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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 #31 : Use The Laws Of Cosines And Sines

Find the length of the missing side,

Possible Answers:

8.10

6.91

5.92

6.37

Correct answer:

6.37

Explanation:

First, use the Law of Sines to find the measurement of angle

$\frac{5}{\sin 50}=\frac{3}{\sin C}$

$5\sin C = 2.30$

$\sin C=0.46$

C= 27.39

Recall that all the angles in a triangle need to add up to 180 degrees.

50+27.39+B=180

B=102.61

Now, use the Law of Sines again to find the length of .

$\frac{d}{\sin 102.61}=\frac{5}{\sin 50}$

$d \sin 50= 4.88$

d=6.37

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Example Question #32 : Use The Laws Of Cosines And Sines

What is the largest possible angle, measured in degrees, in triangle ABC if $\angle A = 30^\circ$ , AB = 10 , and BC = 6 ?

Possible Answers:

$107.56^\circ$

$144.56^\circ$

$156.56^\circ$

$123.56^\circ$

$93.56^\circ$

Correct answer:

$123.56^\circ$

Explanation:

In the ambiguous SSA case, use Law of Sines to solve for the angle opposite the given side.

$\\* \frac {\sin A}{a} = \frac {\sin C}{c} \\* \\* \frac {\sin 30^\circ}{6} = \frac {\sin C}{10} \\* \\* C = \arcsin \left(\frac {10 \sin 30^\circ}{6}\right) \approx 56.44^\circ$

If $C = 56.44^\circ$ , then $B = 93.56^\circ$ .

However! Another possible value of C is $(180 - 56.44)^\circ = 123.56^\circ$ .

In this case the angles will be $30^\circ, 123.56^\circ, 26.44^\circ$ .

This is bigger than $93.56^\circ$ and is consequently the answer.

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Example Question #1 : Sum And Difference Identities

Evaluate the exact value of:

2cos(75)

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to solve 2cos(75) , two special angles will need to be used to solve for the exact values.

The angles chosen are A=30 and B=45 degrees, since:

A+B=30+45 =75

Write the formula for the cosine additive identity.

cos(A+B)= cosAcosB-sinAsinB

Substitute the known variables.

$cos(30)cos(45)-sin(30)sin(45)=\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt2}{2}\right)$

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

$2cos(75)=2\left(\frac{\sqrt6-\sqrt2}{4}\right)=\frac{\sqrt6-\sqrt2}{2}$

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Example Question #21 : 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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 cosine sum formula, we then see:

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

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Example Question #22 : 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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 cosine difference formula, we see:
$\\ \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\\ \\* = \frac {2 \sqrt {6}}{5} \cdot \frac {1}{4} + \frac {1}{5} \cdot \frac {\sqrt {15}}{4} = \frac {2 \sqrt{6} + \sqrt {15}}{20}$

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Example Question #1461 : Pre Calculus

Find $\sin\bigg(\frac{\pi }{3} + \frac{\pi }{6}\bigg)$ using the sum identity.

Possible Answers:

$\frac{\pi }{2}$

1.5

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

Correct answer:

Explanation:

Using the sum formula for sine,

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

where,

$a=\frac{\pi}{3}$ , $b=\frac{\pi}{6}$

yeilds:

$\sin \bigg(\frac{\pi}{3}+\sin\frac{\pi}{6}\bigg)=\sin\bigg(\frac{\pi }{3}\bigg)\cdot\cos\bigg(\frac{\pi }{6}\bigg) +\cos \bigg(\frac{\pi}{3}\bigg)\cdot\sin\bigg(\frac{\pi }{6}\bigg)$

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

$=\frac{3}{4}+\frac{1}{4}=1$ .

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Example Question #1462 : Pre Calculus

Calculate $\sin(75)$ .

Possible Answers:

$\frac{\sqrt{2}+\sqrt{6}}{4}$

$\frac{1}2{}$

$\frac{6+\sqrt{2}}{4}$

Correct answer:

$\frac{\sqrt{2}+\sqrt{6}}{4}$

Explanation:

Notice that $\sin(75)$ is equivalent to $\sin(30+45)$ . With this conversion, the sum formula can be applied using,

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

where

a=30 , b=45 .

Therefore the result is as follows:

$\sin(a+b)=\sin(30)\cdot \cos(45)+\cos(30)\cdot\sin(45)$

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

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

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Example Question #25 : Trigonometric Identities

Find the exact value for: $4\sin (15^{\circ})$

Possible Answers:

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

$\sqrt6-\sqrt2$

$\sqrt2-\frac{\sqrt2}{3}$

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

Correct answer:

$\sqrt6-\sqrt2$

Explanation:

In order to solve this question, it is necessary to know the sine difference identity.

sin(A-B)=sinAcosB-cosAsinB

The values of and must be a special angle, and their difference must be 15 degrees.

A possibility of their values that match the criteria are:

A=45

B=30

Substitute the values into the formula and solve.

sin(45-30)=(sin45)(cos30)-(cos45)(sin30)

$sin(15)=(\frac{\sqrt2}{2})(\frac{\sqrt3}{2})-(\frac{\sqrt2}{2})(\frac{1}{2})= \frac{\sqrt6}{4}-\frac{\sqrt2}{4}$

Evaluate 4sin(15) .

$=4(\frac{\sqrt6}{4}-\frac{\sqrt2}{4})= \sqrt6-\sqrt2$

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Example Question #26 : Trigonometric Identities

Find the exact value of: $\sin(105^{\circ})$

Possible Answers:

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

$\frac{\sqrt2+\sqrt5}{4}$

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

$\frac{\sqrt2+3\sqrt2}{4}$

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

Correct answer:

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

Explanation:

In order to find the exact value of sin(105) , the sum identity of sine must be used. Write the formula.

sin(A+B)=sinAcosB+cosAsinB

The only possibilites of and are 45 and 60 degrees interchangably. Substitute these values into the equation and evaluate.

sin(45+60)=sin45cos60+cos45sin60

$sin(105)=(\frac{\sqrt2}{2})(\frac{1}{2})+(\frac{\sqrt2}{2})(\frac{\sqrt3}{2})=\frac{\sqrt2+\sqrt6}{4}$

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Example Question #27 : 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:

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

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

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

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

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

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 sum 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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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #31 : Use The Laws Of Cosines And Sines

Example Question #32 : Use The Laws Of Cosines And Sines

Example Question #1 : Sum And Difference Identities

Example Question #21 : Trigonometric Identities

Example Question #22 : Trigonometric Identities

Example Question #1461 : Pre Calculus

Example Question #1462 : Pre Calculus

Example Question #25 : Trigonometric Identities

Example Question #26 : Trigonometric Identities

Example Question #27 : Trigonometric Identities

All Precalculus Resources

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