Can you find

(a) a sine graph which touches the lines y = 3 and y = 1?

(b) a cosine graph which crosses the x-axis at x = 1 and x = −1?

(c) a tangent graph which passes through the point (π/3, 0) and for which the line x= π/2 is an asymptote?
Part 2
Write a trigonometric equation that is not an identity. Explain how you know it is not an identity.
Explain why you cannot square each side of the equation when verifying a trigonometric identity.
Explain why both sides of a trigonometric identity are often rewritten in terms of sine and cosine.

(a) To find a sine graph that touches the lines y = 3 and y = 1, we need to determine the amplitude and the vertical shift of the graph.

The amplitude of the sine function represents half the distance between the maximum and minimum values of the graph. In this case, the difference between y = 3 and y = 1 is 2, so the amplitude is 2 divided by 2, which is 1.

The vertical shift represents the amount that the graph is moved up or down. Since the sine graph is centered around y = 0, we need to shift it up by 2 units so that it touches the line y = 3. Therefore, the vertical shift is 2.

Combining the amplitude and the vertical shift, the equation for the sine graph which touches the lines y = 3 and y = 1 is:

y = 1 * sin(x) + 2

(b) To find a cosine graph that crosses the x-axis at x = 1 and x = -1, we need to determine the phase shift of the graph.

The phase shift of the cosine function represents the horizontal displacement of the graph. Since the cosine graph starts at its maximum value and crosses the x-axis at x = 1, it indicates that the graph has shifted to the left by π/2 units.

Combining the phase shift, the equation for the cosine graph which crosses the x-axis at x = 1 and x = -1 is:

y = cos(x - π/2)

(c) To find a tangent graph that passes through the point (π/3, 0) and for which the line x = π/2 is an asymptote, we need to find the vertical shift and the vertical stretching factor of the graph.

The vertical shift represents the amount that the graph is moved up or down. In this case, the tangent graph passes through the point (π/3, 0), so the vertical shift is 0.

The vertical stretching factor represents how the graph is vertically compressed or stretched. Since the line x = π/2 is an asymptote, it means that the tangent function has a vertical stretching factor of 1.

Combining the vertical shift and the vertical stretching factor, the equation for the tangent graph that passes through the point (π/3, 0) and for which the line x = π/2 is an asymptote is:

y = tan(x)

Part 2:
To write a trigonometric equation that is not an identity, we can create an equation where the left side does not equal the right side for all values of the variable.

For example, consider the equation sin(x) = 1. This equation is not an identity because it only holds for specific values of x (for example, x = π/2, 5π/2, etc.), but not for all values of x.

We know it is not an identity because identities hold true for all values of the variable, while this equation only holds true for specific values.

When verifying a trigonometric identity, we cannot square each side of the equation because squaring introduces additional solutions that might not satisfy the original equation. Squaring can create extra solutions that make the equation appear to be an identity when it's not.

Both sides of a trigonometric identity are often rewritten in terms of sine and cosine because the trigonometric functions sine and cosine are fundamental and related to each other. By expressing both sides in terms of sine and cosine, we can simplify and manipulate the equation more easily to prove the identity.

(a) To find a sine graph that touches the lines y=3 and y=1, we need to consider the amplitude and phase shift of the sine function.

The amplitude represents how far the graph stretches vertically from its midline, while the phase shift indicates the horizontal shift. In this case, since the sine graph touches y=3 and y=1, it means the midline should be halfway between these two lines, which is y=2.

The amplitude of the sine function is the distance from the midline to the maximum or minimum value of the graph. Since the midline is y=2 and the graph should touch y=3, the amplitude is 1 (2 to 3) and the maximum is 3.

To determine the phase shift, we can consider the x-values where the maximum and minimum occur. For the sine function, the maximum value is reached at x=0, and the minimum value occurs at x=π. Since we want the graph to touch y=1 at some point, we need to shift the graph horizontally, so the maximum aligns with y=1. This corresponds to shifting the graph right by a half-period, which is π/2.

Using this information, we can construct the equation for the sine graph:
y = amplitude * sin(period * (x - phase shift)) + midline

Plugging in the values, we get:
y = 1 * sin(2π * (x - π/2)) + 2

(b) To find a cosine graph that crosses the x-axis at x=1 and x=-1, we need to determine the amplitude, phase shift, and vertical shift.

The amplitude of the cosine function is the distance from the midline to its highest or lowest point. In this case, since the graph crosses the x-axis at x=1 and x=-1, it means the midline is the x-axis itself, which is y=0. Therefore, the amplitude is 1 (0 to 1), and the maximum value is 1.

To find the phase shift, we look for the x-values where the maximum or minimum occur. For the cosine function, the maximum value happens at x=0, and the minimum occurs at x=π. Since we want the graph to cross the x-axis at x=1, we need to shift it horizontally by one-quarter of a period, which is π/2.

Plugging in the values, we can construct the equation for the cosine graph:
y = amplitude * cos(period * (x - phase shift)) + midline

Using the given values, we get:
y = 1 * cos(2π * (x - π/2)) + 0

(c) To find a tangent graph passing through the point (π/3, 0) with x=π/2 as an asymptote, we need to consider the phase shift and vertical shift.

The tangent function does not have a vertical shift since its midline is the line y=0. Therefore, the equation only needs to consider the phase shift.

For the tangent graph, it has vertical asymptotes at x = (n + 1/2)π, where n is an integer, so x=π/2 (or x=(1/2)π) is already an asymptote in this case.

Since the graph passes through the point (π/3, 0), we need to shift it horizontally to the right so that the point (π/3, 0) aligns with the vertical asymptote at x=π/2. This corresponds to a phase shift of π/6.

Using this information, we can construct the equation for the tangent graph:
y = tangent(period * (x - phase shift))

Plugging in the values, we get:
y = tan(1 * (x - π/6))

Part 2:
To write a trigonometric equation that is not an identity, we can combine trigonometric functions or use non-standard transformations. For example:
sin(x) + cos(x) = 1

To know that it is not an identity, we can try substituting different values of x into the equation and check if it holds true for all values. If we find at least one value where the equation does not hold, then it is not an identity.

When verifying a trigonometric identity, we cannot always square each side of the equation. Squaring both sides can introduce extraneous solutions and may not preserve the equality between the two sides. It can lead to false results for trigonometric identities.

Both sides of a trigonometric identity are often rewritten in terms of sine and cosine because they are fundamental trigonometric functions. They provide a clear representation of the relationships between angles and the unit circle. By rewriting both sides in terms of sine and cosine, we can simplify and manipulate the expressions to establish the identity more easily. Additionally, it allows for better comparison and identification of equivalent expressions.

(a) the center line is at y = (3+1)/2 = 2

the amplitude is (3-1)/2 = 1
So, y = 2+sinx

(b) 1/2 period is 1-(-1) = 2, so the period is 4
cos(kx) has period 2π/k, so k = π/2
y = cos(π/2 x)

(c)(π/3, 0) and for which the line x= π/2 is an asymptote
tan(kx) has period π/k. Your graph has half-period π/2 - π/3 = π/6, so k=3
tanx crosses the x-axis at x=0, so yours is shifted right π/3
That makes the function y = tan(3(x-π/3)) = tan(3x-π)

#2 sinx = cosx
clearly it is not true for x=0, so it is not an identity
why don't you take a stab at the other parts.
consider x^2 = 2^2...