2
2
cos
cos
cos
sin1
22 cossin1
22 cossin1
1+ cot2 θ= csc2θ
22 sin
1
tan
11
2
2
2 sin
1
cossin
11
22
2
sin
1
sin
cos1
22
2
2
2
sin
1
sin
cos
sin
sin
1cossin 22
Section 9.3
Page 572
Given that sin q = and < q < π, find the values of the other five trigonometric functions of q .
45
π2
SOLUTION
STEP 1 Find cos q .
Write Pythagorean identity.
Substitute for sin .q
45( ) + cos q 4
52 2 1=
Subtract ( ) from each side.45
2cos q
2 245
1 – ( )=
Simplify.cos q 2 9
25=
Take square roots of each side.cosq 35
+–=
Because q is in Quadrant II, cos q is negative.
cosq 35–=
STEP 2 Find the values of the other four trigonometric functions of q using theknown values of sin q and cos q.
tan q sin qcos q= =
4535–
= 43–
cot q cos qsin q= =
45
35
–= 3
4–
csc q 1sin q= = 1
45
= 54
sec q 1cos q= = 3
5–
1 = 53–
Find the values of the other five trigonometric functions of q.
16
1. cos q , 0 < q < = π2
SOLUTION
STEP 1 Find sin q .
Write Pythagorean identity.sin q + cos q2 2 = 1
Substitute for cos q .16
Subtract ( ) from each side.16
2
sin q +2 = 1( )216
sin2 q = 1 – ( )216
Take square roots of each side.
Because q is in Quadrant I, sin q is positive.
Simplify.sin2 q = 3536
sin q = 356
sin q = 356
STEP 2 tan q sin qcos q= =
35 616
= 35
csc q 1sin q= = 1
35 6
= 6 35
cot q cos qsin q= =
16
35 6
= 135
sec q 1cos q= = 1
6
1 = 6
= 6 3535
= 3535
Find the values of the other five trigonometric functions of q.
2. sin q = , π < q<
3π 2
–3 7
SOLUTION
STEP 1 Find cos q .
Write Pythagorean identity.sin q + cos q2 2 = 1
Substitute for sin q .– 3 7
( )2 + cos2 q = 1 – 3 7
Simplify.cos q 2 40 49=
cos q 2 = 71 – ( )2– 3
Subtract from each side.
– 3 7
Take square roots of each side.
Because q is in Quadrant lII, cos q 18 is negative.
cos q +–= 2 10 7
cosq –= 2 10 7
cot q cos qsin q= =
37
72 10–
–= 2 10
3
csc q 1sin q= = 1
37
–= 7
3–
sec q 1cos q= =
2 10 7
–
1 = – 72010
10.3 Assignment Day 1
Page 631, 3-9 all
10.3 Verify Trigonometric Identities, day 2
2Simplify the expression csc q cot q + .
1sin q
Reciprocal identity2csc q cot q + 1
sin q csc q cot q + csc q2
=
Pythagorean identity= csc q (csc q – 1) + csc q2
Distributive property= csc q – csc q + csc q3
Simplify.= csc q3
Simplify the expression tan ( – q ) sin q.π2
Cofunction identitytan ( – q ) sin qπ2
cot q sin q=
Cotangent identity= ( ) ( sin q )cos qsin q
Simplify.= cos q
3. sin x cot x sec x
Simplify the expression.
1ANSWER
sin xcos xsin x cos x
1· ·
Substitute identity functionsSimplify
Simplify the expression.
4. tan x csc xsec x
1ANSWER
sin xcos x
cos x1
· sin x1
Substitute identity functions
Simplify
Simplify the expression.
cos –1 π2
– q
1 + sin (– q )5.
– 1ANSWER
Substitute Cofunction identity;Substitute Negative Angle identity1 – sin ( θ )
sin ( θ ) – 1
– 1(sin ( θ ) – 1)sin ( θ ) – 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.
Verify the identity = sin q .sec q – 12
sec q 22
Write as separate fractions.sec q – 12
sec q 2 = sec q2
sec q 2 – 1 sec q 2
Simplify.= 1 – ( )1 sec q
2
Reciprocal identity= 1 – cos q2
Pythagorean identity= sin q2
Verify the identity sec x + tan x = .cos x
1 – sin x
Reciprocal identitysec x + tan x 1cos x + tan x=
Tangent identity1
cos x + sin x cos x=
Add fractions.1 + sin x
cos x =
Multiply by 1 – sin x 1 – sin x
1 + sin xcos x = 1 – sin x
1 – sin x
Simplify.cos x
1 – sin x=
Simplify numerator.1 – sin xcos x (1 – sin x)=
2
Pythagorean identitycos xcos x (1 – sin x)=
2
Shadow LengthA 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 .
sh sin (90° – q )
sin q=
SOLUTION
Simplify the equation.
Write original equation.sh sin (90° – q )
sin q=
Convert 90° to radians.h sin ( – q )
sin q
π2
=
Cofunction identityh cos q sin q=
Cotangent identity= h cot q
Verify the identity.
6. cot (– q ) = – cot q
SOLUTION
Reciprocal identity
Negative angle identity
cot (– q ) = tan (– θ )1
–tan ( θ )1
=
Reciprocal identity= – cot θ
= cot2 x
1sin2 x= cos2 x Reciprocal identity
Tangent identity and cotangent identities
7. csc2 x (1 – sin2 x) = cot2 x
SOLUTION
csc2 x (1 – sin2 x ) = csc2 x cos2x Pythagorean identity
Verify the identity.
Verify the identity.
8. cos x csc x tan x = 1
SOLUTION
cos x csc x tan x = cos x csc x sin xcos x Tangent identity and
cotangent identities
= cos x 1sin x
sin xcos x Reciprocal identity
= 1 Simplify
9. (tan2 x + 1)(cos2 x – 1) = – tan2 x
SOLUTION
(tan2 x + 1)(cos2 x – 1) = – sec2 x (–sin2x) Pythagorean identity
1cos2 x
(–sin2x)= Reciprocal identity
= –tan2 x Tangent identity and cotangent identities
Verify the Identity.
10.3 Assignment, day 3
Page 631, 25-32 all