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Precalculus : Pre-Calculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #28 : Trigonometric Identities

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\sin (\alpha - \beta)$ .

Possible Answers:

$\frac {\sqrt{6} - \sqrt{15}}{10}$

$\frac {1- 6\sqrt{10}}{20}$

$\frac {2\sqrt{6} - \sqrt{15}}{20}$

$\frac {2\sqrt{6} + \sqrt{15}}{20}$

$\frac {1+ 6\sqrt{10}}{20}$

Correct answer:

$\frac {1- 6\sqrt{10}}{20}$

Explanation:

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that y = 1 and r = 5 and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2\sqrt{6}}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that x = 1 and r = 4 and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ .

So $\sin \beta = \frac {\sqrt{15}}{4}$ .

Using the sine difference formula, we see:

$\\ \sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \\* = \frac {1}{5} \cdot \frac {1}{4} - \frac {2 \sqrt {6}}{5} \cdot \frac {\sqrt {15}}{4} = \frac {1 - 6 \sqrt {10}}{20}$

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Example Question #31 : Trigonometric Identities

Evaluate

$\sin \left(\frac{5 \pi}{12}\right)$ .

Possible Answers:

$\frac{-1+\sqrt3}{2 \sqrt 2 }$

$\frac{-1 - \sqrt 3 }{\sqrt 2 }$

$\sqrt{\frac{3}{2}}$

$\frac{1+\sqrt3}{2 \sqrt 2 }$

$\frac{1}{\sqrt 2 }$

Correct answer:

$\frac{1+\sqrt3}{2 \sqrt 2 }$

Explanation:

$\frac{5 \pi }{12}$ is equivalent to $\frac{2 \pi }{12 } + \frac{3 \pi }{12 }$ or more simplified $\frac{\pi}{6 } + \frac{\pi }{4}$ .

We can use the sum identity to evaluate this sine:

$\sin \left(\frac{\pi }{6} + \frac{\pi }{4 } \right) = \sin\left (\frac{\pi }{6} \right) \cos \left(\frac { \pi }{4} \right) + \cos \left(\frac{\pi }{6}\right) \sin \left( \frac{ \pi }{4} \right)$

From the unit circle, we can determine these measures:

$\frac{1}{2} \cdot \frac{1}{\sqrt 2 } + \frac{ \sqrt 3 }{2} \cdot \frac{1}{ \sqrt 2 } = \frac{1}{2 \sqrt 2 }+ \frac{\sqrt 3 }{2 \sqrt 2 } = \frac{ 1 + \sqrt 3 }{ 2 \sqrt 2 }$

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Example Question #32 : Trigonometric Identities

Evaluate

$\sin(285 ^ o)$ .

Possible Answers:

$\frac{\sqrt 3 - 1 }{2 \sqrt 2 }$

$\frac{-1 - \sqrt {3 }}{2 \sqrt 2 }$

$\frac{\sqrt 3 - 1 }{4 \sqrt 2 }$

$\frac{1+ \sqrt 3 }{ 2 \sqrt 2 }$

$\frac { 1 - \sqrt 3 }{ 2 \sqrt 2 }$

Correct answer:

$\frac{-1 - \sqrt {3 }}{2 \sqrt 2 }$

Explanation:

The angle 285 ^o = 225^ o + 60 ^ o or 45^o + 240 ^ o .

Using the first one:

$\sin(60^o + 225^o ) = \sin (60^o) \cos (225^o) + \cos (60^o) \cos(225^o )$

We can find these values in the unit circle:

$\frac{\sqrt 3 }{2 } \cdot \frac{-1}{\sqrt 2 } + \frac{1}{2} \cdot \frac{-1}{\sqrt 2 } = \frac{-\sqrt3 }{2 \sqrt 2 } + \frac{-1}{2\sqrt2} = \frac{-\sqrt 3 - 1 }{2 \sqrt 2 }$

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Example Question #33 : Trigonometric Identities

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\tan (\alpha + \beta)$ .

Possible Answers:

$\frac {25\sqrt{6}-32\sqrt{15}}{9}$

$\frac {4}{5}$

$\frac {25\sqrt{6}+32\sqrt{15}}{9}$

$\frac {32\sqrt{6}+25\sqrt{15}}{9}$

$\frac {32\sqrt{6}-25\sqrt{15}}{9}$

Correct answer:

$\frac {32\sqrt{6}+25\sqrt{15}}{9}$

Explanation:

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that y = 1 and r = 5 and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So
$\tan \alpha = \frac {1}{2\sqrt{6}} = \frac {\sqrt{6}}{12}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that x = 1 and r = 4 and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ .

So $\tan \beta = \sqrt{15}$ .

Using the tangent sum formula, we see:

$\\ \tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \\ \\ \\* = \frac{\frac{\sqrt{6}}{12}+\sqrt{15}}{1-\frac{\sqrt{6}}{12} \cdot \frac{\sqrt{15}}{1}}}} = \frac{\frac{\sqrt{6}+12\sqrt{15}}{12}}{\frac{12-3\sqrt{10}}{12}} \\ \\ \\* = \frac{\sqrt{6}+12\sqrt{15}}{12-3\sqrt{10}} \cdot \frac{12+3\sqrt{10}}{12+3\sqrt{10}} \\ \\ \\* = \frac{12\sqrt{6}+6\sqrt{15}+144\sqrt{15}+180\sqrt{6}}{144-90} \\ \\* = \frac{192\sqrt{6}+150\sqrt{15}}{54} = \frac{32\sqrt{6}+25\sqrt{15}}{9}$

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Example Question #34 : Trigonometric Identities

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\tan(\alpha - \beta)$ .

Possible Answers:

$\frac{1}{9}$

$\frac {32\sqrt{6}-25\sqrt{15}}{9}$

$\frac {32\sqrt{6}+25\sqrt{15}}{9}$

$\frac {25\sqrt{6}+32\sqrt{15}}{9}$

$\frac {25\sqrt{6}-32\sqrt{15}}{9}$

Correct answer:

$\frac {32\sqrt{6}-25\sqrt{15}}{9}$

Explanation:

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that y = 1 and r = 5 and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\tan \alpha = \frac {1}{2\sqrt{6}} = \frac {\sqrt{6}}{12}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that x = 1 and r = 4 and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ .

So $\tan \beta = \sqrt{15}$ .

Using the tangent sum formula, we see:

$\\ \tan(\alpha-\beta) = \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta} \\ \\ \\* = \frac{\frac{\sqrt{6}}{12}-\sqrt{15}}{1+\frac{\sqrt{6}}{12} \cdot \frac{\sqrt{15}}{1}}}} = \frac{\frac{\sqrt{6}-12\sqrt{15}}{12}}{\frac{12+3\sqrt{10}}{12}} \\ \\ \\* = \frac{\sqrt{6}-12\sqrt{15}}{12+3\sqrt{10}} \cdot \frac{12-3\sqrt{10}}{12-3\sqrt{10}} \\ \\ \\* = \frac{12\sqrt{6}-6\sqrt{15}-144\sqrt{15}+180\sqrt{6}}{144-90} \\ \\ \\* = \frac{192\sqrt{6}-150\sqrt{15}}{54} = \frac{32\sqrt{6}-25\sqrt{15}}{9}$

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Example Question #35 : Trigonometric Identities

Given that $\tan x = 2$ and $\tan y = 3$ , find $\tan (x+y)$ .

Possible Answers:

Correct answer:

Explanation:

Jump straight to the tangent sum formula:

$\\ \tan (x+y) = \frac {\tan x + \tan y}{1-\tan x \tan y}$

From here plug in the given values and simplify.

$\\ \\ = \frac {2+3}{1-2\cdot3} \\ \\= \frac {5}{-5} \\ \\= -1$

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Example Question #36 : Trigonometric Identities

Which of the following expressions best represents $\tan\left ( 60^{\circ}\right )$ ?

Possible Answers:

$\frac{\tan(60^\circ)}{1-\tan^2(60^\circ)}$

$\frac{\tan(30^\circ)}{1-\tan^2(30^\circ)}$

$\frac{2\tan(30^{\circ})}{1-\tan^2(30^{\circ})}$

$\frac{2\tan(60^\circ)}{1-\tan^2(60^\circ)}$

$\frac{2\tan(15^\circ)}{1-\tan^2(15^\circ)}$

Correct answer:

$\frac{2\tan(30^{\circ})}{1-\tan^2(30^{\circ})}$

Explanation:

Write the identity for $tan(2\theta)$ .

$tan(2x)=\frac{2tan(x)}{1-tan^2(x)}$

Set the value of the angle equal to .

2x=60

x=30

Substitute the value of into the identity.

$\frac{2tan(30)}{1-tan^2(30)}$

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Example Question #37 : Trigonometric Identities

Find the value of cos(15)-cos(75) .

Possible Answers:

$\sqrt6+\sqrt2$

$\frac{\sqrt2}{2}$

$\frac{\sqrt6+\sqrt2}{2}$

$\sqrt2$

Correct answer:

$\frac{\sqrt2}{2}$

Explanation:

To solve cos(15)-cos(75) , we will need to use both the sum and difference identities for cosine.

Write the formula for these identities.

$cos(A\pm B)= cos(A)cos(B)\pm sin(A)sin(B)$

To solve for cos(15) and cos(75) , find two special angles whose difference and sum equals to the angle 15 and 75, respectively. The two special angles are 45 and 30.

A=45

B=30

Substitute the special angles in the formula.

$cos(45\pm 30)= cos(45)cos(30)\pm sin(45)sin(30)$

Evaluate both conditions.

cos(75)=cos(45+30)=cos(45)cos(30)+sin(45)sin(30)

$=\frac{\sqrt2}{2}\left(\frac{\sqrt3}{2}\right)+\frac{\sqrt2}{2}\left(\frac{1}{2}\right)=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}$

cos(15)=cos(45-30)=cos(45)cos(30)-sin(45)sin(30)

$=\frac{\sqrt2}{2}\left(\frac{\sqrt3}{2}\right)-\frac{\sqrt2}{2}\left(\frac{1}{2}\right)=\frac{\sqrt6}{4}-\frac{\sqrt2}{4}$

Solve for cos(15)-cos(75) .

$=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}-\left[\frac{\sqrt6}{4}-\frac{\sqrt2}{4}\right]$

$=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}-\frac{\sqrt6}{4}+\frac{\sqrt2}{4}$

$= \frac{2\sqrt2}{4}$

$= \frac{\sqrt2}{2}$

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Example Question #1 : Fundamental Trigonometric Identities

Simplify $3sin^2\theta-3$ .

Possible Answers:

$3cos^2\theta$

$3-3cos^2\theta$

$-3tan\theta$

$-3cos^2\theta$

Correct answer:

$-3cos^2\theta$

Explanation:

Write the Pythagorean Identity.

$sin^2\theta+cos^2\theta=1$

Reorganize the left side of this equation so that it matches the form: $3sin^2\theta-3$

Subtract cosine squared theta on both sides.

$sin^2\theta-1=-cos^2\theta$

Multiply both sides by 3.

$3(sin^2\theta-1=-cos^2\theta)$

$3sin^2\theta-3=-3cos^2\theta$

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Example Question #1471 : Pre Calculus

Which of the following statements is false?

Possible Answers:

sec(-x^2) = -sec(x^2)

$sin(x)cos(x)tan(x) = \frac{1}{cot(-x)sec(-x)csc(-x)}$

sin x + cos x = -sin(-x) + cos (-x)

csc x tan(-2x) = csc (-x) tan 2x

tan x + tan(-x) = 0

Correct answer:

sec(-x^2) = -sec(x^2)

Explanation:

Of the six trigonometric functions, four are odd, meaning f(-x) = -f(x) . These four are:

sin x
tan x
cot x
csc x

That leaves two functions which are even, which means that f(-x) = f(x) . These are:

cos x
sec x

Of the aforementioned, only sec(-x^2) = -sec(-x^2) is incorrect, since secant is an even function, which implies that sec(-x^2) = sec(x^2)

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #28 : Trigonometric Identities

Example Question #31 : Trigonometric Identities

Example Question #32 : Trigonometric Identities

Example Question #33 : Trigonometric Identities

Example Question #34 : Trigonometric Identities

Example Question #35 : Trigonometric Identities

Example Question #36 : Trigonometric Identities

Example Question #37 : Trigonometric Identities

Example Question #1 : Fundamental Trigonometric Identities

Example Question #1471 : Pre Calculus

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