1. Find the exact value of the following (Think identity)
Cos(2 Arccos(5/13))
2. Solve the following equation for
0° ≤ Θ < 360°
Sec(Θ)= tan(Θ) + cos(Θ)
Cos(2 Arccos(5/13))
let's look at the arccos 5/13
we would have a right-angled triangle with angle Ø so that cos Ø = 5/13
so then we want cos(2Ø)
but cos 2Ø = 2cos^2 Ø - 1
= 2(25/169) - 1 = -119/169
Sec(Θ)= tan(Θ) + cos(Θ)
1/cosØ = sinØ/cosØ + cosØ
1 = sinØ + cos^2Ø
1 = sinØ + 1 - sin^2Ø
sin^2Ø - sinØ = 0
sinØ(sinØ - 1) = 0
sinØ = 0 ----> Ø = 0,180,360
or sinØ = 1 ---> Ø = 90
so Ø = 0,90,180, and 360°
1. To find the exact value of Cos(2 Arccos(5/13)), we can use the double-angle identity for cosine, which states:
Cos(2θ) = 2Cos^2(θ) - 1.
First, let's find the value of θ by evaluating Arccos(5/13). The arccosine function gives us the angle whose cosine is 5/13.
Cos(θ) = 5/13.
Next, we can substitute this value into our double-angle identity for cosine:
Cos(2 Arccos(5/13)) = 2Cos^2(Arccos(5/13)) - 1.
Since Cos(Arccos(x)) = x, we have:
Cos(2 Arccos(5/13)) = 2(5/13)^2 - 1.
Simplifying further:
Cos(2 Arccos(5/13)) = 2(25/169) - 1.
Multiplying:
Cos(2 Arccos(5/13)) = 50/169 - 1.
Combining the fractions:
Cos(2 Arccos(5/13)) = (50 - 169)/169.
Simplifying:
Cos(2 Arccos(5/13)) = -119/169.
Therefore, the exact value of Cos(2 Arccos(5/13)) is -119/169.
2. To solve the equation Sec(Θ) = tan(Θ) + cos(Θ) for 0° ≤ Θ < 360°, we need to rewrite the equation in terms of one trigonometric function.
Recall the trigonometric identities:
Sec(Θ) = 1/cos(Θ)
Tan(Θ) = sin(Θ)/cos(Θ)
Using these identities, we can rewrite the equation as:
1/cos(Θ) = sin(Θ)/cos(Θ) + cos(Θ)
To simplify further, we can multiply through by the common denominator of cos(Θ):
1 = sin(Θ) + cos^2(Θ)
Next, using the Pythagorean identity sin^2(Θ) + cos^2(Θ) = 1, we can rewrite the equation as:
1 = sin(Θ) + 1 - sin^2(Θ)
Rearranging the terms:
0 = sin^2(Θ) - sin(Θ)
Factoring out sin(Θ):
0 = sin(Θ)(sin(Θ) - 1)
So, either sin(Θ) = 0 or sin(Θ) - 1 = 0.
For sin(Θ) = 0, we have solutions:
Θ = 0° and Θ = 180°.
For sin(Θ) - 1 = 0, we have:
sin(Θ) = 1
Θ = 90°
Therefore, the solutions to the equation Sec(Θ) = tan(Θ) + cos(Θ) are Θ = 0°, 90°, and 180°.
To find the exact value of Cos(2 Arccos(5/13)), we can use the double-angle identity for cosine. This identity states that Cos(2θ) = 2Cos²(θ) - 1.
Let's start by finding the value of Arccos(5/13). Arccos represents the inverse cosine function, which gives us the angle whose cosine is equal to a given value. In this case, we have Arccos(5/13).
To find the value of Arccos(5/13), we can use a calculator or a trigonometric table. Arccos(5/13) is approximately equal to 67.38 degrees, or in radians, about 1.17.
Now, we can substitute the value of Arccos(5/13) into the cosine double-angle identity:
Cos(2 Arccos(5/13)) = 2Cos²(Arccos(5/13)) - 1
Cos(2 Arccos(5/13)) = 2 * (Cos(Arccos(5/13)))² - 1
Cos(2 Arccos(5/13)) = 2 * (5/13)² - 1
Cos(2 Arccos(5/13)) = 2 * (25/169) - 1
Cos(2 Arccos(5/13)) = 50/169 - 1
Cos(2 Arccos(5/13)) = 50/169 - 169/169
Cos(2 Arccos(5/13)) = (50 - 169)/169
Cos(2 Arccos(5/13)) = -119/169
Therefore, the exact value of Cos(2 Arccos(5/13)) is -119/169.
Moving on to the second question:
To solve the equation Sec(Θ) = tan(Θ) + cos(Θ), we need to use trigonometric identities.
Sec(Θ) is the reciprocal of cosine, which means Sec(Θ) = 1/cos(Θ). Similarly, tan(Θ) = sin(Θ)/cos(Θ).
Let's substitute these values into the given equation:
1/cos(Θ) = (sin(Θ)/cos(Θ)) + cos(Θ)
To simplify the equation, we can multiply both sides by cos(Θ):
1 = sin(Θ) + cos²(Θ)
Now, using the Pythagorean identity sin²(Θ) + cos²(Θ) = 1, we can rewrite the equation:
1 = sin(Θ) + 1 - sin²(Θ)
Rearranging terms, we have:
sin²(Θ) - sin(Θ) = 0
Now, we can factor out sin(Θ):
sin(Θ)(sin(Θ) - 1) = 0
So, either sin(Θ) = 0 or sin(Θ) - 1 = 0.
For sin(Θ) = 0, the solutions are Θ = 0°, 180°.
For sin(Θ) - 1 = 0, we have sin(Θ) = 1, which only occurs for Θ = 90°.
Therefore, the solutions for the equation Sec(Θ) = tan(Θ) + cos(Θ) in the given range are Θ = 0°, 90°, 180°.