Questions and Answers Section

Example 1: Degree to Radian Conversion

Question: Convert $75^\circ$ into radian measure.

Solution:

To convert from degrees to radians, we multiply by the conversion factor $\frac{\pi}{180}$.

$$75^\circ = 75 \times \frac{\pi}{180} = \frac{5 \times 15}{12 \times 15}\pi = \mathbf{\frac{5\pi}{12}} \text{ radians}$$.

Example 2: Radian to Degree Conversion

Question: Convert $\frac{11\pi}{12}$ radians into degree measure.

Solution:

To convert from radians to degrees, we multiply by the conversion factor $\frac{180}{\pi}$.

$$\frac{11\pi}{12} \text{ rad} = \frac{11\pi}{12} \times \frac{180}{\pi} = 11 \times \frac{180}{12} = 11 \times 15 = \mathbf{165^\circ}$$.

Example 3: Arc Length

Question: A circular wire of radius 7 cm is cut and bent into an arc of a circle of radius 12 cm. Find the angle subtended by the arc at the center.

Solution:

Step 1: The length of the wire is the circumference of the first circle, $l = 2\pi r_1 = 2\pi(7) = 14\pi$ cm.

Step 2: This length becomes the arc length for the second circle. Using the formula $l=r_2\theta$, where $r_2=12$ cm:

$14\pi = 12\theta \implies \theta = \frac{14\pi}{12} = \mathbf{\frac{7\pi}{6}}$ radians.

Example 4: Clock Angle Problem

Question: Find the angle in degrees between the minute hand and the hour hand of a clock at 3:40 PM.

Solution:

Step 1: At 3:40, the minute hand is at the 8, which is $\frac{8}{12} \times 360^\circ = \mathbf{240^\circ}$ from the 12.

Step 2: The hour hand has moved 3 full hours plus 40/60 = 2/3 of the way through the next hour. Its position is $(3 + \frac{2}{3}) \times 30^\circ = \frac{11}{3} \times 30^\circ = \mathbf{110^\circ}$ from the 12.

Step 3: The difference is $|240^\circ - 110^\circ| = \mathbf{130^\circ}$.

Example 5: Finding Trig Values from a Point

Question: If the terminal side of an angle $\theta$ passes through the point (5, -12), find $\cos\theta$ and $\csc\theta$.

Solution:

Given $x=5, y=-12$. The distance from the origin is $r = \sqrt{x^2+y^2} = \sqrt{5^2+(-12)^2} = \sqrt{25+144}=\sqrt{169}=\mathbf{13}$.

• $\cos\theta = x/r = \mathbf{5/13}$.

• $\csc\theta = r/y = \mathbf{-13/12}$.

Example 6: Finding Trig Values from One Ratio and Quadrant

Question: If $\tan x = -4/3$ and x lies in the second quadrant, find $\sin x + \cos x$.

Solution:

In Q2, $\sin x > 0$ and $\cos x < 0$. From $\tan x = -4/3$, we can form a reference triangle with opposite=4, adjacent=3, hypotenuse=5.

• $\sin x = \text{opposite/hypotenuse} = \mathbf{4/5}$ (positive in Q2).

• $\cos x = \text{adjacent/hypotenuse} = \mathbf{-3/5}$ (negative in Q2).

• The sum is $4/5 + (-3/5) = \mathbf{1/5}$.

Example 7: Evaluating at Large Angles

Question: Find the value of $\cos(-1125^\circ)$.

Solution:

First, use the identity $\cos(-\theta)=\cos\theta$. So we need $\cos(1125^\circ)$.

Next, find the coterminal angle by subtracting multiples of $360^\circ$. $1125^\circ = 3 \times 360^\circ + 45^\circ = 1080^\circ + 45^\circ$.

So, $\cos(1125^\circ) = \cos(45^\circ) = \mathbf{1/\sqrt{2}}$.

Example 8: Simplifying with Allied Angles

Question: Find the value of $\tan(19\pi/3)$.

Solution:

We simplify the angle. $19\pi/3 = (18\pi + \pi)/3 = 6\pi + \pi/3$. Since the period of tangent is $\pi$, we can ignore multiples of $\pi$.

So, $\tan(19\pi/3) = \tan(6\pi+\pi/3) = \tan(\pi/3) = \mathbf{\sqrt{3}}$.

Example 9: Proving a Basic Identity

Question: Prove that $\frac{\sin x}{1+\cos x} + \frac{1+\cos x}{\sin x} = 2\csc x$.

Solution:

Starting with the Left Hand Side (LHS), we find a common denominator:

LHS = $\frac{\sin^2 x + (1+\cos x)^2}{\sin x(1+\cos x)} = \frac{\sin^2 x + 1+2\cos x+\cos^2x}{\sin x(1+\cos x)}$.

Using $\sin^2 x+\cos^2x=1$, the numerator becomes $1+1+2\cos x = 2+2\cos x = 2(1+\cos x)$.

LHS = $\frac{2(1+\cos x)}{\sin x(1+\cos x)} = \frac{2}{\sin x} = \mathbf{2\csc x}$ = RHS.

Example 10: Using Allied Angle Identities

Question: Simplify $\frac{\cos(180^\circ-A)\sin(270^\circ+A)}{\sin(90^\circ-A)\cos(360^\circ-A)}$.

Solution:

We simplify each term:

• Numerator: $\cos(180^\circ-A) = -\cos A$ (Q2). $\sin(270^\circ+A) = -\cos A$ (Q4).

• Denominator: $\sin(90^\circ-A) = \cos A$ (Q1). $\cos(360^\circ-A) = \cos A$ (Q4).

The expression becomes $\frac{(-\cos A)(-\cos A)}{(\cos A)(\cos A)} = \frac{\cos^2A}{\cos^2A} = \mathbf{1}$.

Example 11: Graphing Transformations (Amplitude)

Question: What is the range of the function $f(x) = -3\cos x$?

Solution:

Step 1: The range of the basic function $y=\cos x$ is $[-1,1]$.

Step 2: Multiplying by -3 scales the range by a factor of 3 and reflects it across the x-axis. The new range is $[-3, 3]$. The maximum value is 3 and the minimum value is -3.

Example 12: Graphing Transformations (Period)

Question: What is the period of the function $f(x) = \tan(x/2)$?

Solution:

The period of $\tan(x)$ is $\pi$. The period of $\tan(Bx)$ is given by the formula $\frac{\pi}{|B|}$.

Here, $B=1/2$. The period is $\frac{\pi}{1/2} = \mathbf{2\pi}$.

Example 13: Graphing Transformations (Phase Shift)

Question: The graph of $y=\sin(x+\pi/3)$ is a shift of the standard sine graph by:

Solution:

For a function $y=A\sin(Bx+C)$, the phase shift is given by $-C/B$. Here, $C=\pi/3$ and $B=1$. The phase shift is $-(\pi/3)/1 = -\pi/3$. A negative phase shift corresponds to a shift to the left by $\mathbf{\pi/3}$ units.

Example 14: Solving an Equation Graphically

Question: Find the number of solutions to the equation $\cos x = x^2$.

Solution:

We sketch the graphs of $y=\cos x$ and $y=x^2$. The graph of $y=x^2$ is an upward-opening parabola with its vertex at the origin. The graph of $y=\cos x$ is a wave that passes through (0,1). The parabola is symmetric about the y-axis, as is the cosine function. For $x>0$, they intersect once. By symmetry, they will also intersect once for $x<0$. Total number of solutions is 2.

Example 15: Maximum and Minimum Values

Question: Find the maximum and minimum values of the expression $5\sin x + 12\cos x$.

Solution:

For an expression of the form $a\sin x + b\cos x$, the range is $[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$.

Here, $a=5, b=12$. So, $\sqrt{a^2+b^2} = \sqrt{5^2+12^2} = \sqrt{25+144}=\sqrt{169}=13$. The range is [-13, 13].

Example 16: Compound Angle Formula

Question: Find the value of $\sin(75^\circ)$.

Solution:

We write $75^\circ$ as a sum of standard angles: $75^\circ=45^\circ+30^\circ$. Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$:

$\sin(75^\circ) = \sin(45^\circ)\cos(30^\circ)+\cos(45^\circ)\sin(30^\circ) = (\frac{1}{\sqrt{2}})(\frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}})(\frac{1}{2}) = \mathbf{\frac{\sqrt{3}+1}{2\sqrt{2}}}$.

Example 17: Double Angle Formula

Question: If $\sin\theta = 3/5$ and $\theta$ is in the second quadrant, find the value of $\sin(2\theta)$.

Solution:

In Q2, $\cos\theta$ is negative. Using $\cos^2\theta=1-\sin^2\theta$, we get $\cos\theta = -\sqrt{1-(3/5)^2} = -\sqrt{16/25} = -4/5$. Now, using the double angle formula:

$\sin(2\theta) = 2\sin\theta\cos\theta = 2(3/5)(-4/5) = \mathbf{-24/25}$.

Example 18: Triple Angle Formula

Question: Find the value of $4\cos^3(20^\circ) - 3\cos(20^\circ)$.

Solution:

This expression matches the identity for $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$. Here, $\theta=20^\circ$. The value of the expression is $\cos(3 \times 20^\circ) = \cos(60^\circ) = \mathbf{1/2}$.

Example 19: Product-to-Sum Formula

Question: Express $2\sin(5x)\cos(3x)$ as a sum.

Solution:

Using the identity $2\sin A \cos B = \sin(A+B)+\sin(A-B)$, with $A=5x, B=3x$, we get:

$\sin(5x+3x)+\sin(5x-3x) = \mathbf{\sin(8x)+\sin(2x)}$.

Example 20: Sum-to-Product Formula

Question: Simplify $\frac{\sin(7x)-\sin(5x)}{\cos(7x)+\cos(5x)}$.

Solution:

Using the sum-to-product formulas:

• Numerator: $\sin A - \sin B = 2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2}) = 2\cos(6x)\sin(x)$.

• Denominator: $\cos A + \cos B = 2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2}) = 2\cos(6x)\cos(x)$.

The expression simplifies to $\frac{2\cos(6x)\sin(x)}{2\cos(6x)\cos(x)} = \frac{\sin x}{\cos x} = \mathbf{\tan x}$.

Example 21: Area of a Sector

Question: The area of a sector of a circle is 15 sq cm and the angle of the sector is 1.5 radians. Find the radius of the circle.

Solution:

Using the formula $A = \frac{1}{2}r^2\theta$:

$15 = \frac{1}{2}r^2(1.5) \implies 30 = 1.5r^2 \implies r^2 = \frac{30}{1.5} = 20$. The radius is $r = \sqrt{20} = \mathbf{2\sqrt{5}}$ cm.

Example 22: Simplifying a Complex Identity

Question: Prove that $\sqrt{\frac{1-\sin\theta}{1+\sin\theta}} = \sec\theta - \tan\theta$ for $\theta$ in the first quadrant.

Solution:

To simplify the expression under the square root, we multiply the numerator and denominator by the conjugate of the denominator, which is $(1-\sin\theta)$:

LHS = $\sqrt{\frac{(1-\sin\theta)(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}} = \sqrt{\frac{(1-\sin\theta)^2}{1-\sin^2\theta}} = \sqrt{\frac{(1-\sin\theta)^2}{\cos^2\theta}}$.

Since $\theta$ is in the first quadrant, both $1-\sin\theta$ and $\cos\theta$ are positive, so we can take the square root directly:

LHS = $\frac{1-\sin\theta}{\cos\theta} = \frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta} = \mathbf{\sec\theta - \tan\theta}$ = RHS.

Example 23: Finding Period of a Complex Function

Question: Find the period of the function $f(x) = |\sin x| + |\cos x|$.

Solution:

The period of $|\sin x|$ is $\pi$ and the period of $|\cos x|$ is $\pi$. The period of their sum will be less than or equal to $\pi$. Let's test a smaller value, $T=\pi/2$.

$f(x+\pi/2) = |\sin(x+\pi/2)| + |\cos(x+\pi/2)| = |\cos x| + |-\sin x| = |\cos x| + |\sin x| = f(x)$.

Since the function repeats every $\pi/2$, the fundamental period is $\pi/2$.

Example 24: Finding Domain

Question: Find the domain of the function $f(x) = \frac{1}{1-2\cos x}$.

Solution:

The function is defined everywhere except where the denominator is zero.

$1-2\cos x = 0 \implies \cos x = 1/2$.

The principal values for this are $x=\pi/3$ and $x=-\pi/3$. The general solution, accounting for the periodicity of cosine, is $x = 2n\pi \pm \pi/3$, where n is any integer. The domain is \mathbf{\mathbb{R} - \{2n\pi \pm \pi/3, n \in \mathbb{Z}\}.

Example 25: Proving an Identity with $18^\circ$

Question: Prove that $\sin(18^\circ) = \frac{\sqrt{5}-1}{4}$.

Solution:

Let $\theta=18^\circ$. Then $5\theta=90^\circ \implies 2\theta = 90^\circ-3\theta$. Take sine of both sides:

$\sin(2\theta)=\sin(90^\circ-3\theta)=\cos(3\theta)$.

$2\sin\theta\cos\theta = 4\cos^3\theta-3\cos\theta$. Since $\cos(18^\circ) \n e 0$, we can divide by $\cos\theta$:

$2\sin\theta = 4\cos^2\theta-3 = 4(1-\sin^2\theta)-3$. Let $y=\sin\theta$.

$2y=4-4y^2-3 \implies 4y^2+2y-1=0$. Solving with the quadratic formula gives $y = \frac{-2\pm\sqrt{4-4(4)(-1)}}{8} = \frac{-2\pm\sqrt{20}}{8} = \frac{-1\pm\sqrt{5}}{4}$. Since $18^\circ$ is in the first quadrant, its sine is positive. So, $\sin(18^\circ) = \mathbf{\frac{\sqrt{5}-1}{4}}$.

Example 26: Evaluating a Product

Question: Find the value of $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$.

Solution:

Let $P = \cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$. Multiply and divide by $2\sin(20^\circ)$:

$P = \frac{(2\sin 20^\circ \cos 20^\circ)\cos 40^\circ \cos 80^\circ}{2\sin 20^\circ} = \frac{\sin 40^\circ \cos 40^\circ \cos 80^\circ}{2\sin 20^\circ}$.

Repeat the process: $P = \frac{(1/2)(2\sin 40^\circ \cos 40^\circ) \cos 80^\circ}{2\sin 20^\circ} = \frac{(1/2)\sin 80^\circ \cos 80^\circ}{2\sin 20^\circ}$.

Repeat again: $P = \frac{(1/4)(2\sin 80^\circ \cos 80^\circ)}{2\sin 20^\circ} = \frac{(1/4)\sin 160^\circ}{2\sin 20^\circ}$.

Since $\sin(160^\circ) = \sin(180^\circ-20^\circ)=\sin(20^\circ)$, this becomes:

$P = \frac{\sin(20^\circ)}{8\sin(20^\circ)} = \mathbf{1/8}$.

Example 27: Range of a Transformed Function

Question: Find the range of $f(x) = \frac{1}{3-\cos x}$.

Solution:

Step 1: Start with the range of the basic function: $-1 \le \cos x \le 1$.

Step 2: Multiply by -1, which reverses the inequalities: $1 \ge -\cos x \ge -1$.

Step 3: Add 3 to all parts: $3+1 \ge 3-\cos x \ge 3-1$, which is $4 \ge 3-\cos x \ge 2$.

Step 4: Take the reciprocal, which reverses the inequalities again: $\frac{1}{4} \le \frac{1}{3-\cos x} \le \frac{1}{2}$.

The range is [1/4, 1/2].

Example 28: Eliminating a Parameter

Question: If $x = a\cos^3\theta$ and $y=a\sin^3\theta$, find the relation between x and y.

Solution:

Step 1: Isolate the trigonometric terms.

From the given equations, we have $\cos^3\theta = x/a$ and $\sin^3\theta = y/a$.

Step 2: Raise both to the power of 2/3.

$(\cos^3\theta)^{2/3} = (x/a)^{2/3} \implies \cos^2\theta = (x/a)^{2/3}$.

$(\sin^3\theta)^{2/3} = (y/a)^{2/3} \implies \sin^2\theta = (y/a)^{2/3}$.

Step 3: Use the Pythagorean identity $\cos^2\theta+\sin^2\theta=1$.

$(x/a)^{2/3}+(y/a)^{2/3} = 1$, which simplifies to $\mathbf{x^{2/3}+y^{2/3}=a^{2/3}}$. (This is the equation of an astroid).

Example 29: Simplifying a Sum

Question: Find the value of $\cos^2(\pi/8) + \cos^2(3\pi/8) + \cos^2(5\pi/8) + \cos^2(7\pi/8)$.

Solution:

Use allied angles to relate the larger angles to the smaller ones.

• $\cos(7\pi/8)=\cos(\pi-\pi/8)=-\cos(\pi/8)$, so $\cos^2(7\pi/8) = \cos^2(\pi/8)$.

• $\cos(5\pi/8)=\cos(\pi-3\pi/8)=-\cos(3\pi/8)$, so $\cos^2(5\pi/8) = \cos^2(3\pi/8)$.

The expression becomes $2(\cos^2(\pi/8)+\cos^2(3\pi/8))$.

Now use the co-function identity: $\cos(3\pi/8)=\cos(\pi/2-\pi/8)=\sin(\pi/8)$.

The expression is $2(\cos^2(\pi/8)+\sin^2(\pi/8))=2(1)=\mathbf{2}$.

Example 30: Solving for an Expression

Question: If $\frac{\sin x}{a} = \frac{\cos x}{b}$, then $a\sin(2x)+b\cos(2x)$ is equal to:

Solution:

From the given ratio, let $\frac{\sin x}{a} = \frac{\cos x}{b} = k$. Then $\sin x = ak$ and $\cos x=bk$.

Since $\sin^2x+\cos^2x=1$, we have $(ak)^2+(bk)^2=1 \implies k^2(a^2+b^2)=1$.

Now, expand the target expression using double angle formulas:

$a\sin(2x)+b\cos(2x) = a(2\sin x \cos x) + b(\cos^2x-\sin^2x)$.

Substitute $\sin x = ak$ and $\cos x=bk$:

$= a(2(ak)(bk)) + b((bk)^2-(ak)^2) = 2a^2bk^2 + b(b^2k^2-a^2k^2) = 2a^2bk^2+b^3k^2-a^2bk^2$.

$= a^2bk^2+b^3k^2 = b k^2(a^2+b^2)$.

Since $k^2(a^2+b^2)=1$, the expression simplifies to $\mathbf{b}$.