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Trigonometry : Trigonometry

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

Example Questions

Example Question #2 : Factoring Trigonometric Equations

Factor the following expression:

$sin^{3}x-1$

Possible Answers:

(sinx+1)(sin^2x+sinx+1)

(sinx-1)(sin^2x+2sinx+1)

(sinx-1)(sin^2x+sinx-1)

(2sinx-1)(sin^2x+sinx+1)

(sinx-1)(sin^2x+sinx+1)

Correct answer:

(sinx-1)(sin^2x+sinx+1)

Explanation:

We know that we can write

a^3-1 in the following form

(a^3-1)=(a-1)(a^2+a+1) .

Now taking a=sinx ,

we have:

sin^3 x-1=(sinx-1)(sin^2 x+sinx+1) .

This is the result that we need.

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Example Question #3 : Factoring Trigonometric Equations

We accept that :

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

What is a simple expression of

$cos^{4}x-sin^{4}x$

Possible Answers:

-sin2x

-2cos2x

2sin(2x)

2cos2x

cos2x

Correct answer:

cos2x

Explanation:

First we see that :

a^4-b^4=(a^2+b^2)(a^2-b^2) .

Now letting a=cosx ,b=sinx

we have

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

We know that :

cos^2x+sin^2x=1 and we are given that

cos^2x-sin^2x=cos2x , this gives

cos^4x-sin^4x=cos2x

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Example Question #2 : Factoring Trigonometric Equations

Factor the following expression:

cos^8x-sin^8x

Possible Answers:

We can't factor this expression.

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

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

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

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

Correct answer:

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

Explanation:

Note first that:

a^8-b^8=(a^4-b^4)(a^4+b^4) and :

(a^4-b^4)=(a^2+b^2)(a^2 - b^2) .

Now taking $a=cosx \quad b=sinx$ . We have

cos^8x-sin^8x=(cos^4x+sin^4x)(cos^4-sin^4x) .

Since cos^4x-sin^4x=(cos^2x+sin^2x)(cos^2x-sin^2x) and cos^2x+sin^2x=1 .

We therefore have :

cos^8x-sin^8x=(cos^4x+sin^4x)(cos^2x-sin^2x)

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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 #2 : Factoring Trigonometric Equations

Factor

$1-cos^{11}x$

Possible Answers:

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

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

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

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

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

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 #23 : 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.

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

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

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

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

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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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #2 : Factoring Trigonometric Equations

Example Question #3 : Factoring Trigonometric Equations

Example Question #2 : Factoring Trigonometric Equations

Example Question #6 : Factoring Trigonometric Equations

Example Question #2 : 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 #23 : Trigonometric Equations

All Trigonometry Resources

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