Trigonometry Questions
1. Using a compound angle formula, demonstrate that sin2π/3 is equivalent to sin π/3
2. The expression sinπ is equal to zero, while the expression 1/cscπ is undefined. Why is the identity sin(theta)=1/csc(theta) still an identity?
3. Name two strategies that could be used to solve the equation sec^2x + 8secx + 12 = 0. Why do these strategies make the most sense?
4. A linear trigonometric equations involving cosx has a solution of π/6. Name three other possible solutions. (can you show how you got them please?)
Thanks in advance!
#1. check out sin(π - π/3)
#2. because 1/0 is also undefined.
#3.
just factor the function
(sec+6)(sec+2)
convert to cosines, then factor to get
(6cos+1)(2cos+1)=0
#4. 2π+π/6,2π-π/6
and all multiples of 2π more or less.
1. To demonstrate that sin(2π/3) is equivalent to sin(π/3), we can use the compound angle formula for sine. The formula states that sin(A + B) = sinA*cosB + cosA*sinB.
Let's apply this formula by setting A = π/6 and B = π/2.
Therefore, sin(2π/3) = sin(π/6 + π/2).
Using the compound angle formula, we have:
sin(π/6 + π/2) = sin(π/6)*cos(π/2) + cos(π/6)*sin(π/2).
Since sin(π/2) = 1 and cos(π/2) = 0, the equation simplifies to:
sin(π/6)*0 + cos(π/6)*1 = cos(π/6) = sin(π/3).
Hence, sin(2π/3) is indeed equivalent to sin(π/3) based on the compound angle formula.
2. Although sin(π) is equal to zero and 1/csc(π) is undefined, the identity sin(theta) = 1/csc(theta) still holds true. To understand why, we need to consider the definition of csc(theta) as the reciprocal of sin(theta), which means csc(theta) = 1/sin(theta).
When theta = π, sin(π) = 0, which makes the denominator zero in the expression 1/sin(π) or 1/csc(π). Division by zero is undefined in mathematics.
However, for all other values of theta where sin(theta) does not equal zero, the reciprocal 1/sin(theta) exists. Therefore, for those values, sin(theta) = 1/csc(theta) is a valid identity. It is important to note that identities hold true for valid values of the variables involved.
3. There are a couple of strategies that can be used to solve the equation sec^2(x) + 8sec(x) + 12 = 0.
Strategy 1: Factoring
Let's rewrite the equation as (sec(x) + 2)(sec(x) + 6) = 0 and set each factor equal to zero:
Case 1: sec(x) + 2 = 0
Solving this equation gives sec(x) = -2. However, there are no real solutions for sec(x) = -2 since the range of sec(x) is (-∞, -1] U [1, ∞). Therefore, there are no solutions in this case.
Case 2: sec(x) + 6 = 0
Solving this equation gives sec(x) = -6. Similarly, there are no real solutions for sec(x) = -6 due to the range of sec(x). Therefore, there are no solutions in this case as well.
Hence, the equation sec^2(x) + 8sec(x) + 12 = 0 has no real solutions.
Strategy 2: Quadratic Formula
Alternatively, we can treat the equation as a quadratic equation in terms of sec(x). The equation is in the form ax^2 + bx + c = 0.
Comparing with sec^2(x) + 8sec(x) + 12 = 0, we have a = 1, b = 8, and c = 12.
Using the quadratic formula x = (-b ± √(b^2-4ac)) / (2a) with these values, we get:
x = (-8 ± √(8^2 - 4*1*12)) / (2*1)
x = (-8 ± √(64 - 48)) / 2
x = (-8 ± √16) / 2
x = (-8 ± 4) / 2
This yields two potential solutions:
Solution 1: (-8 + 4) / 2 = -4 / 2 = -2
Solution 2: (-8 - 4) / 2 = -12 / 2 = -6
However, we need to check whether these solutions are valid in the original equation. When substituted back into the equation, we find that neither -2 nor -6 satisfies the equation. Therefore, there are no real solutions for the equation sec^2(x) + 8sec(x) + 12 = 0.
4. If a linear trigonometric equation involving cos(x) has a solution of π/6, we can find three other possible solutions using the periodicity of cosine.
a) Adding the period of cosine (2π) to the solution π/6 gives:
π/6 + 2π = 13π/6
So, π/6 and 13π/6 are two possible solutions.
b) Subtracting the period of cosine (2π) from the solution π/6 gives:
π/6 - 2π = -11π/6
Therefore, -11π/6 is another possible solution.
In summary, the three other possible solutions to a linear trigonometric equation involving cos(x) with a solution of π/6 are 13π/6, -11π/6, and -π/6.