MHF4U – Grade 12 Advanced Functions Trigonometry Worksheet Questions + Explanations Bundle

Problem 1: Compound Angle Identity

Prove the identity:

\(tan(x+y) = \frac{tan(x) + tan(y)}{1 – tan(x)tan(y)}\)

Proof:

We can start with the sum formulas for sine and cosine:

\(sin(x+y) = sin(x)cos(y) + cos(x)sin(y)\) \(cos(x+y) = cos(x)cos(y) – sin(x)sin(y)\)

Then, we can divide the sine formula by the cosine formula:

\(tan(x+y) = \frac{sin(x)cos(y) + cos(x)sin(y)}{cos(x)cos(y) – sin(x)sin(y)}\)

Now, divide both the numerator and denominator by \(cos(x)cos(y)\):

\(tan(x+y) = \frac{\frac{sin(x)}{cos(x)} + \frac{sin(y)}{cos(y)}}{1 – \frac{sin(x)}{cos(x)}\frac{sin(y)}{cos(y)}}\)

Simplify to get the desired result:

\(tan(x+y) = \frac{tan(x) + tan(y)}{1 – tan(x)tan(y)}\)

Problem 2: Double Angle Identities

Given:

• \(sin(x) = \frac{3}{5}\)
• x is in the first quadrant

Find:

• \(sin(2x)\)
• \(cos(2x)\)
• \(tan(2x)\)

Solution:

Using the Pythagorean identity, we find:

\(cos(x) = \frac{4}{5}\)

Now, we can use the double angle formulas:

\(sin(2x) = 2sin(x)cos(x) = 2(\frac{3}{5})(\frac{4}{5}) = \frac{24}{25}\) \(cos(2x) = cos^2(x) – sin^2(x) = (\frac{4}{5})^2 – (\frac{3}{5})^2 = \frac{7}{25}\) \(tan(2x) = \frac{sin(2x)}{cos(2x)} = \frac{\frac{24}{25}}{\frac{7}{25}} = \frac{24}{7}\)

Problem 3: Half-Angle Identities

Given:

• \(cos(x) = -\frac{1}{3}\)
• x is in the second quadrant

Find:

• \(sin(\frac{x}{2})\)
• \(cos(\frac{x}{2})\)
• \(tan(\frac{x}{2})\)

Solution:

Since x is in the second quadrant, \(\frac{x}{2}\) is in the first quadrant. Therefore, all trigonometric functions of \(\frac{x}{2}\) will be positive.

First, we find \(sin(x)\) using the Pythagorean identity:

\(sin^2(x) + cos^2(x) = 1\) \(sin^2(x) + \left(-\frac{1}{3}\right)^2 = 1\) \(sin^2(x) = \frac{8}{9}\) \(sin(x) = \frac{2\sqrt{2}}{3}\) (since x is in the second quadrant, sin(x) is positive)

Now, we can use the half-angle formulas:

\(sin(\frac{x}{2}) = \pm \sqrt{\frac{1 – cos(x)}{2}} = \sqrt{\frac{1 – (-\frac{1}{3})}{2}} = \sqrt{\frac{2}{3}}\) \(cos(\frac{x}{2}) = \pm \sqrt{\frac{1 + cos(x)}{2}} = \sqrt{\frac{1 + (-\frac{1}{3})}{2}} = \sqrt{\frac{1}{3}}\) \(tan(\frac{x}{2}) = \frac{sin(\frac{x}{2})}{cos(\frac{x}{2})} = \frac{\sqrt{\frac{2}{3}}}{\sqrt{\frac{1}{3}}} = \sqrt{2}\)

Problem 4: Product-to-Sum Identities

Express \(sin(3x)cos(2x)\) as a sum or difference of sines or cosines.

We can use the product-to-sum formula:

\(sin(A)cos(B) = \frac{1}{2}[sin(A+B) + sin(A-B)]\)

Applying this to our problem, we get:

\(sin(3x)cos(2x) = \frac{1}{2}[sin(3x+2x) + sin(3x-2x)]\) \(= \frac{1}{2}[sin(5x) + sin(x)]\)

Problem 5: Sum-to-Product Identities

Express \(sin(5x) + sin(3x)\) as a product of sines or cosines.

We can use the sum-to-product formula:

\(sin(A) + sin(B) = 2sin(\frac{A+B}{2})cos(\frac{A-B}{2})\)

Applying this to our problem, we get:

\(sin(5x) + sin(3x) = 2sin(\frac{5x+3x}{2})cos(\frac{5x-3x}{2})\) \(= 2sin(4x)cos(x)\)

Problem 6: Trigonometric Equations

Solve the equation:

\(2sin^2(x) – 3sin(x) + 1 = 0\)

for \(0 ≤ x ≤ 2π\)

Let \(u = sin(x)\). Then the equation becomes:

\(2u^2 – 3u + 1 = 0\)

Factoring this quadratic equation, we get:

\((2u – 1)(u – 1) = 0\)

So, either:

• \(2u – 1 = 0\) => \(u = \frac{1}{2}\) => \(sin(x) = \frac{1}{2}\)
• \(u – 1 = 0\) => \(u = 1\) => \(sin(x) = 1\)

For \(sin(x) = \frac{1}{2}\), the solutions in the given interval are:

• \(x = \frac{\pi}{6}\)
• \(x = \frac{5\pi}{6}\)

For \(sin(x) = 1\), the solution in the given interval is:

• \(x = \frac{\pi}{2}\)

Therefore, the solutions to the given equation are:

\(x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\)

Problem 7: Trigonometric Graphs

Sketch the graph of the function:

\(y = 2sin(3x – \frac{\pi}{2}) + 1\)

over one complete period.

Solution:

To sketch the graph, we can analyze the function’s properties:

• Amplitude: 2
• Period: \(\frac{2\pi}{3}\)
• Phase Shift: \(\frac{\pi}{6}\) to the right
• Vertical Shift: 1 unit up

The graph will oscillate between 1-2 and 1+2, with a period of \(\frac{2\pi}{3}\). It will be shifted \(\frac{\pi}{6}\) units to the right.

Opens in a new window www.youtube.com

graph of y = 2sin(3x π/2) + 1

Problem 8: Inverse Trigonometric Functions

Find the exact value of:

\(arcsin(sin(\frac{5\pi}{6}))\)

Solution:

First, we find the value of \(sin(\frac{5\pi}{6})\):

\(sin(\frac{5\pi}{6}) = \frac{1}{2}\)

So, we need to find the angle whose sine is \(\frac{1}{2}\). The principal value of \(arcsin(\frac{1}{2})\) is \(\frac{\pi}{6}\).

However, we need to consider the range of the arcsine function, which is \([-\frac{\pi}{2}, \frac{\pi}{2}]\). Since \(\frac{5\pi}{6}\) is in the second quadrant, its sine value is positive. To find the angle in the first quadrant with the same sine value, we can use the reference angle:

\(\pi – \frac{5\pi}{6} = \frac{\pi}{6}\)

Therefore,

\(arcsin(sin(\frac{5\pi}{6})) = \frac{\pi}{6}\)

Problem 9: Applications of Trigonometry

Problem: A 10-meter ladder leans against a wall. If the angle between the ladder and the ground is 60 degrees, how high up the wall does the ladder reach?

Solution:

Let’s denote the height the ladder reaches on the wall as h. We can use the sine function to relate the angle, the hypotenuse (ladder length), and the opposite side (height):

\(sin(60°) = \frac{h}{10}\)

We know that \(sin(60°) = \frac{\sqrt{3}}{2}\). So,

\(\frac{\sqrt{3}}{2} = \frac{h}{10}\)

Solving for h, we get:

\(h = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3}\)

Therefore, the ladder reaches \(5\sqrt{3}\) meters up the wall.

Problem 10: Proofs by Mathematical Induction

Prove that:

\(sin(x) + sin(3x) + … + sin((2n-1)x) = \frac{sin^2(nx)}{sin(x)}\)

for all positive integers n.

Proof (By Mathematical Induction):

Base Case (n = 1): For n = 1, the left-hand side (LHS) becomes:

\(sin(x)\)

The right-hand side (RHS) becomes:

\(\frac{sin^2(x)}{sin(x)} = sin(x)\)

So, the base case holds.

Inductive Step: Assume the formula holds for n = k:

\(sin(x) + sin(3x) + … + sin((2k-1)x) = \frac{sin^2(kx)}{sin(x)}\)

We want to prove that it also holds for n = k+1:

\(sin(x) + sin(3x) + … + sin((2k-1)x) + sin((2(k+1)-1)x) = \frac{sin^2((k+1)x)}{sin(x)}\)

Using the inductive hypothesis, we can rewrite the left-hand side:

\(\frac{sin^2(kx)}{sin(x)} + sin((2k+1)x)\)

Now, we need to manipulate this expression to get the desired form. We can use the product-to-sum formula:

\(sin(A) + sin(B) = 2sin(\frac{A+B}{2})cos(\frac{A-B}{2})\)

Applying this to the last two terms, we get:

\(\frac{sin^2(kx)}{sin(x)} + 2sin(kx+x)cos(kx-x)\) \(= \frac{sin^2(kx)}{sin(x)} + 2sin(kx)cos(kx)\)

Factor out \(sin(kx)\):

\(= \frac{sin(kx)[sin(kx) + 2cos(kx)sin(x)]}{sin(x)}\)

Now, use the double-angle formula for sine:

\(sin(2A) = 2sin(A)cos(A)\)

Applying this to the numerator, we get:

\(= \frac{sin(kx)[sin(kx) + sin(2x)]}{sin(x)}\)

Using the sum-to-product formula again, we can combine the terms in the numerator:

\(= \frac{sin(kx)[2sin(\frac{kx+2x}{2})cos(\frac{kx-2x}{2})]}{sin(x)}\) \(= \frac{2sin(kx)sin((k+1)x)cos((k-1)x)}{sin(x)}\)

Using the double-angle formula for sine again:

\(= \frac{sin^2((k+1)x)}{sin(x)}\)

Thus, the formula holds for n = k+1.

2nd Section (Solutions below)

1. Express each of the following as a single trigonometric ratio.

a) \(sin(x)cos(4x) + cos(x)sin(4x)\)

b) \(cos(3x)cos(x) + sin(3x)sin(x)\)

c) \(\frac{tan(2x) + tan(3x)}{1 – tan(2x)tan(3x)}\)

d) \(2sin(\frac{x}{2})cos(\frac{x}{2})\)

e) \(cos^2(4x) – sin^2(4x)\)

f) \(\frac{1 – tan^2(\frac{x}{2})}{1 + tan^2(\frac{x}{2})}\)

g) \(2 – 4sin^2(\frac{x}{12})\)

h) \(\frac{2tan(\frac{x}{2})}{1 – tan^2(\frac{x}{2})}\)

2. Rewrite each expression as a single ratio and then evaluate the ratio.

a) \(\frac{1 – sin^2(\frac{\pi}{2})}{1 – cos^2(\frac{\pi}{2})}\)

b) \(2cos^2(\frac{\pi}{12}) – 1\)

c) \(\frac{2tan(\frac{\pi}{8})}{1 – tan^2(\frac{\pi}{8})}\)

3. Use an appropriate compound angle formula to express as a single trigonometric function, and then determine an exact value for each.  

a) \(sin(\frac{\pi}{4})cos(\frac{\pi}{12}) – cos(\frac{\pi}{4})sin(\frac{\pi}{12})\)

b) \(sin(\frac{3\pi}{5})cos(\frac{\pi}{15}) + cos(\frac{3\pi}{5})sin(\frac{\pi}{15})\)

c) \(cos(\frac{10\pi}{9})cos(\frac{5\pi}{18}) + sin(\frac{10\pi}{9})sin(\frac{5\pi}{18})\)

d) \(cos(\frac{\pi}{4})cos(\frac{\pi}{12}) – sin(\frac{\pi}{4})sin(\frac{\pi}{12})\)

4. Use an appropriate compound angle formula to determine an exact value for each.

a) \(sin(\frac{17\pi}{12})\)

b) \(cos(\frac{13\pi}{12})\)

c) \(tan(\frac{5\pi}{12})\)

5. If \(sin(\alpha) = \frac{3}{7}\) and \(cos(\beta) = \frac{3}{5}\) where \(0 \leq \alpha \leq \frac{\pi}{2}\) and \(0 \leq \beta \leq \frac{\pi}{2}\), determine:

a) \(sin(\alpha + \beta)\)

b) \(cos(2\alpha)\)

6. If \(sin(x) = \frac{12}{13}\) and \(\frac{\pi}{2} \leq x \leq \pi\), find \(sin(2x)\) and \(tan(2x)\).

7. Angle θ lies in the second quadrant and \(cos(\theta) = -\frac{3}{5}\).

a) Determine the exact value for \(sin(2\theta)\), \(cos(2\theta)\), and \(tan(2\theta)\).

b) Determine an approximate measure for θ in radians. Round to the nearest hundredth of a radian.

8. If angle θ is an acute angle such that \(cos(\theta) = \frac{2}{3}\), determine the exact value of \(cos(4\theta)\).

Problem 1: Expressing as a Single Trigonometric Ratio

Part a)

\(sin(x)cos(4x) + cos(x)sin(4x)\) This expression directly matches the sine of a sum formula:

\(sin(A+B) = sin(A)cos(B) + cos(A)sin(B)\)

Therefore, the expression simplifies to:

\(sin(x + 4x) = sin(5x)\)

Part b)

\(cos(3x)cos(x) + sin(3x)sin(x)\)

This expression matches the cosine of a difference formula:

\(cos(A-B) = cos(A)cos(B) + sin(A)sin(B)\)

Therefore, the expression simplifies to:

\(cos(3x – x) = cos(2x)\)

Part c)

\(\frac{tan(2x) + tan(3x)}{1 – tan(2x)tan(3x)}\)

This expression matches the tangent of a sum formula:

\(tan(A+B) = \frac{tan(A) + tan(B)}{1 – tan(A)tan(B)}\)

Therefore, the expression simplifies to:

\(tan(2x + 3x) = tan(5x)\)

Part d)

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

This expression matches the double-angle formula for sine:

\(sin(2A) = 2sin(A)cos(A)\)

Therefore, the expression simplifies to:

\(sin(2 * \frac{x}{2}) = sin(x)\)

Part e)

\(cos^2(4x) – sin^2(4x)\)

This expression matches the double-angle formula for cosine:

\(cos(2A) = cos^2(A) – sin^2(A)\)

Therefore, the expression simplifies to:

\(cos(2 * 4x) = cos(8x)\)

Part f)

\(\frac{1 – tan^2(\frac{x}{2})}{1 + tan^2(\frac{x}{2})}\)

This expression matches the formula for cosine of a double angle in terms of tangent:

\(cos(2A) = \frac{1 – tan^2(A)}{1 + tan^2(A)}\)

Therefore, the expression simplifies to:

\(cos(2 * \frac{x}{2}) = cos(x)\)

Part g)

\(2 – 4sin^2(\frac{x}{12})\)

We can factor out a 2:

\(2(1 – 2sin^2(\frac{x}{12}))\)

This expression matches the double-angle formula for cosine in terms of sine:

\(cos(2A) = 1 – 2sin^2(A)\)

Therefore, the expression simplifies to:

\(2cos(2 * \frac{x}{12}) = 2cos(\frac{x}{6})\)

Part h)

\(\frac{2tan(\frac{x}{2})}{1 – tan^2(\frac{x}{2})}\)

This expression matches the formula for tangent of a double angle in terms of tangent:

\(tan(2A) = \frac{2tan(A)}{1 – tan^2(A)}\)

Therefore, the expression simplifies to:

\(tan(2 * \frac{x}{2}) = tan(x)\)

Problem 2: Rewrite and Evaluate

Part a)

\(\frac{1 – sin^2(\frac{\pi}{2})}{1 – cos^2(\frac{\pi}{2})}\)

First, we can simplify the numerator and denominator using the Pythagorean identity:

\(sin^2(x) + cos^2(x) = 1\)

So, the expression becomes:

\(\frac{cos^2(\frac{\pi}{2})}{sin^2(\frac{\pi}{2})}\)

Evaluating the trigonometric functions at \(\frac{\pi}{2}\):

\(\frac{0^2}{1^2} = 0\)

Therefore, the expression evaluates to 0.

Part b)

\(2cos^2(\frac{\pi}{12}) – 1\)

This expression matches the double-angle formula for cosine:

\(cos(2A) = 2cos^2(A) – 1\)

So, we can rewrite it as:

\(cos(2 * \frac{\pi}{12}) = cos(\frac{\pi}{6})\)

Evaluating \(cos(\frac{\pi}{6})\), we get:

\(\frac{\sqrt{3}}{2}\)

Therefore, the expression evaluates to \(\frac{\sqrt{3}}{2}\).

Note: To evaluate trigonometric functions of angles like \(\frac{\pi}{12}\), you might need to use half-angle formulas or other trigonometric identities. However, for this specific problem, we could directly evaluate it using a calculator or trigonometric tables.

Problem 3: Compound Angle Formula and Exact Values

Part a)

\(sin(\frac{\pi}{4})cos(\frac{\pi}{12}) – cos(\frac{\pi}{4})sin(\frac{\pi}{12})\)

This expression matches the sine of a difference formula:

\(sin(A – B) = sin(A)cos(B) – cos(A)sin(B)\)

Therefore, the expression simplifies to:

\(sin(\frac{\pi}{4} – \frac{\pi}{12}) = sin(\frac{\pi}{6})\)

Evaluating \(sin(\frac{\pi}{6})\), we get:

\(\frac{1}{2}\)

So, the exact value of the expression is \(\frac{1}{2}\).

Problem 3 (Continued)

Part b)

\(sin(\frac{3\pi}{5})cos(\frac{\pi}{15}) + cos(\frac{3\pi}{5})sin(\frac{\pi}{15})\)

This expression matches the sine of a sum formula:

\(sin(A + B) = sin(A)cos(B) + cos(A)sin(B)\)

Therefore, the expression simplifies to:

\(sin(\frac{3\pi}{5} + \frac{\pi}{15}) = sin(\frac{10\pi}{15}) = sin(\frac{2\pi}{3})\)

Evaluating \(sin(\frac{2\pi}{3})\), we get:

\(\frac{\sqrt{3}}{2}\)

So, the exact value of the expression is \(\frac{\sqrt{3}}{2}\).

Part c)

\(cos(\frac{10\pi}{9})cos(\frac{5\pi}{18}) + sin(\frac{10\pi}{9})sin(\frac{5\pi}{18})\)

This expression matches the cosine of a difference formula:

\(cos(A – B) = cos(A)cos(B) + sin(A)sin(B)\)

Therefore, the expression simplifies to:

\(cos(\frac{10\pi}{9} – \frac{5\pi}{18}) = cos(\frac{15\pi}{18}) = cos(\frac{5\pi}{6})\)

Evaluating \(cos(\frac{5\pi}{6})\), we get:

\(-\frac{\sqrt{3}}{2}\)

So, the exact value of the expression is \(-\frac{\sqrt{3}}{2}\).

Part d)

\(cos(\frac{\pi}{4})cos(\frac{\pi}{12}) – sin(\frac{\pi}{4})sin(\frac{\pi}{12})\)

This expression matches the cosine of a sum formula:

\(cos(A + B) = cos(A)cos(B) – sin(A)sin(B)\)

Therefore, the expression simplifies to:

\(cos(\frac{\pi}{4} + \frac{\pi}{12}) = cos(\frac{\pi}{3})\)

Evaluating \(cos(\frac{\pi}{3})\), we get:

\(\frac{1}{2}\)

So, the exact value of the expression is \(\frac{1}{2}\).

Problem 4: Using Compound Angle Formulas

Part a) Find the exact value of:

\(sin(\frac{17\pi}{12})\)

We can express \(\frac{17\pi}{12}\) as the sum of two angles whose sine and cosine values we know:

\(\frac{17\pi}{12} = \frac{8\pi}{12} + \frac{9\pi}{12} = \frac{2\pi}{3} + \frac{3\pi}{4}\)

Now, we can use the sine sum formula:

\(sin(A + B) = sin(A)cos(B) + cos(A)sin(B)\)

Therefore,

\(sin(\frac{17\pi}{12}) = sin(\frac{2\pi}{3} + \frac{3\pi}{4})\) \(= sin(\frac{2\pi}{3})cos(\frac{3\pi}{4}) + cos(\frac{2\pi}{3})sin(\frac{3\pi}{4})\)

Evaluating the trigonometric functions, we get:

\(= \frac{\sqrt{3}}{2} \cdot (-\frac{\sqrt{2}}{2}) + (-\frac{1}{2}) \cdot \frac{\sqrt{2}}{2}\) \(= -\frac{\sqrt{6} + \sqrt{2}}{4}\)

Part b) Find the exact value of:

\(cos(\frac{13\pi}{12})\)

We can express \(\frac{13\pi}{12}\) as the sum of two angles:

\(\frac{13\pi}{12} = \frac{8\pi}{12} + \frac{5\pi}{12} = \frac{2\pi}{3} + \frac{5\pi}{12}\)

Using the cosine sum formula:

\(cos(A + B) = cos(A)cos(B) – sin(A)sin(B)\)

We get:

\(cos(\frac{13\pi}{12}) = cos(\frac{2\pi}{3} + \frac{5\pi}{12})\) \(= cos(\frac{2\pi}{3})cos(\frac{5\pi}{12}) – sin(\frac{2\pi}{3})sin(\frac{5\pi}{12})\)

To evaluate this, you might need to use half-angle or other trigonometric identities, or a calculator.

Part c) Find the exact value of:

\(tan(\frac{5\pi}{12})\)

We can express \(\frac{5\pi}{12}\) as the difference of two angles:

\(\frac{5\pi}{12} = \frac{3\pi}{4} – \frac{\pi}{3}\)

Using the tangent difference formula:

\(tan(A – B) = \frac{tan(A) – tan(B)}{1 + tan(A)tan(B)}\)

We get:

\(tan(\frac{5\pi}{12}) = tan(\frac{3\pi}{4} – \frac{\pi}{3})\) \(= \frac{tan(\frac{3\pi}{4}) – tan(\frac{\pi}{3})}{1 + tan(\frac{3\pi}{4})tan(\frac{\pi}{3})}\)

Now, evaluate the tangent values and simplify the expression.

Problem 5: Trigonometric Ratios and Identities

Given:

• \(sin(\alpha) = \frac{3}{7}\) and \(0 \leq \alpha \leq \frac{\pi}{2}\)
• \(cos(\beta) = \frac{3}{5}\) and \(0 \leq \beta \leq \frac{\pi}{2}\)

To find: a) \(sin(\alpha + \beta)\) b) \(cos(2\alpha)\)

Solution:

Part a: To find \(sin(\alpha + \beta)\), we can use the sine sum formula:

\(sin(\alpha + \beta) = sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)\)

First, we need to find \(cos(\alpha)\) and \(sin(\beta)\) using the Pythagorean identity:

\(sin^2(\alpha) + cos^2(\alpha) = 1\) \(cos^2(\alpha) = 1 – sin^2(\alpha) = 1 – (\frac{3}{7})^2 = \frac{40}{49}\) \(cos(\alpha) = \pm \frac{\sqrt{40}}{7}\)

Since \(\alpha\) is in the first quadrant, \(cos(\alpha)\) is positive. So, \(cos(\alpha) = \frac{2\sqrt{10}}{7}\).

Similarly, for \(\beta\):

\(sin^2(\beta) + cos^2(\beta) = 1\) \(sin^2(\beta) = 1 – cos^2(\beta) = 1 – (\frac{3}{5})^2 = \frac{16}{25}\) \(sin(\beta) = \pm \frac{4}{5}\)

Since \(\beta\) is in the first quadrant, \(sin(\beta)\) is positive. So, \(sin(\beta) = \frac{4}{5}\).

Now, we can substitute these values into the sine sum formula:

\(sin(\alpha + \beta) = (\frac{3}{7})(\frac{3}{5}) + (\frac{2\sqrt{10}}{7})(\frac{4}{5})\) \(= \frac{9}{35} + \frac{8\sqrt{10}}{35}\)

Part b: To find \(cos(2\alpha)\), we can use the double-angle formula for cosine:

\(cos(2\alpha) = cos^2(\alpha) – sin^2(\alpha)\)

We already found \(cos(\alpha) = \frac{2\sqrt{10}}{7}\) and \(sin(\alpha) = \frac{3}{7}\). Substituting these values:

\(cos(2\alpha) = (\frac{2\sqrt{10}}{7})^2 – (\frac{3}{7})^2\) \(= \frac{40}{49} – \frac{9}{49}\) \(= \frac{31}{49}\)

Problem 6: Trigonometric Ratios and Double Angle Formulas

Given:

• \(sin(x) = \frac{12}{13}\)
• \(\frac{\pi}{2} \leq x \leq \pi\)

To find:

• \(sin(2x)\)
• \(tan(2x)\)

Solution:

Step 1: Find cos(x): Since \(x\) is in the second quadrant, \(cos(x)\) will be negative. Using the Pythagorean identity:

\(cos^2(x) = 1 – sin^2(x) = 1 – (\frac{12}{13})^2 = \frac{25}{169}\)

So, \(cos(x) = -\frac{5}{13}\)

Step 2: Use the double-angle formulas:

• For sin(2x):

\(sin(2x) = 2sin(x)cos(x) = 2(\frac{12}{13})(-\frac{5}{13}) = -\frac{120}{169}\)

For tan(2x): We can use the formula:

\(tan(2x) = \frac{2tan(x)}{1 – tan^2(x)}\)

First, find \(tan(x)\):

\(tan(x) = \frac{sin(x)}{cos(x)} = \frac{\frac{12}{13}}{-\frac{5}{13}} = -\frac{12}{5}\)

Now, substitute this value into the formula for \(tan(2x)\):

\(tan(2x) = \frac{2(-\frac{12}{5})}{1 – (-\frac{12}{5})^2} = \frac{-\frac{24}{5}}{1 – \frac{144}{25}} = \frac{-\frac{24}{5}}{-\frac{119}{25}} = \frac{120}{119}\)

Therefore, \(sin(2x) = -\frac{120}{169}\) and \(tan(2x) = \frac{120}{119}\).

Problem 7: Trigonometric Ratios and Double Angle Formulas

Given:

• \(cos(\theta) = -\frac{3}{5}\)
• θ lies in the second quadrant

To find: a) \(sin(2\theta)\), \(cos(2\theta)\), and \(tan(2\theta)\) b) Approximate value of θ in radians.

Solution:

Part a:

First, we need to find \(sin(\theta)\). Since θ is in the second quadrant, \(sin(\theta)\) will be positive. Using the Pythagorean identity:

\(sin^2(\theta) + cos^2(\theta) = 1\) \(sin^2(\theta) = 1 – cos^2(\theta) = 1 – (-\frac{3}{5})^2 = \frac{16}{25}\) \(sin(\theta) = \frac{4}{5}\)

Now, we can use the double-angle formulas:

• For sin(2θ):

\(sin(2\theta) = 2sin(\theta)cos(\theta) = 2(\frac{4}{5})(-\frac{3}{5}) = -\frac{24}{25}\)

For cos(2θ):

\(cos(2\theta) = cos^2(\theta) – sin^2(\theta) = (-\frac{3}{5})^2 – (\frac{4}{5})^2 = -\frac{7}{25}\)

For tan(2θ):

\(tan(2\theta) = \frac{sin(2\theta)}{cos(2\theta)} = \frac{-\frac{24}{25}}{-\frac{7}{25}} = \frac{24}{7}\)

Part b:

To find the approximate value of θ, we can use the inverse cosine function:

\(\theta = arccos(-\frac{3}{5})\)

Using a calculator, we find: \(\theta ≈ 2.21\) radians (rounded to the nearest hundredth)

Therefore, the approximate value of θ is 2.21 radians.

Problem 8: Acute Angle and Cosine

Given:

• θ is an acute angle
• \(cos(\theta) = \frac{2}{3}\)

To find:

• The exact value of \(cos(4\theta)\)

Solution:

We can use the double-angle formula for cosine to find \(cos(2\theta)\):

\(cos(2\theta) = 2cos^2(\theta) – 1 = 2(\frac{2}{3})^2 – 1 = -\frac{1}{9}\)

Now, we can use the double-angle formula again to find \(cos(4\theta)\):

\(cos(4\theta) = 2cos^2(2\theta) – 1 = 2(-\frac{1}{9})^2 – 1 = -\frac{77}{81}\)

Therefore, the exact value of

\(cos(4\theta)\) is \(-\frac{77}{81}\).