Solved Tests for Trigonometry

Trigonometric functions

The trigonometric ratios sine, cosine and tangent are defined as follows:

sine (sin) is opposite over hypotenuse,

cosine (cos) is adjacent over hypotenuse,

tangent (tg) is opposite over adjacent.

SOH, CAH, TOA.

Consider a right-angled triangle ΔOMD.

;

But OM=r=1, MD=y, OD=x

Then

Sin α = y for 0 ≤ y ≤ 1; 0 ≤ Sin α ≤ 1

Cos α=

Cos α = x for –1 ≤ x ≤ 1; –1 ≤ Cos α ≤ 1

Definition:

For any angles α (read: alpha): 0°≤ α≤180°

Sin α (the sine of α) is the ordinate of the point M,

Cos α (the cosine of α) is the abscissa of the point M.

Given:

a circle (O;r),

coordinates of points A(1;0),

B(–1;0), M₁(0;1); M₂(;)

Find:

1) whether these points belong to the circle,

2) sine, cosine, tangent of angles

Solution:

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Given:

The sine of the angle: Sin α =

Find: the cosine of the alpha angle: Cos α = ?

Solution:

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The reduction formulas

Table. Trigonometric formulas of reduction. Sine, cosine in the trigonometry

Given:

The cosine of the angle Cos α =

Find: the sine of the alpha angle Sin α = ?

Solution:

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Given:

The cosine of the angle Cos α =

Find: the sine of the alpha angle Sin α = ?

Solution:

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Given:

The sine of the angle Sin α =

Find: the cosine of the alpha angle Cos α = ?

Solution:

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Given:

The sine of the angle Sin α =0

Find: the cosine of the alpha angle Cos α = ?

Solution:

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Given:

The cosine of the angle Cos α = 1

Find: tangent of the alpha angle tg α = ?

Solution:

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Given:

The cosine of the angle Cos α =

Find: the tangent of the alpha angle tg α = ?

Solution:

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Given:

the angles 120°, 135°, 150°

Find: sine, cosine, tangent of the angles

Sin α =? Cos α = ? tg α = ?

Solution:

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Given:

the coordinates of the point A(2;2)

the circle of center at the origin (0;0)

Find: the angle α

Solution:

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Given: the length of the segment and the angle

Find: the coordinates of the point A(x;y)

Solution:

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Given:

1) A(0;3)

2) B(–;1)

Find: the angle α

Solution:

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Theorem:

The area of the triangle is equal to half the product of the lengths of its two sides by the sine of the angle between them.

Given:

the triangle ΔABC

The sides of the triangle

BC=a, AC=b, AB=c

S - is the area of the triangle

Prove:

S_ΔABC=a • b • Sin C

Proof:

The area of the triangle is equal to half the product of the base of the triangle by its height. Therefore

S_ΔABC=a • h (*), where h – is the height of the triangle, a - is the base

But the height h = b • Sin C (**).

Using the condition (**) in the formula (*), then

The formula of the area of a triangle having used the sine, the base and the height

S_ΔABC=• a • b • Sin C

a = ; b =

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Given:

Find: the area of the triangle S_Δ

Solution:

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Given:

The area of the triangle S_ΔABC= 60 cm²

the side AC = 15 cm

the angle A = 30°

Find: the length of the side AB

Solution:

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Theorem:

The area of the parallelogram is equal to the product of its two adjacent sides by the sine of the angle between them.

Given:

ABCD - is a parallelogram

AB = a

AD = b

S - is the area of the parallelogram

Prove: S = a • b • SinA = a • b • Sin α

Proof:

S = AD • BH

Consider triangles ΔABD and ΔBDC. These triangles are equal by the third theorem.

Then the areas of the triangles are also equal S_ΔABD = S_ΔBDC

S_ΔBDC=• BC • CD • Sin α

Then the area of the parallelogram

S_ABCD= S_ΔABD + S_ΔBDC= 2 • S_ΔBDC

S_ABCD=2 • • a • b • Sin α = a • b • Sin α

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The law of sines

The law of sines for a triangle. Each side of an arbitrary triangle is proportional to the sine of the opposing angle.

Given:

The triangle ΔABC

AB=c, BC=a, AC=b

Prove:

Proof:

By the theorem about the area of a triangle, we obtain

S_ΔABC=c • b • Sin A (1)

S_ΔABC=a • b • Sin C (2)

S_ΔABC=c • a • Sin B (3)

It follows from (1) and (2) that

c • b • Sin A = a • b • Sin C

c • Sin A = a • Sin C

Then we obtain (*)

It follows from (2) and (3) that

a • b • Sin C =c • a • Sin B

b • Sin C = c • Sin B

Then we obtain (**)

From the equalities (*) and (**) we obtain

- is the formula of the sine of the triangle

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Given:

the triangle ΔABC

the side AC = 12 cm

the angle A = 75°

the angle С = 60°

How to find: the side AB of the triangle, and the area of the triangle S_ΔABC= ?

Solution:

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Given:

the triangle ΔABC

the side BC = cm

the angle A = 45°

the angle С = 30°

Find: the side AB of the triangle

Solution:

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Given:

ABCD – is a rectangle

The diagonal AC = 10 cm

The angle BOC = 30°

Find: the area of a rectangle

S_ABCD=?

Solution:

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The law of cosines

The law of cosines for a triangle. The square of either side of the triangle is equal to the sum of the squares of the other sides and minus the doubled product of these sides by the cosine of the angle between them.

Given:

the triangle ΔABC

AB=c, BC=a, AC=b

Prove:

a² = b² + c² - 2bc • Cos A

Proof:

Consider a rectangular coordinate system, where

1) the point A - is the origin of the coordinate plane or A(0;0)

2) the side AB lies on the abscissa, i.e. the x-axis.

Then we get B(c;0), C (b • Cos A; b • Sin A)

Using the formula to calculate the distance between two points , we get

a² = BC² = (x₂ - x₁)² + (y₂ - y₁)² = (b • Cos A - c)² + (b • Sin A - 0)² =

= b² • Cos² A - 2bc • Cos A + c² + b² • Sin² A = b² (Cos² A + Sin² A) - 2bc • Cos A + c² = b² • 1 + c² - 2bc • Cos A = b² + c² - 2bc • Cos A

Formulas of the law of cosines:

is the formula of the cosine of the triangle

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Given:

the triangle ΔABC

the height CH = h_c

the height BH₁ = h_b

the angle A = α

Find: the area of the triangle

S_ΔABC = ?

Solution:

is the formula how to find the area of a triangle through two heights and the sine of an angle.

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