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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

User Sahar
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2 Answers

5 votes

Answer:

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

User Dsb
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4 votes

Answer:


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

Take the arccos of both sides to isolate 2P:


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

Divide both sides by 2:


\begin{aligned}(2P)/(2)&amp;=(150^(\circ)+360^(\circ)n)/(2), \;(210^(\circ)+360^(\circ)n)/(2)\\\\P&amp;=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))&amp;=3 \sin (231.5^(\circ)\\\\&amp;=-2.34782447...\\\\&amp;=-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.)}

User Lee McPherson
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7.4k points