1) Find the exact value of cos 105 degrees by using a half-angle formula.

2)Find the solution of sin 2theta = cos theta if 0 degrees</= theta < 180 degrees.

I have read the book and looked at the examples, but cannot figure out how to work these. Any help would be great!!

Thanks

1)

105 * 2 = 210 is a nice angle, 180 + 30

2)
2 sin t cos t = cos t
2 sin t = 1
sin t = 1/2
t = 30 degrees or pi/6 radians

the other solution t = 150 degrees is no good because the cos t would be negative

Sure! I'm here to help you with these trigonometry problems.

1) To find the exact value of cos 105 degrees using a half-angle formula, we need to use the double angle formula for cosine:

cos 2θ = 2cos^2 θ - 1

First, let's convert 105 degrees to a half-angle by dividing it by 2: 105 degrees / 2 = 52.5 degrees.

Now, we have cos 52.5 degrees. Let's find the exact value of cos 52.5 degrees using the half-angle formula.

cos 105 degrees = cos 2(52.5 degrees) = 2cos^2 (52.5 degrees) - 1

To evaluate this expression, we need to find the exact value of cos (52.5 degrees). Unfortunately, there is no exact value for cos (52.5 degrees). However, we can use a calculator or trigonometric tables to find an approximate value. The approximate value of cos (52.5 degrees) is around 0.616.

Therefore, cos 105 degrees ≈ 2(0.616)^2 - 1 = 0.770.

So, the approximate value of cos 105 degrees using the half-angle formula is 0.770.

2) To find the solution of sin 2θ = cos θ, you can use trigonometric identities and algebraic manipulation.

Let's start by using the double-angle formula for sine:

sin 2θ = 2sin θ cos θ

Now, we have the equation 2sin θ cos θ = cos θ.

Let's bring all terms to one side of the equation and simplify:

2sin θ cos θ - cos θ = 0.

Now, factor out cos θ:

cos θ(2sin θ - 1) = 0.

Now, you have two possibilities for this equation to be true:

1) cos θ = 0: The cosine function is equal to zero when θ is equal to 90 degrees or 270 degrees.

2) 2sin θ - 1 = 0: Solve this equation for sin θ:

2sin θ = 1,
sin θ = 1/2.

The sine function is equal to 1/2 when θ is equal to 30 degrees or 150 degrees.

So, the solutions for sin 2θ = cos θ if 0 degrees ≤ θ < 180 degrees are:
θ = 30 degrees, 90 degrees, 150 degrees, and 270 degrees.

I hope this explanation helps you understand how to approach these problems! Let me know if you have any further questions.

1) To find the exact value of cos 105 degrees using a half-angle formula, we can start by using the double-angle formula for cosine.

The double-angle formula for cosine is:

cos(2θ) = 2cos^2(θ) - 1

Now, we can rewrite the given angle 105 degrees as the sum of two angles: 90 degrees and 15 degrees.

cos(105 degrees) = cos(90 degrees + 15 degrees)

Using the identity cos(A + B) = cos(A)cos(B) - sin(A)sin(B), we can rewrite this as:

cos(90 degrees + 15 degrees) = cos(90 degrees)cos(15 degrees) - sin(90 degrees)sin(15 degrees)

Next, we can use the values of cosine and sine for common angles. We know that cos(90 degrees) = 0 and sin(90 degrees) = 1.

cos(105 degrees) = 0*cos(15 degrees) - 1*sin(15 degrees)

Let's find the exact value of cos(15 degrees). We can apply the half-angle formula for cosine:

cos(15 degrees) = sqrt((1 + cos(30 degrees))/2)

We know that cos(30 degrees) = sqrt(3)/2.

cos(15 degrees) = sqrt((1 + sqrt(3)/2)/2) = sqrt((2 + sqrt(3))/4)

Now, let's substitute this value back into the previous equation:

cos(105 degrees) = 0*sqrt((2 + sqrt(3))/4) - 1*sin(15 degrees)

Since 0 multiplied by anything is 0, we can simplify the equation to:

cos(105 degrees) = -sin(15 degrees)

Now, we need to find the exact value of sin(15 degrees). We can use the half-angle formula for sine:

sin(15 degrees) = sqrt((1 - cos(30 degrees))/2)

sin(15 degrees) = sqrt((1 - sqrt(3)/2)/2) = sqrt((2 - sqrt(3))/4)

Substituting this value back into the previous equation, we have:

cos(105 degrees) = -(sqrt((2 - sqrt(3))/4))

Therefore, the exact value of cos 105 degrees using a half-angle formula is -(sqrt((2 - sqrt(3))/4)).

2) To find the solution of sin(2θ) = cos(θ), we can use the double-angle formula for sine:

sin(2θ) = 2sin(θ)cos(θ)

Now, substitute sin(2θ) = cos(θ) into the equation:

2sin(θ)cos(θ) = cos(θ)

Next, we can rearrange the equation:

2sin(θ)cos(θ) - cos(θ) = 0

Factor out the common term cos(θ):

(2sin(θ) - 1)cos(θ) = 0

Now, we can set each factor equal to zero:

2sin(θ) - 1 = 0 or cos(θ) = 0

Solving the first equation, we have:

2sin(θ) = 1

sin(θ) = 1/2

Since 0 degrees ≤ θ < 180 degrees, we know that the sine function is positive in the first and second quadrants. Therefore, we can find the solutions by considering the angles in those quadrants where sin(θ) = 1/2.

In the first quadrant, θ = 30 degrees.
In the second quadrant, θ = 150 degrees.

For the second equation, we have:

cos(θ) = 0

Again, since 0 degrees ≤ θ < 180 degrees, we know that the cosine function is zero in the second quadrant. Therefore, we have one solution: θ = 90 degrees.

Therefore, the solutions to sin(2θ) = cos(θ), where 0 degrees ≤ θ < 180 degrees, are θ = 30 degrees, 90 degrees, and 150 degrees.