Trigonometry Questions

1.) Find the exaqct value of tan(11π/12)

2.) A linear trig equation involving cosx has a solution of �π/6. Name three other possible solutions

3) Solve 10cosx=-7 where 0≤x≤2π

11π/12 is π/12 short of 2π

so tan 11π/12 = -tan π/12
I know tan 2π/12 = tan π/6 = 1/√3

tan 2π/12 = 2tanπ/12/(1 - tan^2 π/12)
let x = tan π/12 for easier typing

so
1/√3 = 2x/(1 - x^2)
2√3x = 1 - x^2
x^2 + 2√3 - 1 = 0
x = ( -2√3 ± √16)/2
tan π/12 = (4 - 2√3)/2 , since π/12 is in quad I

so tan 11π/12 = -(4-2√3)/2 = (2√3 - 4)/2

2. since cosx is positive, and x = π/6
I know the cosine is positive in I or IV
so other solutions:
2π - π/6= 5π/6
π/6 + 2π = 7π/6

3. 10 cosx = -7
cosx = -.7 , so x is in II or III
x = π - .79534 = 2.3462
x = xπ +.79534 = 3.9370

tan x/2 = (1-cosx)/sinx

So, if you let x=11pi/6, you can easily find sinx and cosx. Then plug them in to find tan x/2.

#2. -pi/6 11pi/6 13pi/6

#3.
cosx = -7/10
find x=arccos(7/10), then take pi±x for solutions.

1.) To find the exact value of tan(11π/12), we can use the fact that tan(x) = sin(x)/cos(x).

First, find the values of sin(11π/12) and cos(11π/12) using the unit circle or trigonometric identities. In this case, we can use the identity cos(x) = sin(π/2 - x) to find the value.

sin(11π/12) can be written as sin(π - π/12), and since sin(π - θ) = sin(θ), we can say that sin(11π/12) = sin(π/12).

Using the unit circle, we can see that sin(π/12) = (√3 - 1) / (2√2).

Next, we can find cos(11π/12) = sin(π/2 - 11π/12) = sin(π/4 + π/12) = sin(π/4)cos(π/12) + cos(π/4)sin(π/12) = (√2/2)(√3 + 1)/(2√2) + (√2/2)((√3 - 1) / (2√2)) = (√3 + 1)/(2√2) + (√3 - 1)/(2√2) = √3/√2 = √6/2.

Now, we can find the value of tan(11π/12) by dividing sin(11π/12) by cos(11π/12):

tan(11π/12) = (sin(11π/12)) / (cos(11π/12)) = ((√3 - 1) / (2√2)) / (√6/2) = (√3 - 1) / √6.

Therefore, the exact value of tan(11π/12) is (√3 - 1) / √6.

2.) If a linear trigonometric equation involving cos(x) has a solution of π/6, we can find three other possible solutions by considering the periodic nature of the cosine function.

The cosine function has a period of 2π, which means that the value of cos(x) repeats itself every 2π. Therefore, we can add any integer multiple of 2π to the original solution to find other possible solutions.

Three other possible solutions would be:

- π/6 + 2π = π/6, since adding 2π does not change the value of the angle.

- π/6 + 4π = 13π/6, since π/6 + 2π + 2π = 13π/6.

- π/6 + 6π = 25π/6, since π/6 + 2π + 2π + 2π = 25π/6.

Therefore, three other possible solutions for the linear trigonometric equation involving cos(x) would be π/6, 13π/6, and 25π/6.

3.) To solve the equation 10cos(x) = -7 where 0 ≤ x ≤ 2π, we can isolate the cosine term by dividing both sides of the equation by 10:

cos(x) = -7/10.

Now, we can use the inverse cosine function (also known as arccos or cos^(-1)) to find the angles in the given range that have a cosine value of -7/10.

Using a calculator or a trigonometric table, find the angle whose cosine is approximately -7/10. Let's denote this angle as θ.

Now, we need to consider that the cosine function has a period of 2π. This means that if θ is a solution, then θ + 2π, θ + 4π, θ + 6π, and so on, will also be solutions.

Since the range given is 0 ≤ x ≤ 2π, we can check if the solutions fall within this range.

Therefore, solve for x:

x = θ or x = θ + 2π, where θ is the approximate angle whose cosine is -7/10, and the solutions for x should satisfy the given range of 0 ≤ x ≤ 2π.

Note: To find the approximate angle using a calculator, you can use the inverse cosine function (cos^(-1)) or the arccos function if available.