10. Practice Problems

1. Prove: 1 + tan²θ = sec²θ

To prove this identity, recall that tan(θ) = sin(θ)/cos(θ) and sec(θ) = 1/cos(θ). Therefore, tan²(θ) = sin²(θ)/cos²(θ). By substituting this into the equation 1 + tan²θ, we get:

1 + tan²θ = 1 + (sin²θ/cos²θ) = (cos²θ/cos²θ) + (sin²θ/cos²θ) = (sin²θ + cos²θ)/cos²θ. Since we know by the Pythagorean identity that sin²θ + cos²θ = 1, we have:

(sin²θ + cos²θ)/cos²θ = 1/cos²θ = sec²θ, thus proving the identity.

Think of this as verifying a recipe for a cake. You have the final result (the cake) and need to show that all the ingredients used (the values of sin and cos) correctly combine to give the final product (the identity). This step-by-step validation ensures that what you've crafted is indeed correct, just like in mathematics.

2. If sin(x) = 3/5 and x is in the first quadrant, find cos(x) and tan(x)

To find cos(x) and tan(x) when sin(x) = 3/5, we can use the Pythagorean identity sin²θ + cos²θ = 1. Substitute sin(x):

(3/5)² + cos²(x) = 1
9/25 + cos²(x) = 1
cos²(x) = 1 - 9/25 = 16/25
Then, cos(x) = √(16/25) = 4/5.
Now, to find tan(x), recall that tan(x) = sin(x)/cos(x):
tan(x) = (3/5) / (4/5) = 3/4.

Imagine you are climbing a ladder. If the height you reach (representing sin) is 3 rungs out of a total of 5, the total distance from the base (representing the hypotenuse) is your ladder. By determining how far horizontally you have stretched (the cosine), you can now figure out your steepness or 'angle' (the tangent) of the ladder—finding how your vertical height relates to your horizontal distance.

3. Sketch the graph of y = sin(x − 90°)

The equation y = sin(x - 90°) indicates a phase shift of 90° to the right. Since the sine graph typically starts at the origin (0,0), beginning from x = 90° shifts the peak of the sine wave to the y-axis point (0,1). This phase shift alters the graph's starting point but maintains the wave's shape and amplitude at a value of 1. Thus, key points on the graph should be drawn considering this shift.

Imagine setting an alarm for 7 AM, but you want to wake up at 8 AM instead. By physically moving the time of your alarm (shifting the graph), the way your morning routine looks (the entire sine wave) doesn’t change in its nature, but it starts at a different point—just as sin(x) shifts with phase changes.

4. Solve the equation: 2 sin²x − 1 = 0 for 0° ≤ x ≤ 360°

To solve 2 sin²x - 1 = 0, first isolate sin²x:

2 sin²x = 1
sin²x = 1/2.
Taking the square root gives us sin(x) = √(1/2) or sin(x) = -√(1/2). Since we need solutions within the first revolution 0° to 360°, we find: sin(x) = √(1/2) ⇒ x = 45°, 135° sin(x) = -√(1/2) ⇒ x = 225°, 315°. Therefore, the solutions are x = 45°, 135°, 225°, 315°.

Think of this as trying to find out at what times during a day the sun is at a specific brightness level (the height of the sine function). You’d look at the times when it’s exactly halfway up before noon or after noon, realizing that there are multiple points in the day when that brightness occurs, similar to finding multiple solutions in your equation.

5. What is the amplitude and period of y = −2 cos(3x)?

In the function y = -2 cos(3x), the amplitude is the absolute value of the coefficient in front of cos, which is |-2| = 2. This value represents how high or low the graph reaches from the center line. The period of a cosine function is calculated using the formula: Period = 360° / B, where B is the coefficient of x. Here, B = 3, so: Period = 360° / 3 = 120°.

Think of a trampoline that can bounce 2 units high and low around the resting point (the center line of the graph). The way you bounce affects how often you complete a cycle (the period)—if you bounce faster (with a higher frequency), you’re completing more bounces in the same timeframe, just like a function completing its wave pattern more quickly.

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