Double Angle & Trig Graphs - IB Mathematics Analysis and Approaches

If $\cos(2x) = \frac{7}{25}$ , and $x$ is an acute angle, what is the value of $\cos x$ ?
Answer: $\frac{4}{5}$ This question tests your understanding of the double angle formula for cosine and how to work backwards from $\cos(2x)$ to find $\cos x$ . The double angle formula states that $\cos(2x) = 2\cos^2 x - 1$ . Since we know $\cos(2x) = \frac{7}{25}$ , we can substitute and solve: $\frac{7}{25} = 2\cos^2 x - 1$ Adding 1 to both sides: $\frac{7}{25} + 1 = 2\cos^2 x$ Converting to a common denominator: $\frac{7}{25} + \frac{25}{25} = \frac{32}{25} = 2\cos^2 x$ Dividing by 2: $\cos^2 x = \frac{16}{25}$ Taking the square root: $\cos x = \pm\frac{4}{5}$ However, since $x$ is an acute angle (between 0° and 90°), cosine must be positive. Therefore, $\cos x = \frac{4}{5}$ . Let's examine why the other answers are incorrect: A) $\frac{3}{5}$ would give $\cos(2x) = 2(\frac{3}{5})^2 - 1 = \frac{18}{25} - 1 = -\frac{7}{25}$ , which has the wrong sign. B) $\pm\frac{4}{5}$ is mathematically correct before considering the constraint, but ignores that $x$ is acute, making cosine positive. C) $\frac{16}{25}$ is the value of $\cos^2 x$ , not $\cos x$ 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.
Let $\sin x = p$ . Which of the following is an expression for $\cos(2x)$ ?
Answer: $1 - 2p^2$ There are three main identities for $\cos(2x)$ : $\cos(2x) = \cos^2 x - \sin^2 x$ , $\cos(2x) = 2\cos^2 x - 1$ , and $\cos(2x) = 1 - 2\sin^2 x$ . Since we are given that $\sin x = p$ , the most direct identity to use is $\cos(2x) = 1 - 2\sin^2 x$ . Substituting $\sin x = p$ into this identity gives $\cos(2x) = 1 - 2p^2$ .
The graph of the function $f(x) = \cos(2x)$ is a horizontal compression of the graph of $g(x) = \cos x$ by a factor of 2. Which of the following functions has a graph which is a vertical translation of the graph of $f(x)$ ?
Answer: $h(x) = 2\cos^2 x$ A vertical translation of $f(x) = \cos(2x)$ would be of the form $y = \cos(2x) + k$ for some non-zero constant $k$ . We need to check which option can be written in this form. Using the double angle identity $\cos(2x) = 2\cos^2 x - 1$ , we can rearrange it to get $2\cos^2 x = \cos(2x) + 1$ . So, the function $h(x) = 2\cos^2 x$ is equivalent to $h(x) = \cos(2x) + 1$ , which is a vertical translation of $f(x)$ by 1 unit upwards. Choice A is $-\cos(2x)$ . Choice C is $\cos(2x)$ , which is not a translation. Choice B is a vertical stretch of $g(x)$ .
The principal axis of the graph of $y = 4\cos^2(x) - 5$ is:
Answer: $y = -3$ 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: $\cos^2(x) = \frac{1 + \cos(2x)}{2}$ . Substituting this into the equation gives $y = 4\left(\frac{1 + \cos(2x)}{2}\right) - 5$ . Simplifying, we get $y = 2(1 + \cos(2x)) - 5 = 2 + 2\cos(2x) - 5 = 2\cos(2x) - 3$ . This is a cosine function with amplitude 2, vertically shifted down by 3 units. The principal axis is the line $y = -3$ .
Find the maximum value of the function $g(x) = 1 - 6\sin x \cos x$ .
Answer: 4 We can simplify the function using the double angle identity $\sin(2x) = 2\sin x \cos x$ . We can write $6\sin x \cos x = 3(2\sin x \cos x) = 3\sin(2x)$ . So, the function becomes $g(x) = 1 - 3\sin(2x)$ . The sine function, $\sin(2x)$ , has a range of $[-1, 1]$ . The maximum value of $g(x)$ will occur when $\sin(2x)$ is at its minimum value, which is -1. Substituting this value: Maximum $g(x) = 1 - 3(-1) = 1 + 3 = 4$ .
Given that $\cos \theta = \frac{2}{3}$ and $\frac{3\pi}{2} < \theta < 2\pi$ , find the exact value of $\sin(2\theta)$ .
Answer: $-\frac{4\sqrt{5}}{9}$ The angle $\theta$ is in the fourth quadrant, where $\cos \theta > 0$ and $\sin \theta < 0$ . We use the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ to find $\sin \theta$ . $\sin^2 \theta = 1 - (\frac{2}{3})^2 = 1 - \frac{4}{9} = \frac{5}{9}$ . Since $\theta$ is in the fourth quadrant, $\sin \theta = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}$ . Now we use the double angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$ . Substituting the values: $\sin(2\theta) = 2(-\frac{\sqrt{5}}{3})(\frac{2}{3}) = -\frac{4\sqrt{5}}{9}$ .
The expression $\frac{\sin(2A)}{1-\cos(2A)}$ is equivalent to:
Answer: $\cot A$ We use the double angle identities for sine and cosine. $\sin(2A) = 2\sin A \cos A$ and $\cos(2A) = 1 - 2\sin^2 A$ . Substitute these into the expression: $\frac{2\sin A \cos A}{1-(1-2\sin^2 A)} = \frac{2\sin A \cos A}{1-1+2\sin^2 A} = \frac{2\sin A \cos A}{2\sin^2 A}$ . Assuming $\sin A \neq 0$ , we can cancel $2\sin A$ from the numerator and denominator, leaving $\frac{\cos A}{\sin A}$ , which is the definition of $\cot A$ .
The function $f(x) = \sin(x)\cos(x)$ is transformed to the function $g(x) = \sin(2x)$ . What transformation maps the graph of $f$ onto the graph of $g$ ?
Answer: A vertical stretch by a factor of 2. First, we express $f(x)$ using the double angle identity $\sin(2x) = 2\sin x \cos x$ . Rearranging this gives $\sin x \cos x = \frac{1}{2}\sin(2x)$ . So, $f(x) = \frac{1}{2}\sin(2x)$ . We are given $g(x) = \sin(2x)$ . To get from $f(x)$ to $g(x)$ , we must multiply $f(x)$ by 2: $g(x) = 2 \cdot f(x)$ . A transformation of the form $y = kf(x)$ is a vertical stretch by a factor of $k$ . Here, $k=2$ , so the transformation is a vertical stretch by a factor of 2.
How many solutions does the equation $\sin(2x) = 2\sin x$ have in the interval $[0, 2\pi]$ ?
Answer: 3 First, use the double angle identity $\sin(2x) = 2\sin x \cos x$ . The equation becomes $2\sin x \cos x = 2\sin x$ . It is a common mistake to divide by $2\sin x$ ; instead, we should rearrange and factor. $2\sin x \cos x - 2\sin x = 0$ , which gives $2\sin x (\cos x - 1) = 0$ . This equation is satisfied if $\sin x = 0$ or $\cos x - 1 = 0$ (i.e., $\cos x = 1$ ). In the interval $[0, 2\pi]$ , $\sin x = 0$ has solutions $x = 0, \pi, 2\pi$ . In the interval $[0, 2\pi]$ , $\cos x = 1$ has solutions $x = 0, 2\pi$ . The set of distinct solutions is $\{0, \pi, 2\pi\}$ . There are 3 solutions.
The graph of $y=\sin(x)$ is transformed to the graph of $y=\sin(2x)$ . Which of the following describes this transformation?
Answer: A horizontal compression by a factor of 2. 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 $y = \sin(x)$ to $y = \sin(2x)$ , examine what happens to the period. The original sine function has period $2\pi$ , meaning it completes one full cycle every $2\pi$ units. For $y = \sin(2x)$ , when $x = \pi$ , we get $\sin(2\pi) = 0$ , completing one full cycle in just $\pi$ units. The new period is $\pi$ , 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 $f(kx)$ , if $k > 1$ , you get horizontal compression by factor $k$ . If $0 < k < 1$ , you get horizontal stretch by factor $\frac{1}{k}$ . The key insight: changes inside the function parentheses affect horizontal transformations in the opposite way you might initially expect.
The graph of $y = a \cos(2x) + d$ passes through the points $( \frac{\pi}{2}, -5 )$ and $(\pi, 3)$ . Find the value of $a$ .
Answer: 4 We substitute the coordinates of the two points into the equation to create a system of linear equations for $a$ and $d$ . For $( \frac{\pi}{2}, -5 )$ : $-5 = a \cos(2 \cdot \frac{\pi}{2}) + d = a \cos(\pi) + d = a(-1) + d = -a + d$ . This gives the equation $d - a = -5$ . For $(\pi, 3)$ : $3 = a \cos(2 \cdot \pi) + d = a(1) + d = a + d$ . This gives the equation $d + a = 3$ . We now have a system of two equations: 1) $d - a = -5$ and 2) $d + a = 3$ . Adding the two equations gives $(d - a) + (d + a) = -5 + 3$ , which simplifies to $2d = -2$ , so $d = -1$ . Substituting $d = -1$ into the second equation gives $-1 + a = 3$ , which means $a = 4$ .
Which of the following functions has the same graph as $f(x) = 1 - 2\sin^2(\frac{x}{2})$ ?
Answer: $g(x) = \cos(x)$ 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 $f(x) = 1 - 2\sin^2(\frac{x}{2})$ , recall the cosine double angle identity: $\cos(2\theta) = 1 - 2\sin^2(\theta)$ . This identity is derived from $\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$ combined with the Pythagorean identity $\cos^2(\theta) + \sin^2(\theta) = 1$ . In our function, if we let $\theta = \frac{x}{2}$ , then $2\theta = x$ . Applying the identity: $f(x) = 1 - 2\sin^2(\frac{x}{2}) = \cos(x)$ . Therefore, choice A is correct. Let's examine why the other options don't work. Choice B, $g(x) = \cos(\frac{x}{2})$ , would result from misidentifying which angle to use in the identity. Choice C, $g(x) = \cos(2x)$ , represents the opposite error—doubling the angle instead of recognizing that $2 \cdot \frac{x}{2} = x$ . Choice D, $g(x) = \sin(x)$ , 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: $\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = 2\cos^2(\theta) - 1 = 1 - 2\sin^2(\theta)$ . The third form is particularly useful for simplifying expressions with $1 - 2\sin^2$ patterns.
What is the period of the function $f(x) = \sin^2(3x)$ ?
Answer: $\frac{\pi}{3}$ To find the period of $f(x) = \sin^2(3x)$ , we use the power-reducing identity which is derived from a double angle identity: $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$ . Let $\theta = 3x$ . Then $f(x) = \frac{1 - \cos(2 \cdot 3x)}{2} = \frac{1}{2} - \frac{1}{2}\cos(6x)$ . The period of a function of the form $y = a \cos(bx) + d$ is $\frac{2\pi}{|b|}$ . In this case, $b=6$ , so the period is $\frac{2\pi}{6} = \frac{\pi}{3}$ .
The line $y=k$ intersects the graph of $y = \sin(2x)$ exactly twice in the interval $0 \le x \le \pi$ . Which of the following is a possible value for $k$ ?
Answer: 0.5 The graph of $y = \sin(2x)$ has a period of $\frac{2\pi}{2} = \pi$ . In the interval $0 \le x \le \pi$ , this graph completes one full cycle. It starts at $y=0$ , increases to a maximum of $y=1$ at $x=\pi/4$ , decreases through $y=0$ at $x=\pi/2$ , reaches a minimum of $y=-1$ at $x=3\pi/4$ , and returns to $y=0$ at $x=\pi$ . For the line $y=k$ to intersect the graph exactly twice, $k$ must be between -1 and 1, but not equal to 0, 1, or -1. If $k=1$ or $k=-1$ , there is one intersection (at the peak or trough). If $k=0$ , there are three intersections (at $x=0, \pi/2, \pi$ ). If $-1 < k < 0$ or $0 < k < 1$ , there will be exactly two intersections. Of the choices given, only $k=0.5$ falls into this range.
Consider the function $f(x) = 1 - \cos(2x)$ . What are the amplitude and period of the graph of $y=f(x)$ ?
Answer: Amplitude is 1, Period is $\pi$ The function can be written as $f(x) = -\cos(2x) + 1$ . In the standard form $y=a\cos(b(x-c))+d$ , we have $a=-1$ , $b=2$ , and $d=1$ . The amplitude is $|a| = |-1| = 1$ . The period is $\frac{2\pi}{|b|} = \frac{2\pi}{2} = \pi$ . Alternatively, using the identity $\cos(2x) = 1 - 2\sin^2 x$ , the function is $f(x) = 1 - (1 - 2\sin^2 x) = 2\sin^2 x$ . 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 $y=1$ . The amplitude is the distance from the principal axis to the maximum, which is $2-1=1$ . The period of $\sin^2 x$ is $\pi$ , and multiplying by 2 vertically does not change the period.
The function $f(x) = \frac{\sin(4x)}{\sin(2x)}$ is defined for $\sin(2x) \neq 0$ . Which of the following expressions is equivalent to $f(x)$ ?
Answer: $2\cos(2x)$ We can rewrite the numerator using the double angle identity for sine, $\sin(2\theta) = 2\sin\theta\cos\theta$ . In this case, let $\theta = 2x$ . Then $\sin(4x) = \sin(2(2x)) = 2\sin(2x)\cos(2x)$ . Substituting this into the function's expression gives $f(x) = \frac{2\sin(2x)\cos(2x)}{\sin(2x)}$ . Since the domain specifies that $\sin(2x) \neq 0$ , we can cancel the $\sin(2x)$ terms. This simplifies the expression to $f(x) = 2\cos(2x)$ .
The function $f(x) = (\sin x + \cos x)^2$ can be written in the form $y = a \sin(2x) + b$ . What is the period of $f(x)$ ?
Answer: $\pi$ First, expand the expression for $f(x)$ : $f(x) = (\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x$ . Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ and the double angle identity $2\sin x \cos x = \sin(2x)$ , we can rewrite the function as $f(x) = 1 + \sin(2x)$ . The period of a function of the form $y = a \sin(b(x-c)) + d$ is given by $\frac{2\pi}{|b|}$ . For $f(x) = 1 + \sin(2x)$ , we have $b=2$ . Therefore, the period is $\frac{2\pi}{2} = \pi$ .
The equation $\cos(2x) - \sin x = 0$ has two solutions in the interval $0 \le x \le \pi$ . Find the sum of these two solutions.
Answer: $\pi$ To solve the equation, we use the double angle identity $\cos(2x) = 1 - 2\sin^2 x$ to express the equation in terms of $\sin x$ only. The equation becomes $1 - 2\sin^2 x - \sin x = 0$ , which can be rearranged to $2\sin^2 x + \sin x - 1 = 0$ . Let $u = \sin x$ . We have the quadratic equation $2u^2 + u - 1 = 0$ , which factors as $(2u-1)(u+1) = 0$ . So, $u = \frac{1}{2}$ or $u = -1$ . This means $\sin x = \frac{1}{2}$ or $\sin x = -1$ . For $\sin x = \frac{1}{2}$ in the interval $0 \le x \le \pi$ , the solutions are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ . For $\sin x = -1$ , there are no solutions in the interval $0 \le x \le \pi$ . The two solutions are $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ . Their sum is $\frac{\pi}{6} + \frac{5\pi}{6} = \frac{6\pi}{6} = \pi$ .

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