Why does a trigonometric equation usually come with a specified solution interval?

Because trigonometric functions are periodic , so across all real numbers there are usually infinitely many solutions. Restricting the interval is what makes the answer a finite list of values.

How do you solve two trigonometric inequalities that must both hold?

Find the solution set of each one separately , then take their intersection . Only the values satisfying every inequality belong in the final answer, and a number line makes the overlap easier to see.

Which of the inverse trigonometric functions are increasing, and which decreasing?

Arcsine and arctangent are increasing everywhere on their domains. Arccosine is decreasing everywhere on its domain, which is the odd one out of the three.

How does knowing whether an inverse trigonometric function increases help with an inequality?

Solve the corresponding equation first to get the boundary value, then the direction of the function tells you which side of it works. For an increasing inverse function a greater-than holds to the right of the boundary, and for a decreasing one it holds to the left.

Why do reciprocal trigonometric inequalities often have solution sets in several pieces?

Because the reciprocal functions have vertical asymptotes , and those cut the interval into separate parts. An inequality involving secant can therefore come out as a union of two or more disjoint intervals rather than as a single one.

What sign pattern distinguishes the sine identities from the cosine ones?

The sine identities have matching signs, a plus on the left going with a plus on the right and a minus with a minus. The cosine identities have opposite signs, a plus on the left giving a minus on the right and a minus giving a plus.

Define trigonometric identity .

A trigonometric identity is an equation involving trigonometric functions that is true for all values of the variable in the common domain of both sides. That is what separates it from an equation to be solved, which holds only at particular values.

How does verifying a trigonometric identity differ from solving an equation?

You transform one side step by step until it matches the other, and you never move terms across the equals sign. Moving terms across would assume the two sides are already equal, which is the very thing being proved.

When verifying a trigonometric identity, which side should you start from?

Usually the more complicated side, because simplifying is easier than complicating. Then look at the target side to see what form you are aiming at, which often tells you which identity will get you there.

True or False? Both sides of a proposed trigonometric identity may be worked on separately.

True. Reducing each side to the same simpler expression is a perfectly legitimate way of finding the route when it is not obvious how to proceed from one side alone. The final write-up should still show one side being transformed into the other.

When is a trigonometric identity needed in order to solve an equation?

When the equation cannot be reduced to a simple form directly, because it contains more than one trigonometric function or arguments of different sizes . An identity rewrites it so that only a single function, or a single angle, is left.

How do you choose which trigonometric identity to apply to an equation?

Identify the obstacle first, then pick the identity that removes it: several different angles call for a sum, difference or double-angle identity several different functions call for the Pythagorean identity or a quotient relationship

True or False? Once an identity has rewritten an equation, the solving method is the same as for a simple trigonometric equation.

True. The identity's only job is to remove the obstacle, and after that you factor, find an initial solution, and use symmetry and periodicity as usual. You still have to check that every solution lies inside the given interval.

AP®MathsCollege BoardPrecalculusFlashcardsTrigonometric & Polar FunctionsTrigonometric Equations, Inequalities & Identities

Trigonometric Equations, Inequalities & Identities (College Board AP® Precalculus): Flashcards

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Why does a trigonometric equation usually come with a specified solution interval?

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Why does a trigonometric equation usually come with a specified solution interval?
Because trigonometric functions are periodic, so across all real numbers there are usually infinitely many solutions.
Restricting the interval is what makes the answer a finite list of values.
What is the first move in solving $\sin θ = k$ on a given interval?
Find an initial solution, meaning any one value of $θ$ that satisfies the equation.
Use a known exact value if $k$ is one of the special values, and an inverse trigonometric function on a calculator otherwise.
One solution of a trigonometric equation is $θ_{0}$ . Fill in the two gaps in the rules for finding a second one.
For a sine equation a second solution is given by $______$ and for a cosine equation by $______$ instead.
The completed sentence is:
For a sine equation a second solution is given by $π - θ_{0}$ and for a cosine equation by $- θ_{0}$ instead.
Both come from the symmetry of the unit circle, sine values being symmetric about $θ = \frac{π}{2}$ and cosine values about $θ = 0$ .
Once two solutions of a sine or cosine equation are known, how do you find the rest?
Add or subtract whole multiples of $2 π$ , the period, to each of them.
Then keep only the values that land inside the specified interval and discard the others.
True or False?
Solving $\tan θ = k$ needs two initial solutions, just as $\sin θ = k$ does.
False.
Tangent takes each of its values once per period and its period is $π$ , so one initial solution is enough.
Every other solution is that one plus a whole multiple of $π$ , so there is no separate symmetry rule to apply.
An equation such as $2 \cos^{2} θ = \cos θ$ has the trigonometric function appearing twice. How is it solved?
Rearrange so that one side is zero, then factor, giving $\cos θ (2 \cos θ - 1) = 0$ .
That splits it into the two simple equations $\cos θ = 0$ and $\cos θ = \frac{1}{2}$ , each solved in the usual way.
Solving $2 \cos^{2} x + 7 \cos x + 3 = 0$ gives $\cos x = - \frac{1}{2}$ and $\cos x = - 3$ . What happens next?
Reject $\cos x = - 3$ , because the range of cosine is $[- 1, 1]$ and no angle has a cosine of $- 3$ .
Then solve $\cos x = - \frac{1}{2}$ in the given interval as usual, and those are the only solutions.
True or False?
To solve an equation on $0 \leq θ < 2 π$ , the value an inverse trigonometric function returns is enough.
False.
That gives only one solution, and symmetry and periodicity usually produce others inside the interval.
It can also return a value outside the interval altogether, as $\sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ does, which then needs $2 π$ added to it.
How is a trigonometric inequality such as $\sin θ < k$ solved?
Solve the corresponding equation first, which gives the boundary values.
Then use a graph or the unit circle to decide which intervals between or outside those boundaries satisfy the inequality.
Include the endpoints only if the inequality is non-strict.
How do you solve two trigonometric inequalities that must both hold?
Find the solution set of each one separately, then take their intersection.
Only the values satisfying every inequality belong in the final answer, and a number line makes the overlap easier to see.
An equation has the form $\sin^{- 1} (2 x - 1) = \frac{π}{6}$ with the inverse function on one side. What is the first move?
Apply sine to both sides, which cancels the arcsine and leaves $2 x - 1 = \sin \frac{π}{6}$ .
That turns it into an ordinary equation, here giving $2 x - 1 = \frac{1}{2}$ and so $x = \frac{3}{4}$ .
After solving an inverse trigonometric equation, what two checks are needed?
Check that the expression inside the inverse function lies in its domain, so within $[- 1, 1]$ for arcsine and arccosine.
Then check that the value on the other side lies in the function's range, because if it does not then the equation has no solution at all.
Which of the inverse trigonometric functions are increasing, and which decreasing?
Arcsine and arctangent are increasing everywhere on their domains.
Arccosine is decreasing everywhere on its domain, which is the odd one out of the three.
How does knowing whether an inverse trigonometric function increases help with an inequality?
Solve the corresponding equation first to get the boundary value, then the direction of the function tells you which side of it works.
For an increasing inverse function a greater-than holds to the right of the boundary, and for a decreasing one it holds to the left.
Fill in the two gaps in the method for solving $\csc θ = k$ .
First take the reciprocal of both sides to get $\sin θ =______$ and then solve that equation, which has no solutions at all when $|k|$ is less than $______$ to begin with.
The completed sentence is:
First take the reciprocal of both sides to get $\sin θ = \frac{1}{k}$ and then solve that equation, which has no solutions at all when $|k|$ is less than $1$ to begin with.
The same conversion works for secant and cotangent, giving equations in cosine and tangent respectively.
True or False?
The equation $\sec θ = 0.5$ has no solution.
True.
Secant never takes a value strictly between $- 1$ and $1$ , so $0.5$ is simply not in its range.
Reciprocating shows the same thing another way, since it would require $\cos θ = 2$ and cosine never exceeds $1$ .
The equation $3 - 2 \csc x = 7$ is to be solved on $0 \leq x < 2 π$ . How is it done?
Isolate the reciprocal function to get $\csc x = - 2$ , then take reciprocals to get $\sin x = - \frac{1}{2}$ .
Solving that in the interval gives $x = \frac{7 π}{6}$ and $x = \frac{11 π}{6}$ .
Why do reciprocal trigonometric inequalities often have solution sets in several pieces?
Because the reciprocal functions have vertical asymptotes, and those cut the interval into separate parts.
An inequality involving secant can therefore come out as a union of two or more disjoint intervals rather than as a single one.
An equation such as $\sin (2 x + \frac{π}{3}) = \frac{1}{2}$ has a whole expression inside the function. Why does that complicate the solving?
Because the interval the solutions must lie in applies to $x$ , not to the expression inside the function.
That inside expression runs over a different and usually wider interval, so the number of solutions changes.
What substitution makes a transformed trigonometric equation easier to solve?
Let a new variable stand for the whole argument of the trigonometric function.
For $\sin (2 x + \frac{π}{3}) = \frac{1}{2}$ putting $u = 2 x + \frac{π}{3}$ reduces it to $\sin u = \frac{1}{2}$ .
How do you convert the solution interval to the new variable?
Apply exactly the same operations to every part of the inequality.
If $0 \leq x \leq 2 π$ and $u = 2 x + \frac{π}{3}$ , then doubling gives $0 \leq 2 x \leq 4 π$ and adding $\frac{π}{3}$ gives $\frac{π}{3} \leq u \leq \frac{13 π}{3}$ .
After solving for the new variable, what still remains to be done?
Convert every solution back by reversing the substitution.
From $u = 2 x + \frac{π}{3}$ a solution of $u = \frac{5 π}{6}$ gives $2 x = \frac{π}{2}$ and therefore $x = \frac{π}{4}$ .
An equation has been reduced to $\sin u = p$ with initial solution $u_{0}$ , and no interval is given. Fill in the two gaps in its general solution.
$u = u_{0} +______n or u = (______- u_{0}) + 2 π n$
The completed general solution is:
$u = u_{0} + 2 π n or u = (π - u_{0}) + 2 π n$
Here $n$ stands for any integer, and the two families together give every solution.
A tangent equation needs only one family, $u = u_{0} + π n$ , because tangent repeats twice as often.
True or False?
A contextual restriction can turn infinitely many solutions into a finite list.
True.
An equation with no stated interval has infinitely many solutions, but a context supplies one: if $t$ is hours after midnight then only $0 \leq t < 24$ makes sense.
Solutions outside that are discarded even though they satisfy the equation.
An equation composes an inverse function with a trigonometric one, as in $\cos^{- 1} (\tan 2 x) = 0$ . How is it solved?
Work from the outside in, applying cosine to both sides to remove the arccosine and leaving $\tan 2 x = 1$ .
That is then an ordinary transformed trigonometric equation, solved by the substitution method.
Fill in the two gaps in the Pythagorean trigonometric identity.
$\sin^{2} θ +______=______$
The completed identity is:
$\sin^{2} θ + \cos^{2} θ = 1$
It holds for every real value of $θ$ , not just for particular ones.
Where does the Pythagorean trigonometric identity come from?
From the Pythagorean Theorem applied to the unit circle.
The terminal ray meets the circle at $(\cos θ, \sin θ)$ , giving a right triangle with legs $|\cos θ|$ and $|\sin θ|$ and hypotenuse $1$ , which is the radius.
True or False?
$\sin^{2} θ$ means $\sin (θ^{2})$ .
False.
$\sin^{2} θ$ is shorthand for ${(\sin θ)}^{2}$ , so the sine is taken first and the result is then squared.
$\sin (θ^{2})$ squares the angle before taking the sine, and the two give different values.
The Pythagorean identity can be divided through by a squared trigonometric function. What two further identities does that produce?
Dividing by $\cos^{2} θ$ gives $\tan^{2} θ + 1 = \sec^{2} θ$ .
Dividing by $\sin^{2} θ$ gives $1 + \cot^{2} θ = \csc^{2} θ$ .
How is the Pythagorean identity used to rewrite a trigonometric expression?
Look for a combination matching one of its rearranged forms, such as $1 - \sin^{2} x$ or $\sec^{2} x - 1$ , and replace it.
Then convert any remaining reciprocal functions into sine and cosine and simplify algebraically until the expression reaches the requested form.
Rewrite $\frac{1 - \cos^{2} x}{\sin x}$ using $\sin x$ only.
The numerator matches the rearranged identity $1 - \cos^{2} x = \sin^{2} x$ , so the expression becomes $\frac{\sin^{2} x}{\sin x}$ .
Cancelling one factor of $\sin x$ leaves $\sin x$ .
Rewrite $(\csc^{2} x - 1) \sin^{2} x$ as a single trigonometric term.
Use $\csc^{2} x - 1 = \cot^{2} x$ to turn it into $\cot^{2} x \cdot \sin^{2} x$ .
Writing $\cot^{2} x = \frac{\cos^{2} x}{\sin^{2} x}$ then cancels the $\sin^{2} x$ and leaves $\cos^{2} x$ .
Fill in the two gaps in the sum identities for sine and cosine.
$\sin (α + β) = \sin α \cos β +______, \cos (α + β) = \cos α \cos β -______$
The completed identities are:
$\sin (α + β) = \sin α \cos β + \cos α \sin β, \cos (α + β) = \cos α \cos β - \sin α \sin β$
Each one breaks a compound angle into the sines and cosines of its two separate parts.
True or False?
$\sin (α + β) = \sin α + \sin β$
False.
A trigonometric function cannot be distributed across a sum inside its argument.
The correct expansion mixes sines and cosines of both angles, which is precisely why a compound angle needs an identity of its own.
How are the difference identities obtained from the sum identities?
Replace $β$ by $- β$ , then use the fact that sine is odd and cosine is even.
That flips the sign of every term containing $\sin β$ , giving $\sin (α - β) = \sin α \cos β - \cos α \sin β$ and $\cos (α - β) = \cos α \cos β + \sin α \sin β$ .
What sign pattern distinguishes the sine identities from the cosine ones?
The sine identities have matching signs, a plus on the left going with a plus on the right and a minus with a minus.
The cosine identities have opposite signs, a plus on the left giving a minus on the right and a minus giving a plus.
How can a sum or difference identity give an exact value at a non-standard angle?
Write the angle as a sum or a difference of two standard angles, then apply the matching identity.
Since $\frac{π}{12} = \frac{π}{3} - \frac{π}{4}$ , the difference identity gives $\sin \frac{π}{12} = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$ .
How are the double-angle identities obtained, and what is the one for sine?
Set $α = β = θ$ in the sum identities, so that the compound angle becomes a double angle.
For sine that gives $\sin (θ + θ) = \sin θ \cos θ + \cos θ \sin θ$ , which simplifies to $\sin (2 θ) = 2 \sin θ \cos θ$ .
True or False?
There is only one correct form of the double-angle identity for cosine.
False.
Three forms are equal to one another: $\cos^{2} θ - \sin^{2} θ$ , then $2 \cos^{2} θ - 1$ , and then $1 - 2 \sin^{2} θ$ .
The Pythagorean identity is what converts between them, so all three are available whenever one of them is.
When is the $1 - 2 \sin^{2} θ$ form of the cosine double-angle identity the useful one?
When the expression being worked on already contains $\sin^{2} θ$ , since that form avoids bringing cosine into it at all.
Choosing whichever form matches what is already present is what keeps the algebra short.
The function $g$ is given by $g (x) = 5 \cos (2 x)$ . Write an equivalent form using single angles.
Apply the double-angle identity to get $g (x) = 5 (\cos^{2} x - \sin^{2} x) = 5 \cos^{2} x - 5 \sin^{2} x$ .
Multiplying the $5$ through the bracket is what produces the two separate coefficients.
Define trigonometric identity.
A trigonometric identity is an equation involving trigonometric functions that is true for all values of the variable in the common domain of both sides.
That is what separates it from an equation to be solved, which holds only at particular values.
How does verifying a trigonometric identity differ from solving an equation?
You transform one side step by step until it matches the other, and you never move terms across the equals sign.
Moving terms across would assume the two sides are already equal, which is the very thing being proved.
When verifying a trigonometric identity, which side should you start from?
Usually the more complicated side, because simplifying is easier than complicating.
Then look at the target side to see what form you are aiming at, which often tells you which identity will get you there.
Fill in the two gaps in the common strategies for verifying a trigonometric identity.
A good first move is to rewrite every function in terms of $______$ and cosine, and where one side holds several fractions it usually helps to combine them over a common $______$ first.
The completed sentence is:
A good first move is to rewrite every function in terms of sine and cosine, and where one side holds several fractions it usually helps to combine them over a common denominator first.
Factoring, and watching for Pythagorean patterns, are the two other moves that most often open a verification up.
True or False?
Both sides of a proposed trigonometric identity may be worked on separately.
True.
Reducing each side to the same simpler expression is a perfectly legitimate way of finding the route when it is not obvious how to proceed from one side alone.
The final write-up should still show one side being transformed into the other.
Verify $\frac{\sin x}{1 - \cos x} + \frac{\sin x}{1 + \cos x} = 2 \csc x$ by transforming its left-hand side.
Combine the two fractions over the common denominator $(1 - \cos x) (1 + \cos x)$ , which is a difference of two squares and equals $1 - \cos^{2} x$ .
The numerator simplifies to $2 \sin x$ and the Pythagorean identity turns the denominator into $\sin^{2} x$ , leaving $\frac{2}{\sin x}$ , which is $2 \csc x$ .
When is a trigonometric identity needed in order to solve an equation?
When the equation cannot be reduced to a simple form directly, because it contains more than one trigonometric function or arguments of different sizes.
An identity rewrites it so that only a single function, or a single angle, is left.
How do you choose which trigonometric identity to apply to an equation?
Identify the obstacle first, then pick the identity that removes it:
- several different angles call for a sum, difference or double-angle identity
- several different functions call for the Pythagorean identity or a quotient relationship
The equation $\sin (2 x) - \cos x = 0$ mixes two different angles. What identity clears it, and what does the equation become?
The double-angle identity $\sin (2 x) = 2 \sin x \cos x$ brings everything down to a single angle.
The equation becomes $2 \sin x \cos x - \cos x = 0$ , which factors as $\cos x (2 \sin x - 1) = 0$ .
True or False?
Once an identity has rewritten an equation, the solving method is the same as for a simple trigonometric equation.
True.
The identity's only job is to remove the obstacle, and after that you factor, find an initial solution, and use symmetry and periodicity as usual.
You still have to check that every solution lies inside the given interval.
The equation $2 \cos^{2} x + \sin x - 1 = 0$ mixes two different functions. How is it made solvable?
Use the Pythagorean identity as $\cos^{2} x = 1 - \sin^{2} x$ to put everything in terms of sine.
That gives $2 \sin^{2} x - \sin x - 1 = 0$ , a hidden quadratic which factors as $(2 \sin x + 1) (\sin x - 1) = 0$ .
The equation $\sec^{2} x = 3 \tan x - 1$ is to be rewritten in terms of tangent alone. Fill in the two gaps.
Substituting $\sec^{2} x =______+ \tan^{2} x$ turns it into $\tan^{2} x - 3 \tan x +______= 0$ after rearranging.
The completed working is:
Substituting $\sec^{2} x = 1 + \tan^{2} x$ turns it into $\tan^{2} x - 3 \tan x + 2 = 0$ after rearranging.
That factors as $(\tan x - 1) (\tan x - 2) = 0$ , giving two simple tangent equations to solve.

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