Question 1

Find the radian measures corresponding to the following degree measures:
(i) 25°
(ii) -47°30'

Solution:

(i) For 25°:
We know that 180° = π radians.
Therefore, 1° = (π / 180) radians.
25° = 25 × (π / 180) = (5π / 36) radians.

(ii) For -47°30':
First, convert minutes into degrees: 30' = (30 / 60)° = (1 / 2)°.
So, -47°30' = -(47 + 1/2)° = -(95 / 2)°.
Converting to radians:
-(95 / 2)° = -(95 / 2) × (π / 180) = -(19π / 72) radians.

Question 2

Find the degree measure corresponding to 11/16 radians (Use π = 22/7).

Solution:

We know that 1 radian = (180 / π) degrees.
(11 / 16) radians = (11 / 16) × (180 / π)
= (11 / 16) × 180 × (7 / 22)
= (1 × 45 × 7) / (4 × 2) = 315 / 8 degrees.

Converting to degrees, minutes, and seconds:
315 / 8 = 39° + (3 / 8)°
(3 / 8)° = (3 / 8) × 60' = (45 / 2)' = 22' + (1 / 2)'
(1 / 2)' = (1 / 2) × 60" = 30"

Thus, 11/16 radians = 39° 22' 30".

Question 3

A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Solution:

Number of revolutions in 1 minute (60 seconds) = 360.
Number of revolutions in 1 second = 360 / 60 = 6.

In 1 complete revolution, the wheel turns through an angle of 2π radians.
Therefore, in 6 revolutions, it turns through an angle of:
6 × 2π = 12π radians.

Question 4

Find the exact values of:
(i) sin 75°
(ii) tan 15°

Solution:

(i) sin 75°:
Write 75° as (45° + 30°):
sin 75° = sin(45° + 30°)
Using sin(A + B) = sin A cos B + cos A sin B:
sin 75° = sin 45° cos 30° + cos 45° sin 30°
= (1 / √2) × (√3 / 2) + (1 / √2) × (1 / 2)
= (√3 + 1) / (2√2).

(ii) tan 15°:
Write 15° as (45° - 30°):
tan 15° = tan(45° - 30°)
Using tan(A - B) = (tan A - tan B) / (1 + tan A tan B):
tan 15° = (tan 45° - tan 30°) / (1 + tan 45° tan 30°)
= (1 - 1/√3) / (1 + 1 × 1/√3)
= (√3 - 1) / (√3 + 1)
Rationalizing the denominator:
= ((√3 - 1)²) / (3 - 1) = (3 + 1 - 2√3) / 2 = (4 - 2√3) / 2 = 2 - √3.

Question 5

Prove that: cos(π/4 - x) cos(π/4 - y) - sin(π/4 - x) sin(π/4 - y) = sin(x + y).

Solution:

Let A = (π/4 - x) and B = (π/4 - y).
Left Hand Side (LHS) = cos A cos B - sin A sin B.
Using the identity cos(A + B) = cos A cos B - sin A sin B:
LHS = cos(A + B)
= cos[(π/4 - x) + (π/4 - y)]
= cos[π/2 - (x + y)]

We know that cos(π/2 - θ) = sin θ.
Therefore, cos[π/2 - (x + y)] = sin(x + y) = RHS.
Hence proved!

Question 6

Prove that: (sin 5x + sin 3x) / (cos 5x + cos 3x) = tan 4x.

Solution:

Using sum to product formulas:
• sin C + sin D = 2 sin((C+D)/2) cos((C-D)/2)
• cos C + cos D = 2 cos((C+D)/2) cos((C-D)/2)

Numerator: sin 5x + sin 3x = 2 sin((5x + 3x)/2) cos((5x - 3x)/2) = 2 sin(4x) cos(x).
Denominator: cos 5x + cos 3x = 2 cos((5x + 3x)/2) cos((5x - 3x)/2) = 2 cos(4x) cos(x).

LHS = [2 sin(4x) cos(x)] / [2 cos(4x) cos(x)]
= sin(4x) / cos(4x)
= tan 4x = RHS.
Hence proved!

Question 7

Find the general solution for the equation: cos 4x = cos 2x.

Solution:

Rearranging the equation:
cos 4x - cos 2x = 0

Using the formula cos C - cos D = -2 sin((C+D)/2) sin((C-D)/2):
-2 sin((4x + 2x)/2) sin((4x - 2x)/2) = 0
-2 sin(3x) sin(x) = 0

This gives two possibilities:
1) sin 3x = 0 ⇒ 3x = nπ ⇒ x = nπ / 3, where n ∈ Z.
2) sin x = 0 ⇒ x = mπ, where m ∈ Z.

Since the values x = mπ are already included in x = nπ/3 (when n is a multiple of 3), the general solution is:
x = nπ / 3, where n ∈ Z.