Exercises: Trigonometric ratios of double angles

Given that $\sin(\alpha) = \frac{4}{5}$ and that $\alpha$ is an angle in the first quadrant, calculate the exact value of $\sin(2\alpha)$.

Given that $\cos(\alpha) = -\frac{1}{3}$ and that $\alpha$ belongs to the second quadrant, determine the exact value of $\cos(2\alpha)$.

If $\tan(\alpha) = \frac{3}{2}$, calculate the value of $\tan(2\alpha)$, expressing the result as an irreducible fraction.

Simplify the following trigonometric expression to a single exact numerical value:

$4\sin(15^\circ)\cos(15^\circ)$

Calculate the exact value of the following expression without using a calculator:

$\cos^2\left(\frac{\pi}{8}\right) - \sin^2\left(\frac{\pi}{8}\right)$

Write the following expression in terms of a single trigonometric ratio of the angle $x$, simplifying as much as possible:

$\frac{\sin(2x)}{2\sin(x)}$

Find the exact value of the following operation using double-angle identities:

$\frac{2\tan(22,5^\circ)}{1 - \tan^2(22,5^\circ)}$

Given that $\cos(x) = 0,6$, determine the value of $\cos(2x)$ using exclusively the identity relating the cosine of a double angle to the cosine of a single angle.

Marcos has to calculate the value of $\sin(2\alpha)$ knowing that $\alpha \in \left(0, \frac{\pi}{2}\right)$ and that $\sin\alpha = \frac{3}{5}$. On his exam, he writes the following solution:

"Since this is the double angle, the value of the sine simply doubles: $\sin(2\alpha) = 2 \cdot \sin\alpha = 2 \cdot \frac{3}{5} = \frac{6}{5} = 1,2$. However, since the sine of any real angle can never be greater than $1$, this problem has no solution and the proposed angle is impossible."

1. a) Identify the two mistakes made by Marcos (a conceptual one regarding the supposed linearity of trigonometric ratios and another in his final conclusion).
2. b) Correct the exercise by calculating the exact value of $\sin(2\alpha)$ using the double-angle identity and determine with reasoning in which quadrant $2\alpha$ lies by finding the sign of $\cos(2\alpha)$.

Consider an acute angle $\alpha \in \left(0, \frac{\pi}{2}\right)$ for which it is known that $\cos\alpha = 0,8$.

1. a) Before performing any algebraic calculation, estimate with reasoning the sign that $\cos(2\alpha)$ will have. To do this, compare $\cos\alpha = 0,8$ with the reference value $\cos(45^\circ) = \frac{\sqrt{2}}{2} \approx 0,707$, deduce whether $\alpha$ is greater or smaller than $45^\circ$, and reason in which quadrant the double angle $2\alpha$ must fall.
2. b) Select the variant of the formula for $\cos(2\alpha)$ that depends solely on the cosine to calculate its exact numerical value in decimal form. Check whether the sign obtained matches your previous estimate.

In the construction of an industrial warehouse, two adjacent roofs have been designed. The first roof has an inclination $\alpha$ to the horizontal such that $\tan(\alpha) = 0,4$. For water drainage reasons, the second roof has been designed with an angle of inclination that is exactly twice that of the first, that is, $2\alpha$. Calculate the exact slope (tangent) of the second roof.

In physics, the maximum range $R$ of a projectile launched over horizontal ground is modelled using the formula $R = \frac{v_0^2 \sin(2\theta)}{g}$, where $v_0$ is the initial velocity, $\theta$ is the launch angle, and $g \approx 10 \text{ m/s}^2$ is the acceleration due to gravity. If a test cannon fires a projectile with an initial velocity of $30 \text{ m/s}$ and it is desired to hit a target situated $45 \text{ m}$ away, find the possible launch angles $\theta$ in degrees.

Solve the following trigonometric equation, finding all solutions in the interval $[0, 2\pi)$:

$\sin(2x) + \sqrt{3}\cos(x) = 0$

A scissor mechanism for a hydraulic lift has a central pivot forming an opening angle $2\alpha$. The shear force on the central pin is given by an expression proportional to $F(\alpha) = \frac{1 - \cos(2\alpha)}{\sin(2\alpha)}$. Prove that this expression can be simplified to a single trigonometric ratio of the angle $\alpha$.

Solve the following trigonometric equation and give the solutions in degrees for the interval $[0^\circ, 360^\circ)$:

$\cos(2x) - 3\sin(x) = 2$

A ray of light is incident on a mirror system such that its final deviation angle is $2\theta$. If laboratory measurements indicate that $\cos(2\theta) = 0,28$ and we know that the angle of incidence $\theta$ is acute, determine the exact value of $\sin(\theta)$ and $\cos(\theta)$.

Prove the following trigonometric identity, transforming the left-hand side to obtain the right-hand side:

$\frac{\sin(2x)}{1 + \cos(2x)} = \tan(x)$

In rhombus $ABCD$ shown in the figure, the diagonals measure $AC = 16 \text{ cm}$ and $BD = 12 \text{ cm}$. Let $\alpha$ be half of the interior angle at vertex $A$. Calculate the exact value of $\sin(2\alpha)$ and $\cos(2\alpha)$ to determine the sine and cosine of the full angle $\widehat{A}$.

To solve the trigonometric equation $\sin(2x) - \sqrt{3}\cos(x) = 0$ in the interval $[0, 2\pi)$, two classmates propose different strategies:

• Alba proposes: "I divide both sides by $\cos(x)$ to get $\frac{\sin(2x)}{\cos(x)} - \sqrt{3} = 0$, substitute $\sin(2x) = 2\sin(x)\cos(x)$, simplify the factor $\cos(x)$, and solve directly for $\sin(x) = \frac{\sqrt{3}}{2}$."

• Bruno proposes: "I first substitute $\sin(2x) = 2\sin(x)\cos(x)$ into the equation, factor out $\cos(x)$ without dividing by any unknown, and set each factor equal to zero."

1. a) Critically analyze Alba's method: explain what algebraic violation she commits by dividing by $\cos(x)$ and which specific solutions in the interval $[0, 2\pi)$ are lost if her procedure is applied.
2. b) Develop Bruno's method fully and rigorously, and write all exact solutions of the equation in the interval $[0, 2\pi)$ expressed in radians.

At the High Performance Centre (CAR) of Sierra Nevada (Granada), a biomechanics team analyzes the horizontal range of shots on goal in soccer. According to the simplified physics model over flat ground, the horizontal range is given by the expression $d(\alpha) = \frac{v_0^2}{g}\sin(2\alpha)$, where $v_0$ is the initial striking velocity, $g \approx 9,8\text{ m/s}^2$, and $\alpha$ is the elevation angle with respect to the horizontal ($0^\circ < \alpha < 90^\circ$).

In a first test with a fixed velocity $v_0$, a soccer player strikes the ball at an angle $\alpha_1$ such that $\tan(\alpha_1) = \frac{3}{4}$.

1. a) Calculate the exact value of $\sin(2\alpha_1)$ to determine the range coefficient of this first shot.
2. b) If in a second shot with identical initial velocity the player doubles the elevation angle to $\alpha_2 = 2\alpha_1$, calculate the exact value of $\sin(2\alpha_2) = \sin(4\alpha_1)$ by successively applying the double-angle ratios.
3. c) Compare both ranges ($d(\alpha_1)$ versus $d(\alpha_2)$) and reason physically why doubling the initial inclination of the strike does not double the distance reached, but instead produces the opposite effect.

Prove the following trigonometric identity, valid for any angle $x$ where the expressions are defined:

$\frac{1 + \sin(2x) - \cos(2x)}{1 + \sin(2x) + \cos(2x)} = \tan(x)$

A planar robotic arm consists of two links, both of length $L$. The first link forms an angle $\alpha$ with the positive horizontal axis. The second link forms an angle $2\alpha$ to the horizontal. The $x$-coordinate of the end effector $P$ is given by the expression $x_P = L\cos(\alpha) + L\cos(2\alpha)$. Determine the exact value of the angle $\alpha \in [0, \pi/2]$ for which the end effector is situated exactly on the y-axis ($x_P = 0$).

Solve the following trigonometric equation, determining all solutions in the interval $[0, 2\pi)$:

$\tan(2x) \cdot \tan(x) = 1$

Find the values of the parameter $k \in \mathbb{R}$ for which the equation $\cos(2x) - 4\sin(x) + k = 0$ has exactly one solution in the interval $\left[0, \frac{\pi}{2}\right]$.

First prove the identity $\cos^4(x) - \sin^4(x) = \cos(2x)$. Then use this result to solve the equation $\cos^4(x) - \sin^4(x) = \frac{1}{2}$ in the interval $[0^\circ, 360^\circ)$.

In a circle with center $O$ and radius $R$, the central angle $\widehat{AOB}$ measures $2\alpha$ and subtends the chord $AB$. A point $P$ located at the midpoint of the major arc subtends an inscribed angle $\widehat{APB}$ measuring $\alpha$. Knowing that $\cos(\alpha) = \frac{4}{5}$, calculate the ratio (the quotient) between the area of the central triangle $\triangle OAB$ and the area of the inscribed triangle $\triangle PAB$.

Solve the following system of trigonometric equations for $x, y \in [0, \pi/2]$:

$\begin{cases} \sin(x)\cos(y) + \cos(x)\sin(y) = 1 \\ \cos(2x) - \cos(2y) = 0 \end{cases}$

A company's revenue function from the sale of a seasonal product is given by the expression $I(x) = 5\sin(x)\cos(x) - 2\sin^2(x) + 1$, where $x$ is the elapsed time in months ($x \ge 0$). Express $I(x)$ in terms of $\sin(2x)$ and $\cos(2x)$, and determine the maximum revenue the company can achieve.

The goal is to design and reconstruct trigonometric equations from their geometric conditions and their solutions.

1. a) It is known that a certain angle $\alpha$ belonging to the second quadrant $\left(\frac{\pi}{2} < \alpha < \pi\right)$ satisfies $\cos(2\alpha) = -\frac{7}{25}$. Uniquely deduce the exact values of $\cos\alpha$, $\sin\alpha$, and $\tan\alpha$, justifying the choice of sign at each step based on the quadrant of origin.
2. b) Construct a trigonometric equation of the form $\cos(2x) + b\cos(x) + c = 0$ (with $b, c \in \mathbb{R}$) whose only solutions in the closed interval $[0, \pi]$ are exactly $x = \frac{\pi}{3}$ and $x = \pi$. Show step by step how you determine the coefficients $b$ and $c$ by first setting up a quadratic equation in the auxiliary variable $u = \cos(x)$, and check that no extraneous solution is introduced in that interval.

Consider the function $f(x) = \sin(2x) - 2\sin(x)$ defined on the interval $[0, 2\pi]$.

1. a) Apply the double-angle formula to express $f(x)$ as a product of the form $2\sin(x)(\cos(x) - 1)$ and determine exactly all the points in the interval $[0, 2\pi]$ where the graph intersects the horizontal axis ($f(x) = 0$).
2. b) Perform an analytical study of the sign of each factor in the different quadrants and prove, without resorting to derivatives or computer graphs, that $f(x) \le 0$ for all $x \in [0, \pi]$ and that $f(x) \ge 0$ for all $x \in [\pi, 2\pi]$. Can $f(x)$ take strictly positive values in the first quadrant? Justify your answer.
3. c) A student states: "Given that the minimum value of $\sin(2x)$ is $-1$ and the minimum value of $-2\sin(x)$ is $-2$, the absolute minimum value that $f(x)$ can reach is $-1 + (-2) = -3$." Refute this reasoning by explaining why both terms cannot be optimized independently, and verify that for $x = \frac{2\pi}{3}$ the function reaches a value lower than $-2,5$ by calculating its exact value.