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Trigonometry : Trigonometric Equations

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6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #6 : Factoring Trigonometric Equations

Factor the following expression

$sin^{2n}(x)+2sin^n(x)+3$ where is assumed to be a positive integer.

Possible Answers:

(sin^nx+1)(sin^nx+2)

(sin^nx+1)(sin^nx-2)

(2sin^nx+1)(sin^nx-2)

We cannot factor the above expression.

(sin^nx-1)(sin^nx-2)

Correct answer:

We cannot factor the above expression.

Explanation:

Letting sin^n(x)=X , we have the equivalent expression:

X^2+2X+3 .

We cant factor X^2+2X+3 since $X^2+2X+3=(x+1)^2+2 >0 \forall \quadx\in\mathbb{R}$ .

This shows that we cannot factor the above expression.

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Example Question #3 : Factoring Trigonometric Equations

Factor

$1-cos^{11}x$

Possible Answers:

$(1-cosx)(cos^{10}x+cos^{9}x+...cosx-1)$

$(1-cosx)(cos^{10}x+cos^{9}x+...cosx+1)$

$(1-cosx)(cos^{11}x+cos^{9}x+...cosx+1)$

$(-1-cosx)(cos^{10}x+cos^{9}x+...cosx+1)$

$(2-cosx)(cos^{10}x+cos^{9}x+...cosx+1)$

Correct answer:

$(1-cosx)(cos^{10}x+cos^{9}x+...cosx+1)$

Explanation:

We first note that we have:

$1-a^{11}=(1-a)(a^{10}+a^9+...a+1)$

Then taking a=cos(x) , we have the result.

$(1-cosx)(cos^{10}x+cos^{9}x+...cosx+1)$

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Example Question #4 : Factoring Trigonometric Equations

Find a simple expression for the following :

$cos^{2n}x-cos^{2n+2} x$

Possible Answers:

$-cos^{2n}xsin^{2}x$

3sinx

$cos^{2n}xsin^{2}x$

$cos^{2n}x (1-sin^{2}x)$

$2cos^{2n}xsin^{2}x$

Correct answer:

$cos^{2n}xsin^{2}x$

Explanation:

First of all we know that :

cos^2x+sin^2x=1 and this gives:

$1-cos^2{x}=sin^{2}x$ .

Now we need to see that: $cos^{2n}x-cos^{2n+2} x$ can be written as

$cos^{2n}x(1-cos^2x)$ and since $1-cos^2{x}=sin^{2}x$

we have then:

$cos^{2n}x(1-cos^2x)=cos^{2n}xsin^2x$ .

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Example Question #21 : Trigonometric Equations

What is a simple expression for the formula:

$cos^{3}x-sin^{3}x$

Possible Answers:

(cosx-sinx)(1+cosx sinx)

(cosx-sinx)(2-cosx sinx)

2(cosx-sinx)(1+cosx sinx)

(cosx+sinx)(1+cosx sinx)

(cosx-sinx)(1+3cosx sinx)

Correct answer:

(cosx-sinx)(1+cosx sinx)

Explanation:

From the expression :

a^3-b^3=(a-b)(a^2+ab+b^2)

we have:

$cos^{3}x-sin^{3}x=(cosx-sinx)(cos^2x+cosxsinx+sin^2x)$

Now since we know that :

cos^2x+sin^2x=1 . This expression becomes:

$cos^{3}x-sin^{3}x=(cosx-sinx)(1+cosxsinx)$ .

This is what we need to show.

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Example Question #22 : Trigonometric Equations

Factor: (sin^4x-cos^4x)

Possible Answers:

(sin^2x+cos^2x)(sinx+cosx)

(sinx-cosx)(sinx+cosx)

-(sin^2x+cos^2x)(sinx-cosx)

(sin^2x-cos^2x)(sin^2x+cos^2x)

Correct answer:

(sinx-cosx)(sinx+cosx)

Explanation:

Step 1: Recall the difference of squares (or powers of four) formula:

x^4-y^4=(x^2+y^2)(x^2-y^2)=(x^2+y^2)(x-y)(x+y)

Step 2: Factor the question:

(sin^4x-cos^4x)=(sin^2x+cos^2x)(sin^2x-cos^2x)

Factor more:

(sin^2x+cos^2x)(sinx+cosx)(sinx-cosx)

Step 3: Recall a trigonometric identity:

sin^2x+cos^2x=1 .. Replace this

Final Answer:

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Example Question #185 : Trigonometry

For this question, we will denote by max the maximum value of the function and min the minimum value of the function.

What is the maximum and minimum values of

cos^4x+sin^3x, where is a real number.

Possible Answers:

max=0, min=-1

max=1, min=0

max=2, min=-2

max=2, min=-1

max=1, min=-1

Correct answer:

max=2, min=-1

Explanation:

To find the maximum and the minimum , we can view the above function as

a system where $-1\le cosx\le 1$ and $-1\le sinx\le 1$ . Using these two conditions we find the maximum and the minimum.

$0\le cos^2x\le 1$ means also that $0\le cos^4x\le 1$ ( $\star$ ) We also have:

$-1\le sinx\le 1$ implies that :

$-1\le sin^{3}x\le 1$ ( $\star \star$ ) Therefore we have by adding ( $\star$ ) and( $\star \star$ )

$-1\le cos^4x+sin^3x\le 2$

This means that max=2 and min=-1

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Example Question #1 : Solving Trigonometric Equations

Find the values of that satisfy the following system:

$\left\{\begin{matrix} 2cosx\le \sqrt{2}\\2sinx\le \sqrt{2} \\ cosx< sinx\end{matrix}\right.$

where is assumed to be $[0,\frac{\pi}{2}]$

Possible Answers:

This system does not have a solution.

$[\frac{\pi}{4},\frac{\pi}{2}]$

$[0,\frac{\pi}{2}]$

$[-\frac{\pi}{4},\frac{\pi}{2}]$

$[0,\frac{\pi}{4}]$

Correct answer:

This system does not have a solution.

Explanation:

We can write the system in the equivalent form:

$cosx\le \frac{\sqrt{2}}{2}$

$sinx\le \frac{\sqrt{2}}{2}$

cosx< sinx

The solution to the first equation is $x\in \[0,\frac{\pi}{4}]$

cosx< sinx means that $x\in (\frac{\pi}{4},\frac{\pi}{2}]$

This means that there is no x that satisfies the system.

Therefore there is no x that solves the 3 inequalities simultaneously.

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Example Question #1 : Systems Of Trigonometric Equations

Which of the following systems of trigonometric equations have a solution with an -coordinate of $\frac{\pi}{4}$ ?

Possible Answers:

y=sin(x)

y=cos(7x)

y=sin(x)

y=cos(3x)

More than one of these answers has a solutions at $x=\frac{\pi}{4}$ .

y=sin(x)

y=cos(5x)

Correct answer:

y=sin(x)

y=cos(7x)

Explanation:

The solution to the correct answer would be $\left(\frac{\pi}{4}, \frac{1}{\sqrt2}\right)$ .

For all of the other answers, plugging in $\frac{\pi}{4}$ for the second equation gives a y value of $-\frac{1}{\sqrt2}$ .

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Example Question #1 : Solving Trigonometric Equations

Solve the system for $\theta$ :

$\begin{matrix} y = 6 \sin \theta - 1 \\ y = \sin ^2 \theta + 4 \end{matrix}$

Possible Answers:

no solution

$\pi$

$2 \pi$

$\frac{ \pi }{2 }$

$\frac{3 \pi }{2}$

Correct answer:

$\frac{ \pi }{2 }$

Explanation:

First, set both equations equal to each other:

$6 \sin \theta - 1 = \sin ^2 \theta + 4$ subtract $6 \sin \theta$ from both sides

$-1 = \sin ^2 \theta - 6 \sin \theta + 4$ add 1 to both sides

$0 = \sin ^2 \theta - 6 \sin \theta + 5$

Now we can solve this as a quadratic equation, where "x" is $\sin \theta$ . Using the quadratic formula:

$\sin \theta = \frac{ 6 \pm \sqrt{ 36 - 4(1)(5)}}{2} = \frac{6 \pm \sqrt{16 }}{2} = \frac{ 6 \pm 4 }{2}$

This gives us 2 potential solutions for $\sin \theta$ :

$\sin \theta = \frac{10}{2} = 5$ the sine of an angle cannot be greater than 1

$\sin \theta = \frac{2}{2 } = 1$

$\theta = \sin ^ {-1} (1) = \frac{ \pi }{2}$

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Example Question #1 : Systems Of Trigonometric Equations

Solve this system for $\theta$ :

$\begin{matrix} y = 5 \sin (\frac{ \theta }{2}) + 3 \\ y = 3 \sin (\frac{\theta }{2} ) + 2 \end{matrix}$

Possible Answers:

$\frac{ \pi }{3} , \frac{ 5 \pi }{3}$

$\frac{ 11 \pi }{6} , \frac{ 7 \pi }{6}$

$\frac{ 11 \pi }{3} , \frac{ 7 \pi }{3}$

$\frac{ 11 \pi }{12} , \frac{ 7 \pi }{12}$

$\frac{ \pi }{6} , \frac{ 5 \pi }{6}$

Correct answer:

$\frac{ 11 \pi }{3} , \frac{ 7 \pi }{3}$

Explanation:

First, set the two equations equal to each other

$5 \sin (\frac{ \theta }{2}) + 3 = 3 \sin (\frac{ \theta }{2} ) + 2$ subtract the sine term from the right

$2 \sin (\frac{ \theta } { 2 }) + 3 = 2$ subtract 3 from both sides

$2 \sin (\frac{ \theta }{2} ) = -1$ divide by 2

$\sin (\frac{ \theta }{2} ) = -\frac{1}{2}$

$\frac{ \theta } {2} = \sin ^{-1} (- \frac{1}{2}) = \frac{ 11 \pi }{6} , \frac{ 7 \pi }{6}$ multiply by 2

$\theta = \frac{ 11 \pi }{3} , \frac{ 7 \pi }{3}$

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Trigonometry : Trigonometric Equations

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #6 : Factoring Trigonometric Equations

Example Question #3 : Factoring Trigonometric Equations

Example Question #4 : Factoring Trigonometric Equations

Example Question #21 : Trigonometric Equations

Example Question #22 : Trigonometric Equations

Example Question #185 : Trigonometry

Example Question #1 : Solving Trigonometric Equations

Example Question #1 : Systems Of Trigonometric Equations

Example Question #1 : Solving Trigonometric Equations

Example Question #1 : Systems Of Trigonometric Equations

All Trigonometry Resources

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