IB Mathematics: Analysis and Approaches Quiz: Double Angle And Trig Graphs
20 questions · exam conditions
0:00
Double Angle And Trig GraphsQuestion 1 of 20

If cos⁡(2x)=725\cos(2x) = \frac{7}{25}, and xx is an acute angle, what is the value of cos⁡x\cos x?

35\frac{3}{5}
±45\pm \frac{4}{5}
1625\frac{16}{25}
45\frac{4}{5}
← Back to quizzes

IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Double Angle And Trig Graphs

Practice Double Angle And Trig Graphs in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Double Angle And Trig Graphs, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

If cos⁡(2x)=725\cos(2x) = \frac{7}{25}, and xx is an acute angle, what is the value of cos⁡x\cos x?

  1. 35\frac{3}{5}
  2. ±45\pm \frac{4}{5}
  3. 1625\frac{16}{25}
  4. 45\frac{4}{5} (correct answer)
Explanation: This question tests your understanding of the double angle formula for cosine and how to work backwards from cos⁡(2x)\cos(2x) to find cos⁡x\cos x. The double angle formula states that cos⁡(2x)=2cos⁡2x−1\cos(2x) = 2\cos^2 x - 1. Since we know cos⁡(2x)=725\cos(2x) = \frac{7}{25}, we can substitute and solve: 725=2cos⁡2x−1\frac{7}{25} = 2\cos^2 x - 1 Adding 1 to both sides: 725+1=2cos⁡2x\frac{7}{25} + 1 = 2\cos^2 x Converting to a common denominator: 725+2525=3225=2cos⁡2x\frac{7}{25} + \frac{25}{25} = \frac{32}{25} = 2\cos^2 x Dividing by 2: cos⁡2x=1625\cos^2 x = \frac{16}{25} Taking the square root: cos⁡x=±45\cos x = \pm\frac{4}{5} However, since xx is an acute angle (between 0° and 90°), cosine must be positive. Therefore, cos⁡x=45\cos x = \frac{4}{5}. Let's examine why the other answers are incorrect: A) 35\frac{3}{5} would give cos⁡(2x)=2(35)2−1=1825−1=−725\cos(2x) = 2(\frac{3}{5})^2 - 1 = \frac{18}{25} - 1 = -\frac{7}{25}, which has the wrong sign. B) ±45\pm\frac{4}{5} is mathematically correct before considering the constraint, but ignores that xx is acute, making cosine positive. C) 1625\frac{16}{25} is the value of cos⁡2x\cos^2 x, not cos⁡x\cos x itself—a common mistake when students forget to take the square root. Remember: when working with double angle formulas, always check the quadrant constraints given in the problem to determine the correct sign for your final answer.

Question 2

Which of the following functions has the same graph as f(x)=1−2sin⁡2(x2)f(x) = 1 - 2\sin^2(\frac{x}{2})?

  1. g(x)=cos⁡(x)g(x) = \cos(x) (correct answer)
  2. g(x)=cos⁡(x2)g(x) = \cos(\frac{x}{2})
  3. g(x)=cos⁡(2x)g(x) = \cos(2x)
  4. g(x)=sin⁡(x)g(x) = \sin(x)
Explanation: When you encounter trigonometric expressions with squared terms, look for opportunities to apply double angle identities. These identities often simplify complex expressions into basic trigonometric functions. To transform f(x)=1−2sin⁡2(x2)f(x) = 1 - 2\sin^2(\frac{x}{2}), recall the cosine double angle identity: cos⁡(2θ)=1−2sin⁡2(θ)\cos(2\theta) = 1 - 2\sin^2(\theta). This identity is derived from cos⁡(2θ)=cos⁡2(θ)−sin⁡2(θ)\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) combined with the Pythagorean identity cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1. In our function, if we let θ=x2\theta = \frac{x}{2}, then 2θ=x2\theta = x. Applying the identity: f(x)=1−2sin⁡2(x2)=cos⁡(x)f(x) = 1 - 2\sin^2(\frac{x}{2}) = \cos(x). Therefore, choice A is correct. Let's examine why the other options don't work. Choice B, g(x)=cos⁡(x2)g(x) = \cos(\frac{x}{2}), would result from misidentifying which angle to use in the identity. Choice C, g(x)=cos⁡(2x)g(x) = \cos(2x), represents the opposite error—doubling the angle instead of recognizing that 2⋅x2=x2 \cdot \frac{x}{2} = x. Choice D, g(x)=sin⁡(x)g(x) = \sin(x), might tempt you because the original function contains sine, but the double angle transformation produces cosine, not sine. Study tip: Memorize all three forms of the cosine double angle identity: cos⁡(2θ)=cos⁡2(θ)−sin⁡2(θ)=2cos⁡2(θ)−1=1−2sin⁡2(θ)\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 2\cos^2(\theta) - 1 = 1 - 2\sin^2(\theta). The third form is particularly useful for simplifying expressions with 1−2sin⁡21 - 2\sin^2 patterns.

Question 3

Which of the following functions has the same graph as f(x)=1−2sin⁡2(x2)f(x) = 1 - 2\sin^2(\frac{x}{2})?

  1. g(x)=cos⁡(x)g(x) = \cos(x) (correct answer)
  2. g(x)=cos⁡(x2)g(x) = \cos(\frac{x}{2})
  3. g(x)=cos⁡(2x)g(x) = \cos(2x)
  4. g(x)=sin⁡(x)g(x) = \sin(x)
Explanation: When you encounter trigonometric expressions with squared terms, look for opportunities to apply double angle identities. These identities often simplify complex expressions into basic trigonometric functions. To transform f(x)=1−2sin⁡2(x2)f(x) = 1 - 2\sin^2(\frac{x}{2}), recall the cosine double angle identity: cos⁡(2θ)=1−2sin⁡2(θ)\cos(2\theta) = 1 - 2\sin^2(\theta). This identity is derived from cos⁡(2θ)=cos⁡2(θ)−sin⁡2(θ)\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) combined with the Pythagorean identity cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1. In our function, if we let θ=x2\theta = \frac{x}{2}, then 2θ=x2\theta = x. Applying the identity: f(x)=1−2sin⁡2(x2)=cos⁡(x)f(x) = 1 - 2\sin^2(\frac{x}{2}) = \cos(x). Therefore, choice A is correct. Let's examine why the other options don't work. Choice B, g(x)=cos⁡(x2)g(x) = \cos(\frac{x}{2}), would result from misidentifying which angle to use in the identity. Choice C, g(x)=cos⁡(2x)g(x) = \cos(2x), represents the opposite error—doubling the angle instead of recognizing that 2⋅x2=x2 \cdot \frac{x}{2} = x. Choice D, g(x)=sin⁡(x)g(x) = \sin(x), might tempt you because the original function contains sine, but the double angle transformation produces cosine, not sine. Study tip: Memorize all three forms of the cosine double angle identity: cos⁡(2θ)=cos⁡2(θ)−sin⁡2(θ)=2cos⁡2(θ)−1=1−2sin⁡2(θ)\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 2\cos^2(\theta) - 1 = 1 - 2\sin^2(\theta). The third form is particularly useful for simplifying expressions with 1−2sin⁡21 - 2\sin^2 patterns.

Question 4

If cos⁡(2x)=725\cos(2x) = \frac{7}{25}, and xx is an acute angle, what is the value of cos⁡x\cos x?

  1. 35\frac{3}{5}
  2. ±45\pm \frac{4}{5}
  3. 1625\frac{16}{25}
  4. 45\frac{4}{5} (correct answer)
Explanation: This question tests your understanding of the double angle formula for cosine and how to work backwards from cos⁡(2x)\cos(2x) to find cos⁡x\cos x. The double angle formula states that cos⁡(2x)=2cos⁡2x−1\cos(2x) = 2\cos^2 x - 1. Since we know cos⁡(2x)=725\cos(2x) = \frac{7}{25}, we can substitute and solve: 725=2cos⁡2x−1\frac{7}{25} = 2\cos^2 x - 1 Adding 1 to both sides: 725+1=2cos⁡2x\frac{7}{25} + 1 = 2\cos^2 x Converting to a common denominator: 725+2525=3225=2cos⁡2x\frac{7}{25} + \frac{25}{25} = \frac{32}{25} = 2\cos^2 x Dividing by 2: cos⁡2x=1625\cos^2 x = \frac{16}{25} Taking the square root: cos⁡x=±45\cos x = \pm\frac{4}{5} However, since xx is an acute angle (between 0° and 90°), cosine must be positive. Therefore, cos⁡x=45\cos x = \frac{4}{5}. Let's examine why the other answers are incorrect: A) 35\frac{3}{5} would give cos⁡(2x)=2(35)2−1=1825−1=−725\cos(2x) = 2(\frac{3}{5})^2 - 1 = \frac{18}{25} - 1 = -\frac{7}{25}, which has the wrong sign. B) ±45\pm\frac{4}{5} is mathematically correct before considering the constraint, but ignores that xx is acute, making cosine positive. C) 1625\frac{16}{25} is the value of cos⁡2x\cos^2 x, not cos⁡x\cos x itself—a common mistake when students forget to take the square root. Remember: when working with double angle formulas, always check the quadrant constraints given in the problem to determine the correct sign for your final answer.

Question 5

The graph of the function f(x)=cos⁡(2x)f(x) = \cos(2x) is a horizontal compression of the graph of g(x)=cos⁡xg(x) = \cos x by a factor of 2. Which of the following functions has a graph which is a vertical translation of the graph of f(x)f(x)?

  1. h(x)=1−2cos⁡2xh(x) = 1 - 2\cos^2 x
  2. h(x)=2cos⁡xh(x) = 2\cos x
  3. h(x)=cos⁡2x−sin⁡2xh(x) = \cos^2 x - \sin^2 x
  4. h(x)=2cos⁡2xh(x) = 2\cos^2 x (correct answer)
Explanation: A vertical translation of f(x)=cos⁡(2x)f(x) = \cos(2x) would be of the form y=cos⁡(2x)+ky = \cos(2x) + k for some non-zero constant kk. We need to check which option can be written in this form. Using the double angle identity cos⁡(2x)=2cos⁡2x−1\cos(2x) = 2\cos^2 x - 1, we can rearrange it to get 2cos⁡2x=cos⁡(2x)+12\cos^2 x = \cos(2x) + 1. So, the function h(x)=2cos⁡2xh(x) = 2\cos^2 x is equivalent to h(x)=cos⁡(2x)+1h(x) = \cos(2x) + 1, which is a vertical translation of f(x)f(x) by 1 unit upwards. Choice A is −cos⁡(2x)-\cos(2x). Choice C is cos⁡(2x)\cos(2x), which is not a translation. Choice B is a vertical stretch of g(x)g(x).

Question 6

Given that cos⁡θ=23\cos \theta = \frac{2}{3} and 3π2<θ<2π\frac{3\pi}{2} < \theta < 2\pi, find the exact value of sin⁡(2θ)\sin(2\theta).

  1. −459-\frac{4\sqrt{5}}{9} (correct answer)
  2. −259-\frac{2\sqrt{5}}{9}
  3. 253\frac{2\sqrt{5}}{3}
  4. 459\frac{4\sqrt{5}}{9}
Explanation: The angle θ\theta is in the fourth quadrant, where cos⁡θ>0\cos \theta > 0 and sin⁡θ<0\sin \theta < 0. We use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to find sin⁡θ\sin \theta. sin⁡2θ=1−(23)2=1−49=59\sin^2 \theta = 1 - (\frac{2}{3})^2 = 1 - \frac{4}{9} = \frac{5}{9}. Since θ\theta is in the fourth quadrant, sin⁡θ=−59=−53\sin \theta = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}. Now we use the double angle identity sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\theta. Substituting the values: sin⁡(2θ)=2(−53)(23)=−459\sin(2\theta) = 2(-\frac{\sqrt{5}}{3})(\frac{2}{3}) = -\frac{4\sqrt{5}}{9}.

Question 7

How many solutions does the equation sin⁡(2x)=2sin⁡x\sin(2x) = 2\sin x have in the interval [0,2π][0, 2\pi]?

  1. 1
  2. 2
  3. 3 (correct answer)
  4. 4
Explanation: First, use the double angle identity sin⁡(2x)=2sin⁡xcos⁡x\sin(2x) = 2\sin x \cos x. The equation becomes 2sin⁡xcos⁡x=2sin⁡x2\sin x \cos x = 2\sin x. It is a common mistake to divide by 2sin⁡x2\sin x; instead, we should rearrange and factor. 2sin⁡xcos⁡x−2sin⁡x=02\sin x \cos x - 2\sin x = 0, which gives 2sin⁡x(cos⁡x−1)=02\sin x (\cos x - 1) = 0. This equation is satisfied if sin⁡x=0\sin x = 0 or cos⁡x−1=0\cos x - 1 = 0 (i.e., cos⁡x=1\cos x = 1). In the interval [0,2π][0, 2\pi], sin⁡x=0\sin x = 0 has solutions x=0,π,2πx = 0, \pi, 2\pi. In the interval [0,2π][0, 2\pi], cos⁡x=1\cos x = 1 has solutions x=0,2πx = 0, 2\pi. The set of distinct solutions is {0,π,2π}\{0, \pi, 2\pi\}. There are 3 solutions.

Question 8

What is the period of the function f(x)=sin⁡2(3x)f(x) = \sin^2(3x)?

  1. π6\frac{\pi}{6}
  2. π3\frac{\pi}{3} (correct answer)
  3. 2π3\frac{2\pi}{3}
  4. 2π2\pi
Explanation: To find the period of f(x)=sin⁡2(3x)f(x) = \sin^2(3x), we use the power-reducing identity which is derived from a double angle identity: sin⁡2θ=1−cos⁡(2θ)2\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}. Let θ=3x\theta = 3x. Then f(x)=1−cos⁡(2⋅3x)2=12−12cos⁡(6x)f(x) = \frac{1 - \cos(2 \cdot 3x)}{2} = \frac{1}{2} - \frac{1}{2}\cos(6x). The period of a function of the form y=acos⁡(bx)+dy = a \cos(bx) + d is 2π∣b∣\frac{2\pi}{|b|}. In this case, b=6b=6, so the period is 2π6=π3\frac{2\pi}{6} = \frac{\pi}{3}.

Question 9

The expression sin⁡(2A)1−cos⁡(2A)\frac{\sin(2A)}{1-\cos(2A)} is equivalent to:

  1. tan⁡A\tan A
  2. cot⁡A\cot A (correct answer)
  3. tan⁡(2A)\tan(2A)
  4. sin⁡A\sin A
Explanation: We use the double angle identities for sine and cosine. sin⁡(2A)=2sin⁡Acos⁡A\sin(2A) = 2\sin A \cos A and cos⁡(2A)=1−2sin⁡2A\cos(2A) = 1 - 2\sin^2 A. Substitute these into the expression: 2sin⁡Acos⁡A1−(1−2sin⁡2A)=2sin⁡Acos⁡A1−1+2sin⁡2A=2sin⁡Acos⁡A2sin⁡2A\frac{2\sin A \cos A}{1-(1-2\sin^2 A)} = \frac{2\sin A \cos A}{1-1+2\sin^2 A} = \frac{2\sin A \cos A}{2\sin^2 A}. Assuming sin⁡A≠0\sin A \neq 0, we can cancel 2sin⁡A2\sin A from the numerator and denominator, leaving cos⁡Asin⁡A\frac{\cos A}{\sin A}, which is the definition of cot⁡A\cot A.

Question 10

The equation cos⁡(2x)−sin⁡x=0\cos(2x) - \sin x = 0 has two solutions in the interval 0≤x≤π0 \le x \le \pi. Find the sum of these two solutions.

  1. π2\frac{\pi}{2}
  2. 5π6\frac{5\pi}{6}
  3. π\pi (correct answer)
  4. 3π2\frac{3\pi}{2}
Explanation: To solve the equation, we use the double angle identity cos⁡(2x)=1−2sin⁡2x\cos(2x) = 1 - 2\sin^2 x to express the equation in terms of sin⁡x\sin x only. The equation becomes 1−2sin⁡2x−sin⁡x=01 - 2\sin^2 x - \sin x = 0, which can be rearranged to 2sin⁡2x+sin⁡x−1=02\sin^2 x + \sin x - 1 = 0. Let u=sin⁡xu = \sin x. We have the quadratic equation 2u2+u−1=02u^2 + u - 1 = 0, which factors as (2u−1)(u+1)=0(2u-1)(u+1) = 0. So, u=12u = \frac{1}{2} or u=−1u = -1. This means sin⁡x=12\sin x = \frac{1}{2} or sin⁡x=−1\sin x = -1. For sin⁡x=12\sin x = \frac{1}{2} in the interval 0≤x≤π0 \le x \le \pi, the solutions are x=π6x = \frac{\pi}{6} and x=5π6x = \frac{5\pi}{6}. For sin⁡x=−1\sin x = -1, there are no solutions in the interval 0≤x≤π0 \le x \le \pi. The two solutions are π6\frac{\pi}{6} and 5π6\frac{5\pi}{6}. Their sum is π6+5π6=6π6=π\frac{\pi}{6} + \frac{5\pi}{6} = \frac{6\pi}{6} = \pi.

Question 11

The graph of the function f(x)=cos⁡(2x)f(x) = \cos(2x) is a horizontal compression of the graph of g(x)=cos⁡xg(x) = \cos x by a factor of 2. Which of the following functions has a graph which is a vertical translation of the graph of f(x)f(x)?

  1. h(x)=1−2cos⁡2xh(x) = 1 - 2\cos^2 x
  2. h(x)=2cos⁡xh(x) = 2\cos x
  3. h(x)=cos⁡2x−sin⁡2xh(x) = \cos^2 x - \sin^2 x
  4. h(x)=2cos⁡2xh(x) = 2\cos^2 x (correct answer)
Explanation: A vertical translation of f(x)=cos⁡(2x)f(x) = \cos(2x) would be of the form y=cos⁡(2x)+ky = \cos(2x) + k for some non-zero constant kk. We need to check which option can be written in this form. Using the double angle identity cos⁡(2x)=2cos⁡2x−1\cos(2x) = 2\cos^2 x - 1, we can rearrange it to get 2cos⁡2x=cos⁡(2x)+12\cos^2 x = \cos(2x) + 1. So, the function h(x)=2cos⁡2xh(x) = 2\cos^2 x is equivalent to h(x)=cos⁡(2x)+1h(x) = \cos(2x) + 1, which is a vertical translation of f(x)f(x) by 1 unit upwards. Choice A is −cos⁡(2x)-\cos(2x). Choice C is cos⁡(2x)\cos(2x), which is not a translation. Choice B is a vertical stretch of g(x)g(x).

Question 12

The function f(x)=(sin⁡x+cos⁡x)2f(x) = (\sin x + \cos x)^2 can be written in the form y=asin⁡(2x)+by = a \sin(2x) + b. What is the period of f(x)f(x)?

  1. π2\frac{\pi}{2}
  2. π\pi (correct answer)
  3. 2π2\pi
  4. 4π4\pi
Explanation: First, expand the expression for f(x)f(x): f(x)=(sin⁡x+cos⁡x)2=sin⁡2x+2sin⁡xcos⁡x+cos⁡2xf(x) = (\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x. Using the Pythagorean identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 and the double angle identity 2sin⁡xcos⁡x=sin⁡(2x)2\sin x \cos x = \sin(2x), we can rewrite the function as f(x)=1+sin⁡(2x)f(x) = 1 + \sin(2x). The period of a function of the form y=asin⁡(b(x−c))+dy = a \sin(b(x-c)) + d is given by 2π∣b∣\frac{2\pi}{|b|}. For f(x)=1+sin⁡(2x)f(x) = 1 + \sin(2x), we have b=2b=2. Therefore, the period is 2π2=π\frac{2\pi}{2} = \pi.

Question 13

The principal axis of the graph of y=4cos⁡2(x)−5y = 4\cos^2(x) - 5 is:

  1. y=−5y = -5
  2. y=−3y = -3 (correct answer)
  3. y=−1y = -1
  4. y=2y = 2
Explanation: The principal axis of a trigonometric function is its central horizontal line, which corresponds to the vertical shift. To find it, we rewrite the function using the identity derived from the double angle formula for cosine: cos⁡2(x)=1+cos⁡(2x)2\cos^2(x) = \frac{1 + \cos(2x)}{2}. Substituting this into the equation gives y=4(1+cos⁡(2x)2)−5y = 4\left(\frac{1 + \cos(2x)}{2}\right) - 5. Simplifying, we get y=2(1+cos⁡(2x))−5=2+2cos⁡(2x)−5=2cos⁡(2x)−3y = 2(1 + \cos(2x)) - 5 = 2 + 2\cos(2x) - 5 = 2\cos(2x) - 3. This is a cosine function with amplitude 2, vertically shifted down by 3 units. The principal axis is the line y=−3y = -3.

Question 14

The function f(x)=sin⁡(x)cos⁡(x)f(x) = \sin(x)\cos(x) is transformed to the function g(x)=sin⁡(2x)g(x) = \sin(2x). What transformation maps the graph of ff onto the graph of gg?

  1. A horizontal compression by a factor of 2.
  2. A vertical stretch by a factor of 12\frac{1}{2}.
  3. A vertical stretch by a factor of 2. (correct answer)
  4. A vertical translation of 2 units upwards.
Explanation: First, we express f(x)f(x) using the double angle identity sin⁡(2x)=2sin⁡xcos⁡x\sin(2x) = 2\sin x \cos x. Rearranging this gives sin⁡xcos⁡x=12sin⁡(2x)\sin x \cos x = \frac{1}{2}\sin(2x). So, f(x)=12sin⁡(2x)f(x) = \frac{1}{2}\sin(2x). We are given g(x)=sin⁡(2x)g(x) = \sin(2x). To get from f(x)f(x) to g(x)g(x), we must multiply f(x)f(x) by 2: g(x)=2⋅f(x)g(x) = 2 \cdot f(x). A transformation of the form y=kf(x)y = kf(x) is a vertical stretch by a factor of kk. Here, k=2k=2, so the transformation is a vertical stretch by a factor of 2.

Question 15

Consider the function f(x)=1−cos⁡(2x)f(x) = 1 - \cos(2x). What are the amplitude and period of the graph of y=f(x)y=f(x)?

  1. Amplitude is 1, Period is π\pi (correct answer)
  2. Amplitude is 1, Period is 2π2\pi
  3. Amplitude is 2, Period is π\pi
  4. Amplitude is 2, Period is 2π2\pi
Explanation: The function can be written as f(x)=−cos⁡(2x)+1f(x) = -\cos(2x) + 1. In the standard form y=acos⁡(b(x−c))+dy=a\cos(b(x-c))+d, we have a=−1a=-1, b=2b=2, and d=1d=1. The amplitude is ∣a∣=∣−1∣=1|a| = |-1| = 1. The period is 2π∣b∣=2π2=π\frac{2\pi}{|b|} = \frac{2\pi}{2} = \pi. Alternatively, using the identity cos⁡(2x)=1−2sin⁡2x\cos(2x) = 1 - 2\sin^2 x, the function is f(x)=1−(1−2sin⁡2x)=2sin⁡2xf(x) = 1 - (1 - 2\sin^2 x) = 2\sin^2 x. This function oscillates between a minimum value of 0 and a maximum value of 2. The principal axis is halfway between the min and max, at y=1y=1. The amplitude is the distance from the principal axis to the maximum, which is 2−1=12-1=1. The period of sin⁡2x\sin^2 x is π\pi, and multiplying by 2 vertically does not change the period.

Question 16

The function f(x)=(sin⁡x+cos⁡x)2f(x) = (\sin x + \cos x)^2 can be written in the form y=asin⁡(2x)+by = a \sin(2x) + b. What is the period of f(x)f(x)?

  1. π2\frac{\pi}{2}
  2. π\pi (correct answer)
  3. 2π2\pi
  4. 4π4\pi
Explanation: First, expand the expression for f(x)f(x): f(x)=(sin⁡x+cos⁡x)2=sin⁡2x+2sin⁡xcos⁡x+cos⁡2xf(x) = (\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x. Using the Pythagorean identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 and the double angle identity 2sin⁡xcos⁡x=sin⁡(2x)2\sin x \cos x = \sin(2x), we can rewrite the function as f(x)=1+sin⁡(2x)f(x) = 1 + \sin(2x). The period of a function of the form y=asin⁡(b(x−c))+dy = a \sin(b(x-c)) + d is given by 2π∣b∣\frac{2\pi}{|b|}. For f(x)=1+sin⁡(2x)f(x) = 1 + \sin(2x), we have b=2b=2. Therefore, the period is 2π2=π\frac{2\pi}{2} = \pi.

Question 17

The function f(x)=sin⁡(x)cos⁡(x)f(x) = \sin(x)\cos(x) is transformed to the function g(x)=sin⁡(2x)g(x) = \sin(2x). What transformation maps the graph of ff onto the graph of gg?

  1. A horizontal compression by a factor of 2.
  2. A vertical stretch by a factor of 12\frac{1}{2}.
  3. A vertical stretch by a factor of 2. (correct answer)
  4. A vertical translation of 2 units upwards.
Explanation: First, we express f(x)f(x) using the double angle identity sin⁡(2x)=2sin⁡xcos⁡x\sin(2x) = 2\sin x \cos x. Rearranging this gives sin⁡xcos⁡x=12sin⁡(2x)\sin x \cos x = \frac{1}{2}\sin(2x). So, f(x)=12sin⁡(2x)f(x) = \frac{1}{2}\sin(2x). We are given g(x)=sin⁡(2x)g(x) = \sin(2x). To get from f(x)f(x) to g(x)g(x), we must multiply f(x)f(x) by 2: g(x)=2⋅f(x)g(x) = 2 \cdot f(x). A transformation of the form y=kf(x)y = kf(x) is a vertical stretch by a factor of kk. Here, k=2k=2, so the transformation is a vertical stretch by a factor of 2.

Question 18

The graph of y=sin⁡(x)y=\sin(x) is transformed to the graph of y=sin⁡(2x)y=\sin(2x). Which of the following describes this transformation?

  1. A horizontal compression by a factor of 2. (correct answer)
  2. A horizontal stretch by a factor of 2.
  3. A vertical stretch by a factor of 2.
  4. A vertical compression by a factor of 2.
Explanation: When you encounter function transformations, focus on how changes inside versus outside the function affect the graph differently. Changes to the input (inside the function) create horizontal transformations, while changes to the output (outside the function) create vertical transformations. To understand the transformation from y=sin⁡(x)y = \sin(x) to y=sin⁡(2x)y = \sin(2x), examine what happens to the period. The original sine function has period 2π2\pi, meaning it completes one full cycle every 2π2\pi units. For y=sin⁡(2x)y = \sin(2x), when x=πx = \pi, we get sin⁡(2π)=0\sin(2\pi) = 0, completing one full cycle in just π\pi units. The new period is π\pi, which is half the original period. Since the function now completes its cycle in half the horizontal distance, the graph is compressed horizontally by a factor of 2. This confirms answer A is correct. Let's examine why the other options are wrong: B suggests a horizontal stretch by factor 2, but the graph actually gets squeezed together, not stretched apart. C proposes a vertical stretch by factor 2, but the amplitude remains 1 (the coefficient outside the sine function is still 1). D suggests vertical compression by factor 2, but again, the amplitude is unchanged at 1. Study tip: For transformations of the form f(kx)f(kx), if k>1k > 1, you get horizontal compression by factor kk. If 0<k<10 < k < 1, you get horizontal stretch by factor 1k\frac{1}{k}. The key insight: changes inside the function parentheses affect horizontal transformations in the opposite way you might initially expect.

Question 19

Given that cos⁡θ=23\cos \theta = \frac{2}{3} and 3π2<θ<2π\frac{3\pi}{2} < \theta < 2\pi, find the exact value of sin⁡(2θ)\sin(2\theta).

  1. −459-\frac{4\sqrt{5}}{9} (correct answer)
  2. −259-\frac{2\sqrt{5}}{9}
  3. 253\frac{2\sqrt{5}}{3}
  4. 459\frac{4\sqrt{5}}{9}
Explanation: The angle θ\theta is in the fourth quadrant, where cos⁡θ>0\cos \theta > 0 and sin⁡θ<0\sin \theta < 0. We use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to find sin⁡θ\sin \theta. sin⁡2θ=1−(23)2=1−49=59\sin^2 \theta = 1 - (\frac{2}{3})^2 = 1 - \frac{4}{9} = \frac{5}{9}. Since θ\theta is in the fourth quadrant, sin⁡θ=−59=−53\sin \theta = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}. Now we use the double angle identity sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\theta. Substituting the values: sin⁡(2θ)=2(−53)(23)=−459\sin(2\theta) = 2(-\frac{\sqrt{5}}{3})(\frac{2}{3}) = -\frac{4\sqrt{5}}{9}.

Question 20

Find the maximum value of the function g(x)=1−6sin⁡xcos⁡xg(x) = 1 - 6\sin x \cos x.

  1. -5
  2. -2
  3. 4 (correct answer)
  4. 7
Explanation: We can simplify the function using the double angle identity sin⁡(2x)=2sin⁡xcos⁡x\sin(2x) = 2\sin x \cos x. We can write 6sin⁡xcos⁡x=3(2sin⁡xcos⁡x)=3sin⁡(2x)6\sin x \cos x = 3(2\sin x \cos x) = 3\sin(2x). So, the function becomes g(x)=1−3sin⁡(2x)g(x) = 1 - 3\sin(2x). The sine function, sin⁡(2x)\sin(2x), has a range of [−1,1][-1, 1]. The maximum value of g(x)g(x) will occur when sin⁡(2x)\sin(2x) is at its minimum value, which is -1. Substituting this value: Maximum g(x)=1−3(−1)=1+3=4g(x) = 1 - 3(-1) = 1 + 3 = 4.