Unlocking Sine and Cosine of Complementary Angles

	$sin θ$	$cos θ$
$θ = 1 5^{\circ}$	$0.258819 \dots$	$0.965925 \dots$
$θ = 3 0^{\circ}$	$0.5$	$0.866025 \dots$
$θ = 4 5^{\circ}$	$0.707106 \dots$	$0.707106 \dots$
$θ = 6 0^{\circ}$	$0.866025 \dots$	$0.5$
$θ = 7 5^{\circ}$	$0.965925 \dots$	$0.258819 \dots$

sin θ

cos θ

θ = 1 5^{\circ}

0.258819 \dots

0.965925 \dots

θ = 3 0^{\circ}

0.5

0.866025 \dots

θ = 4 5^{\circ}

0.707106 \dots

0.707106 \dots

θ = 6 0^{\circ}

0.866025 \dots

0.5

θ = 7 5^{\circ}

0.965925 \dots

0.258819 \dots

Begin by labeling points on the diagram.

External credits: @brgfx

Notice that

B,

D,

and

C

are collinear. Therefore, the measures of

∠ F D E,

∠ F D A,

and

∠ A D B

add up to

18 0^{\circ} .

m ∠ F D E + m ∠ F D A + m ∠ A D B = 18 0^{\circ}

SubstituteII

$m ∠ F D E = α$ , $m ∠ F D A = 9 0^{\circ}$

α + 9 0^{\circ} + m ∠ A D B = 18 0^{\circ}

▼

Solve for

m ∠ A D B

SubEqn

$LHS - α + 9 0^{\circ} = RHS - α + 9 0^{\circ}$

m ∠ A D B = 18 0^{\circ} - α - 9 0^{\circ}

SubTerm

Subtract term

m ∠ A D B = 9 0^{\circ} - α

This means that

∠ F D E

and

∠ A D B

are complementary angles. Consequently, the cotangent of

α

is equal to the tangent of its complementary angle.

cot α = tan (90 - α)

From here, the distance between

A

and

D

can be calculated using tangent ratio of

9 0^{\circ} - α .

tan (90 - α) = \frac{3 0}{A D} ⇓ cot α = \frac{3 0}{A D}

Substitute the given value for

cot α

and solve for

A D .

cot α = \frac{3 0}{A D}

Substitute

$cot α = 0.75$

0.75 = \frac{3 0}{A D}

▼

Solve for

A D

MultEqn

$LHS \cdot A D = RHS \cdot A D$

A D \cdot 0.75 = 30

DivEqn

$LHS / 0.75 = RHS / 0.75$

A D = 40

Now that

A D

is found,

A C

can be calculated. By the Alternate Interior Angles Theorem,

β

and

∠ A C D

are congruent.

External credits: @brgfx

The cotangent ratio of

β

is then equal to

\frac{A C}{4 0} .

Since

cot β

is known,

A C

can be found.

cot β = \frac{A C}{4 0}

Substitute

$cot α = 1.73$

1.73 = \frac{A C}{4 0}

▼

Solve for

A C

MultEqn

$LHS \cdot 40 = RHS \cdot 40$

69.2 = A C

RearrangeEqn

Rearrange equation

A C = 69.2

RoundInt

Round to nearest integer

A C \approx 69

The whale is located at a depth of about

69

feet. Therefore, Ignacio who is located at

E

sees the whale about

39

feet closer to the surface.

A C - A B = 69 - 30 = 39

What a cool phenomenon made sense by complementary angles!

Cofunction Identities
$sin θ = cos (9 0^{\circ} - θ)$	$cos θ = sin (9 0^{\circ} - θ)$
$tan θ = cot (9 0^{\circ} - θ)$	$cot θ = tan (9 0^{\circ} - θ)$
$sec θ = csc (9 0^{\circ} - θ)$	$csc θ = sec (9 0^{\circ} - θ)$

Cofunction Identities

sin θ = cos (9 0^{\circ} - θ)

cos θ = sin (9 0^{\circ} - θ)

tan θ = cot (9 0^{\circ} - θ)

cot θ = tan (9 0^{\circ} - θ)

sec θ = csc (9 0^{\circ} - θ)

csc θ = sec (9 0^{\circ} - θ)

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Catch-Up and Review

Investigating the Relationship Between Sine and Cosine

Sine and Cosine of Complementary Angles

Proof

Convert Between Sine and Cosine

Hint

Solution

Practice Converting Between Sine and Cosine

Investigating Properties of Right Triangles

Answer

Hint

Solution

Solving Problems Using the Relationship Between Sine and Cosine

Hint

Solution

Using the Relationship Between Sine and Cosine to Solve Problems

Hint

Solution

Tangent and Cotangent of Complementary Angles

Proof

Solving Problems Using the Relationship Between Tangent and Cotangent

Hint

Solution

Calculating Distances Using Trigonometry

Hint

Solution

Secant and Cosecant Ratios

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