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. Why do the solutions to the equation cos^2x = 3/4 lie in all four quadrants?
sin(2π/3) = sin(π-π/3) = sinπ cosπ/3 - cosπ sinπ/3 = 0+sinπ/3
1/0 is undefined.
cosx = ±√3/2
cosx > 0 in QI,QIV
cosx < 0 in QII,QIII
1. To demonstrate that sin(2π/3) is equivalent to sin(π/3), we'll use the compound angle formula. According to the formula, sin(A + B) = sin(A)cos(B) + cos(A)sin(B).
Let's substitute A = π/3 and B = π/3 into the formula:
sin(π/3 + π/3) = sin(2π/3) = sin(π/3)cos(π/3) + cos(π/3)sin(π/3)
Now let's simplify the right side of the equation:
sin(2π/3) = (√3/2)(1/2) + (1/2)(√3/2)
= (√3 + √3) / 4
= (2√3) / 4
= √3 / 2
We can see that √3 / 2 is the same as sin(π/3), so sin(2π/3) is equivalent to sin(π/3).
2. The expression sin(π) is equal to zero because sin(π) represents the y-coordinate of a point on the unit circle that is diametrically opposite to the point (1, 0). The y-coordinate of that point is 0, hence sin(π) = 0.
On the other hand, 1/csc(π) is undefined because csc(π) represents the reciprocal of sin(π), and since sin(π) is zero, its reciprocal (1/0) is undefined.
Although sin(π) and 1/csc(π) are not mathematically equal, the identity sin(theta) = 1/csc(theta) still holds true. This is because the identity is valid for all values of theta except for those that make the denominator of 1/csc(theta) equal to zero. In this case, π is an exception, making the expression undefined. However, for all other values of theta, sin(theta) and 1/csc(theta) are equal.
3. The equation cos^2(x) = 3/4 is a quadratic equation in terms of cos(x). When solving this equation, we need to find values of x that satisfy the equation.
Taking the square root of both sides, we get cos(x) = ±√(3/4). Simplifying further, we have cos(x) = ±√3/2.
Recall that the cosine function is positive in quadrants I and IV, while it is negative in quadrants II and III.
Therefore, in quadrant I, cos(x) = √3/2, giving us one solution.
In quadrant II, cos(x) = -√3/2, providing us with another solution.
In quadrant III, cos(x) = -√3/2, giving us an additional solution.
In quadrant IV, cos(x) = √3/2, providing us with the final solution.
Hence, the solutions to the equation cos^2(x) = 3/4 lie in all four quadrants.
1. To demonstrate that sin(2π/3) is equivalent to sin(π/3) using a compound angle formula, we can use the formula for the sine of a sum of angles:
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
In this case, let a = π/3 and b = π/3. Plugging these values into the formula, we get:
sin(2π/3) = sin(π/3 + π/3) = sin(2π/3)cos(π/3) + cos(2π/3)sin(π/3)
Now, let's recall some key trigonometric values for π/3 and 2π/3:
sin(π/3) = √3/2
cos(π/3) = 1/2
sin(2π/3) = √3/2
cos(2π/3) = -1/2
Plugging these values into the equation above, we get:
sin(2π/3) = (√3/2)(1/2) + (-1/2)(√3/2)
= √3/4 - √3/4
= 0
Since sin(π/3) is also equal to 0, we can conclude that sin(2π/3) is indeed equivalent to sin(π/3).
2. The expression sin(π) is equal to zero because the sine function has a periodic behavior with a period of 2π. This means that after every 2π radians (or 360 degrees), the sine function repeats its values. Therefore, sin(π) equals sin(π + 2π), sin(π + 4π), sin(π - 2π), and so on, all of which are equal to zero.
On the other hand, 1/csc(π) is undefined because the cosecant function is the reciprocal of the sine function. The cosecant function becomes undefined whenever the sine function equals zero (i.e., at integer multiples of π). Since csc(π) would be the reciprocal of sin(π), it is undefined.
Despite sin(θ) being equal to zero and 1/csc(θ) being undefined at certain points, the identity sin(θ) = 1/csc(θ) still holds. This is because the identity is valid for all values of θ except where sin(θ) is zero, as dividing by zero is undefined in mathematics.
3. The equation cos^2(x) = 3/4 indicates that the square of the cosine of angle x is equal to 3/4. To find the solutions to this equation, we need to consider the unit circle and the values of cosine in different quadrants.
First, let's find the solutions for the principal range of x between 0 and 2π. Taking the square root of both sides of the equation, we get:
cos(x) = ±√(3/4)
cos(x) = ±√3/2
The cosine function is positive in the first and fourth quadrants, so the solutions in the principal range are:
x = π/6 in the first quadrant
x = 11π/6 in the fourth quadrant
Now, to find solutions in all four quadrants, we need to add or subtract integer multiples of 2π to these principal solutions. This is because the cosine function has a period of 2π, so it repeats its values after every 2π radians. Thus, the solutions in all four quadrants are:
x = π/6 + 2πn, where n is an integer (first quadrant)
x = 11π/6 + 2πn, where n is an integer (fourth quadrant)
These solutions cover all possible angles where cos^2(x) = 3/4, which is why the solutions lie in all four quadrants.