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Trigonometry : Understanding Trigonometric Equations

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Example Questions

Example Question #171 : Trigonometry

Which of these functions fulfills these criteria?

Amplitude of
Period of
-intercept of
No phase shift
Minimum of

Possible Answers:

$f(x) = 4tan(2\pi x)$

$f(x) = 4sin(2\pi x) + 4$

$f(x) = 4sin(2\pi x)$

$f(x) = 4cos(2\pi x)$

$f(x) = 4cos(2\pi x) + 4$

Correct answer:

$f(x) = 4cos(2\pi x)$

Explanation:

Combining a good deal of our information - an amplitude of , a -intercept of and a minimum of with no phase shift - means we are looking for a cosine function. In other words, we can start right at...

f(x) = 4cos(bx)

We also know that $b = 2\pi$ because there is a period of (since $\frac{2\pi}{b} = 1$ when $b = 2\pi$ . In other words, we can conclude that the function we are looking for is...

$f(x) = 4cos(2\pi x)$

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Example Question #171 : Trigonometry

Which of these functions includes the following points?

(0, 0), (1, 1), (2, 0), (3, -1), (4, 0)

Possible Answers:

$f(x) = cos(\frac{x}{2})$

$f(x) = cos(\frac{\pi x}{2})$

$f(x) = tan(\frac{\pi x}{2})$

$f(x) = sin(\frac{x}{2})$

$f(x) = sin(\frac{\pi x}{2})$

Correct answer:

$f(x) = sin(\frac{\pi x}{2})$

Explanation:

Judging from those five given points, we can draw the following clues:

Amplitude of
Period of

Also, note that none of the answer choices have a phase shift, meaning that you can instantly start looking for a sine function because of the zero-rise-zero-fall-zero period that is marked by those five points. Furthermore, we can also realize that we are looking for...

$f(x) = sin(\frac{\pi x}{2})$

...because we need a period of 4 - i.e,

$\frac{2\pi}{b} = 4$ when $b = \frac{\pi}{2}$ .

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Example Question #173 : Trigonometry

A sine function f(t) where is time measured in seconds has the following properties:

Amplitude of
Minimum of
No phase shift
Frequency of Hz (cycles per second)

f(t) is which of these functions?

Possible Answers:

$f(t) = 5sin(8\pi t) + 5$

$f(t) = 10sin(8\pi t) + 10$

f(t) = 5sin(4t) + 10

$f(t) = 5sin(8\pi t)$

f(t) = 5sin(8t) + 5

Correct answer:

$f(t) = 5sin(8\pi t) + 5$

Explanation:

One important thing to realize is that the frequency is the reciprocal of the period. So if the function has a frequency of Hertz (or cycles per second), the period has to be $\frac{1}{4}$ or 0.25 seconds. Because $\frac{2\pi}{b} = \frac{1}{4}$ when $b = 2\pi * 4 = 8\pi$ , we know we are looking for a equation that includes $8\pi t$ .

Also, because we have an amplitude of but a minimum of , there must be a shift upwards by units. Only one function fulfills those two criteria and the period criteria:

$f(t) = 5sin(8\pi t) + 5$

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Example Question #1 : Factoring Trigonometric Equations

Factor $1-2\cos x+\cos^2 x$ .

Possible Answers:

$(1-\cos x)^2$

$\sin^4 x$

$\cos^4 x$

$\sin x \tan x$

$\sin^2x$

Correct answer:

$(1-\cos x)^2$

Explanation:

Don't get scared off by the fact we're doing trig functions! Factor as you normally would. Because our middle term is negative ( $-2\cos x$ ), we know that the signs inside of our parentheses will be negative.

This means that $1-2\cos x+\cos^2 x$ can be factored to $(1-\cos x)(1-\cos x)$ or $(1-\cos x)^2$ .

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Example Question #1 : Factoring Trigonometric Equations

Which of the following values of in radians satisfy the equation

tan^2(x)-tan(x)=0

$\frac{\pi}{4}$
$\frac{\pi}{2}$
$\frac{3\pi}{4}$

Possible Answers:

3 only

1 only

2 only

1 and 2

1, 2, and 3

Correct answer:

1 only

Explanation:

The fastest way to solve this equation is to simply try the three answers. Plugging in $\frac{\pi}{4}$ gives

$tan^2(\frac{\pi}{4})-tan(\frac{\pi}{4})=(1)^2-1=1-1=0$

Our first choice is valid.

Plugging in $\frac{\pi}{2}$ gives

$tan^2(\frac{\pi}{2})-tan(\frac{\pi}{2})$

However, since $tan(\frac{\pi}{}2)$ is undefined, this cannot be a valid answer.

Finally, plugging in $\frac{3\pi}{4}$ gives

$tan^2(\frac{3\pi}{4})-tan(\frac{3\pi}{4})=(-1)^2-(-1)=1+1=2$

Therefore, our third answer choice is not correct, meaning only 1 is correct.

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Example Question #11 : Trigonometric Equations

$f(x)=\sin^2(x)+\cos^2(x)-\sin(2x)$

Find the zeros of the above equation in the interval

$[0,\pi]$ .

Possible Answers:

$\frac{\pi}{2}$

$2\pi$

$\frac{\pi}{4}$

$\pi$

Correct answer:

$\frac{\pi}{4}$

Explanation:

$f(x)=\sin^2(x)-\sin(2x)+\cos^2(x)=1-2\sin(x)cos(x)$

$=1-2\sin(x)cos(x)=0$

Therefore,

1=2sin(x)cos(x)

$\frac{1}{2}=sin(x)cos(x)$

and that only happens once in the given interval, at $\frac{\pi}{4}$ , or 45 degrees.

$\sin\bigg(\frac{\pi}{4}\bigg)\cos\bigg(\frac{\pi}{4}\bigg)=\frac{1}{\sqrt2}\cdot \frac{1}{\sqrt2}=\frac{1}{2}$

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Example Question #3 : Factoring Trigonometric Equations

Factor the expression

1+2cos(x)sin(x)

Possible Answers:

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

$(cos(x)+sin(x))^{2}$

$(cos(x)+2sin(x))^{2}$

$2(cos(x)+sin(x))^{2}$

$(cos(x)+sin(x))^{3}$

Correct answer:

$(cos(x)+sin(x))^{2}$

Explanation:

We have cos^2x+sin^2x=1 .

Now since 1+2cos(x)sin(x)=cos^2(x)+sin^2(x)+2sin(x)cos(x)

This last expression can be written as :

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

This shows the required result.

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Example Question #2 : Factoring Trigonometric Equations

Factor the following expression:

$sin^{3}x-1$

Possible Answers:

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

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

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

(sinx-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 #5 : 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:

2sin(2x)

-2cos2x

cos2x

2cos2x

-sin2x

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 #5 : Factoring Trigonometric Equations

Factor the following expression:

cos^8x-sin^8x

Possible Answers:

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

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

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

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

We can't factor this expression.

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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Trigonometry : Understanding Trigonometric Equations

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #171 : Trigonometry

Example Question #171 : Trigonometry

Example Question #173 : Trigonometry

Example Question #1 : Factoring Trigonometric Equations

Example Question #1 : Factoring Trigonometric Equations

Example Question #11 : Trigonometric Equations

Example Question #3 : Factoring Trigonometric Equations

Example Question #2 : Factoring Trigonometric Equations

Example Question #5 : Factoring Trigonometric Equations

Example Question #5 : Factoring Trigonometric Equations

All Trigonometry Resources

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