Answer:
a) 
b)

c)

d)
cos 330° = 1- 2 sin² (165°)
Step-by-step explanation:
<u><em>Step(i):-</em></u>
By using trigonometry formulas
a)
cos2∝ = 2 cos² ∝-1
cos∝ = 2 cos² ∝/2 -1
1+ cos∝ = 2 cos² ∝/2

b)
cos2∝ = 1- 2 sin² ∝
cos∝ = 1- 2 sin² ∝/2

<u><em>Step(i):-</em></u>
Given

we know that trigonometry formulas

1- cos∝ = 2 sin² ∝/2
Given

put ∝ = 315

multiply with ' 2 sin (∝/2) both numerator and denominator

Apply formulas

1- cos∝ = 2 sin² ∝/2
now we get

b)

put ∝ = 330° above formula



c )

put ∝ = 315° above formula


d)
cos∝ = 1- 2 sin² ∝/2
put ∝ = 330°

cos 330° = 1- 2 sin² (165°)
This question is not complete
Complete Question
A boat sails 4km on a bearing of 038 degree and then 5km on a bearing of 067 degree.(a)how far is the boat from its starting point.(b) calculate the bearing of the boat from its starting point
Answer:
a)8.717km
b) 54.146°
Step-by-step explanation:
(a)how far is the boat from its starting point.
We solve this question using resultant vectors
= (Rcos θ, Rsinθ + Rcos θ, Rsinθ)
Where
Rcos θ = x
Rsinθ = y
= (4cos38,4sin38) + (5cos67,5sin67)
= (3.152, 2.4626) + (1.9536, 4.6025)
= (5.1056, 7.065)
x = 5.1056
y = 7.065
Distance = √x² + y²
= √(5.1056²+ 7.065²)
= √75.98137636
= √8.7167296826
Approximately = 8.717 km
Therefore, the boat is 8.717km its starting point.
(b)calculate the bearing of the boat from its starting point.
The bearing of the boat is calculated using
tan θ = y/x
tan θ = 7.065/5.1056
θ = arc tan (7.065/5.1056)
= 54.145828196°
θ ≈ 54.146°
ANSWER

EXPLANATION
If f(x) is continuous at

Then,

The given function is





Since,

The function is continuous at

The graph of G(x) is the graph of F(x) shifted 4 units down. B. The graph of G(x) is the graph of F(x) shifted 4 units to the left C. The graph of G(x)...
Distribute
to every term inside the parentheses.

Simplify with multiplication.

You can't simplify further by adding, so leave the answer as-is.