**Question 1.Find the trigonometric functions of 0°, 30°, 45°, 60°, 150°, 180°, 210°, 300°, 330°, – 30°, – 45°, – 60°, – 90°, – 120°, – 225°, – 240°, – 270°, – 315°Solution:**

Angle of measure 0°:

Let m∠XOA = 0° = 0^{c}

Its terminal arm (ray OA) intersects the standard

unit circle in P(1, 0).

Hence,x = 1 and y = 0

sin 0° = y = 0,

cos 0° = x = 1,

Angle of measure 30°:

Let m∠XOA = 30°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y)

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP= 1

Since point P lies in 1st quadrant, x > 0, y > 0

Angle of measure 45°:

Let m∠XOA = 45°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 45° – 45° – 90° triangle.

OP = 1,

Since point P lies in the 1st quadrant, x > 0, y > 0

Angle of measure 60°:

Let m∠XOA = 60°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP= 1,

Angle of measure 150°:

Let m∠XOA = 150°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP= 1,

Since point P lies in the 2nd quadrant, x < 0, y > 0

Angle of measure 180°:

Let m∠XOA = 180°

Its terminal arm (ray OA) intersects the standard unit circle at P(-1, 0).

∴ x = – 1 and y = 0

sin 180° =y = 0

cos 180° = x = -1

Angle of measure 210°:

Let m∠XOA = 210°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP= 1,

Since point P lies in the 3rd quadrant, x < 0,y < 0

Angle of measure 300°:

Let m∠XOA = 300°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP = 1,

Since point P lies in the 1st quadrant, x > 0,y > 0

Angle of measure 330°:

Let m∠XOA = 330°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP= 1,

Since point P lies in the 4th quadrant, x > 0, y < 0

Angle of measure 30°

Let m∠XOA = -30°

Its terminal arm (ray OA) intersects the standard unit circle at P(x,y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60 — 90° triangle.

op = 1,

Since point P lies in the 4th quadrant x > 0, y < 0

Angle of measure 45°:

Let m∠XOA = 45°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 45° – 45° – 90° triangle.

OP = 1,

Since point P lies in the 4th quadrant x > 0, y < 0

[Note : Answer given in the textbook of sin (45°) = – 1/2. However, as per our

Angle of measure (-60°):

Let m∠XOA = -60°

Its terminal arm (ray OA) intersects the standard

unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

ΔOMP is a 30° – 60° – 90° triangle.

OP = 1,

Since point P lies in the 4’ quadrant,

x > 0, y < 0

Angle of measure (-90°):

Let m∠XOA = -90°

It terminal arm (ray OA) intersects the standard unit circle at P(0, -1)

∴ x = 0 and y = -1

sin (-90°) = y = -1

cos (-90°) = s = 0

Angle of measure (-120°):

Let m∠XOA = – 120°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP = 1,

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure (- 225°):

Let m∠XOA = – 225°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

ΔOMP is a 45° – 45° – 90° triangle.

OP = 1,

Since point P lies in the 2nd quadrant, x < 0, y > 0

Angle of measure 2400):

Let m∠XOA = 240°

Its terminal arm (ray OA) intersects the standard

unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 900 triangle.

Since point P lies in the 2nd quadrant, x<0, y>0

Angle of measure (- 270°):

Let m∠XOA = – 270°

Its terminal arm (ray OA)

intersects the standard unit,

circle at P(0, 1).

∴ x = 0 and y = 1

sin (- 270°) = y = 1

cos (- 270°) = x = 0

which is not defined.

Angle of measure ( 315°):

Let m∠XOA 315°

Its terminal arm (ray OA) intersects the standard unit circle at P(x,y).

Draw seg PM perpendicular to the X-axis.

ΔOMP is a 45° – 45° – 90° triangle.

OP = 1,

**Question 2.State the signs of:i. tan 380°ii. cot 230°iii 468°Solution:**

1. 380° = 360° + 20°

∴ 380° and 20° are co-terminal angles.

Since 0° < 20° <90°0,

20° lies in the l quadrant.

∴ 380° lies in the 1st quadrant,

∴ tan 380° is positive.

ii. Since, 180° <230° <270°

∴ 230° lies in the 3rd quadrant.

∴ cot 230° is positive.

iii. 468° = 360°+108°

∴ 468° and 108° are co-terminal angles.

Since 90° < 108° < 180°,

108° lies in the 2nd quadrant.

∴ 468° lies in the 2nd quadrant.

∴ sec 468° is negative.

**Question 3.State the signs of cos 4 ^{c} and cos 4°. Which of these two functions is greater?Solution:**

Since 0° < 4° < 90°, 4° lies in the first quadrant. ∴ cos4° >0 …(i)

Since 1

^{c}= 57° nearly,

180° < 4

^{c}< 270°

∴ 4

^{c}lies in the third quadrant.

∴ cos 4

^{c}< 0 ………(ii)

From (i) and (ii),

cos 4° is greater.

**Question 4.State the quadrant in which 6 lies ifi. sin θ < 0 and tan θ > 0ii. cos θ < 0 and tan θ > 0Solution:**

i. sin θ < 0 sin θ is negative in 3rd and 4th quadrants, tan 0 > 0

tan θ is positive in 1st and 3rd quadrants.

∴ θ lies in the 3rd quadrant.

ii. cos θ < 0 cos θ is negative in 2nd and 3rd quadrants, tan 0 > 0

tan θ is positive in 1st and 3rd quadrants.

∴ θ lies in the 3rd quadrant.

**Question 5.Evaluate each of the following:i. sin 30° + cos 45° + tan 180°ii. cosec 45° + cot 45° + tan 0°iii. sin 30° x cos 45° x lies tan 360°Solution:**

i. We know that,

**Question 6.Find all trigonometric functions of angle in standard position whose terminal arm passes through point (3, – 4).Solution:**

Let θ be the measure of the angle in standard position whose terminal arm passes through P(3, -4).

∴ x = 3 and y = -4

r = OP

**Question 7.**

**Question 8.Using tables evaluate the following:**

ii. We know that,

**Question 9.**

Since A lies in the 2nd quadrant,

tan A < 0

Since x lies in the 3rd quadrant, cosec x < 0

Since x lies in the 4th quadrant,

sec x > 0