Question

1. determine P if 2cos2P = -(square root of 3) and tan 2P < 0 (general solution)

2. then, write down all the values of P [180°; 360°]
3. hence, determine the value of 3sin(P- 23,5°), correct to 3 decimal places

Answer 1

To determine the value of P, let's start by solving the equation 2cos(2P) = -√3. Since we know that cos(π/6) = √3/2, we can write:

2cos(2P) = 2cos(π/6)

2P = π/6

P = π/12

Next, let's consider the condition tan(2P) < 0. The tangent function is negative in the second and fourth quadrants. Since 2P = π/6, which is in the first quadrant, we can ignore the fourth quadrant. Thus, the solution lies in the second quadrant.

In the second quadrant, the angle is π + π/6 = 7π/6.

Now, let's find all the values of P in the interval [180°, 360°]:

P = π/12 (30°) in the first quadrant

P = 7π/6 (210°) in the second quadrant

Finally, we can determine the value of 3sin(P - 23.5°):

P - 23.5° = π/12 - 23.5°

≈ 15.651°

3sin(15.651°) ≈ 0.802

Therefore, the value of 3sin(P - 23.5°), correct to 3 decimal places, is approximately 0.802.

hope it helps. pls mark me brain list :D

Answer 2

`\textsf{1)} \quad P=75^(\circ)+180^(\circ)n`

`\textsf{2)} \quad P=255^(\circ)`

`\textsf{3)} \quad -2.348\; \sf (3\;d.p.)`

Explanation:

Question 1

We are given the equation 2cos(2P) = -√3 and the condition tan(2P) < 0.

To solve for P, begin by dividing both sides of the equation by 2:

`\begin{aligned}(2 \cos (2P))/(2)&=(-√(3))/(2)\\\\\cos(2P)&=-(√(3))/(2)\end{aligned}`

Take the arccos of both sides to isolate 2P:

`\begin{aligned}\arccos\left(\cos(2P)\right)&=\arccos\left(-(√(3))/(2)\right)\\\\2P&=150^(\circ)+360^(\circ)n, \;210^(\circ)+360^(\circ)n\end{aligned}`

Divide both sides by 2:

`\begin{aligned}(2P)/(2)&=(150^(\circ)+360^(\circ)n)/(2), \;(210^(\circ)+360^(\circ)n)/(2)\\\\P&=75^(\circ)+180^(\circ)n,\;105^(\circ)+180^(\circ)n\end{aligned}`

Given the condition tan(2P) < 0, and the fact that tan is negative in quadrants II and IV, the only valid value of P is:

`\large\boxed{P=75^(\circ)+180^(\circ)n}`

where n is an integer.

`\hrulefill`

Question 2

The value of P in the interval [180°, 360°] is:

`\large\boxed{P=255^(\circ)}`

`\hrulefill`

Question 3

To find the value of 3sin(P - 23.5°), substitute P = 255° into the equation:

`\begin{aligned}3 \sin (255^(\circ)-23.5^(\circ))&=3 \sin (231.5^(\circ)\\\\&=-2.34782447...\\\\&=-2.348\; \sf (3\;d.p.)\end{aligned}`

Therefore, the value of 3sin(P - 23.5°) is:

`\large\boxed{-2.348\; \sf (3\;d.p.)}`