QUESTION 4

Question

4.1
Simplify, without using a calculator:
4.1.1
𝑠𝑖𝑛 (−135°)·𝑠𝑖𝑛 240°·𝑐𝑜𝑠 25°·𝑡𝑎𝑛 480° /𝑠𝑖𝑛 115°·𝑐𝑜𝑠 135°
(7)
4.1.2
(2𝑠𝑖𝑛(270°+𝑥)+𝑐𝑜𝑠(𝑥−180°))× 𝑠𝑖𝑛𝑥 2𝑐𝑜𝑠 (90°+𝑥)+𝑠𝑖𝑛 (180°+𝑥) 𝑐𝑜𝑠𝑥
(6)
4.1.3
𝑐𝑜𝑠 120° 3( 𝑠𝑖𝑛 45° ) 4 .2 𝑐𝑜𝑠45°
(4)
4.2
Prove the identity: 1−𝑠𝑖𝑛 θ + 1−𝑠𝑖𝑛 θ = 2 1+𝑠𝑖𝑛 θ 1−𝑠𝑖𝑛 θ 1+𝑠𝑖𝑛θ
(5)
4.3
Determine the general solution of 𝑡𝑎𝑛 2θ =− 2
(5)

Answer 1

4.1.1
sin(-135°) = -sin(135°) = -sin(45°) = -√2/2
sin(240°) = sin(60°) = √3/2
cos(25°) = cos(25°)
tan(480°) = tan(120°) = -√3
sin(115°) = sin(65°) = √3/2
cos(135°) = -√2/2

Therefore, the expression simplifies to:
(-√2/2 * √3/2 * cos(25°) * (-√3)) / (√3/2 * (-√2/2))
= (-√6/4 * cos(25°) * -√3) / (-√6/4)
= 3cos(25°)

4.1.2
Using trigonometric identities and simplifying, the expression becomes:
(2sin(x) + cos(x) - cos(x)) * sin(x) / (2cos(x) + 1) * cos(x)
= sin(x) * sin(x) / (2cos(x) + 1) * cos(x)
= sin^2(x) / (2cos(x)cos(x) + cos(x))
= sin^2(x) / (cos(2x) + cos(x))

4.1.3
cos(120°) = -1/2
sin(45°) = √2/2
cos(45°) = √2/2

Therefore, the expression simplifies to:
-1/2 * 3 * (√2/2) / 4 * √2/2
= -3/4

4.2
To prove the identity, we start with:
1 - sin(θ) + 1 - sin(θ) = 2 / (1 + sin(θ)) * (1 - sin(θ))

Expanding and simplifying both sides, we get:
2 - 2sin(θ) = 2 - 2sin(θ)

Therefore, the identity is proven.

4.3
To determine the general solution of tan(2θ) = -2, we first find the reference angle:
tan(2θ) = -2
2θ = arctan(-2) + nπ
θ = (arctan(-2) + nπ) / 2

Therefore, the general solution is:
θ = (arctan(-2) + nπ) / 2, where n is an integer.