Trigonometry - Math

Card 0 of 500

Question

Simplify the function below:

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

Answer

We need to use the following formulas:

a) $cos^2x=\frac{1+cos2x}{2}$

and

b) $cos(cos^{-1}x)=x$

We can simplify as follows:

$f(x)=cos^2(\frac{1}{2}cos^{-1}x)=\frac{1+cos(cos^{-1}x)}{2}=\frac{1+x}{2}$

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Question

What is the length of CB?

Answer

$\tan X= \frac{OPPOSITE}{ADJACENT}$

$\tan C=\frac{AB}{CB}$

$CB= \frac{AB}{\tan C}$

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Question

If $\theta$ equals $55^{\circ}$ and is , how long is ?

Answer

This problem can be easily solved using trig identities. We are given the hypotenuse and $\theta (55^{\circ})$ . We can then calculate side using the $\sin$ .

$\sin=\frac{opposite}{hypotenuse}$

$\sin55^{\circ}=\frac{o}{15ft}$

Rearrange to solve for .

$o=\sin55^{\circ}*15ft$

$\dpi{100} \rightarrow 12.3ft$

If you calculated the side to equal then you utilized the $\cos$ function rather than the $\sin$ .

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Question

Solve for .

(Figure not drawn to scale).

Answer

The side-angle-side (SAS) postulate can be used to determine that the triangles are similar. Both triangles share the angle farthest to the right. In the smaller triangle, the upper edge has a length of , and in the larger triangle is has a length of . In the smaller triangle, the bottom edge has a length of , and in the larger triangle is has a length of . We can test for comparison.

$\frac{Upper\ edge_1}{Upper\ edge_2}=\frac{Lower\ edge_1}{Lower\ edge_2}$

$\frac{8}{16}=\frac{6}{12}=\frac{1}{2}$

The statement is true, so the triangles must be similar.

We can use this ratio to solve for the missing side length.

$\frac{Upper\ edge_1}{Upper\ edge_2}=\frac{Lower\ edge_1}{Lower\ edge_2}=\frac{x}{Left\ edge_2}$

To simplify, we will only use the lower edge and left edge comparison.

$\small \frac{6}{12}=\frac{x}{5}$

Cross multiply.

$\small 12x=30$

$\small x=\frac{30}{12}=\frac{5}{2}=2.5$

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Question

Solve for .

(Figure not drawn to scale).

Answer

We can solve using the trigonometric definition of tangent.

$\small tan\Theta =\frac{opp}{adj}$

We are given the angle and the adjacent side.

$\small tan(22^o)=\frac{z}{18}$

We can find $\small tan(22^o)$ with a calculator.

$\small tan(22^o)=0.4$

$0.4=\frac{z}{18}$

$\small 18*0.4=z$

$\small z=7.3$

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Question

Simplify the function below:

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

Answer

We need to use the following formulas:

a) $cos^2x=\frac{1+cos2x}{2}$

and

b) $cos(cos^{-1}x)=x$

We can simplify as follows:

$f(x)=cos^2(\frac{1}{2}cos^{-1}x)=\frac{1+cos(cos^{-1}x)}{2}=\frac{1+x}{2}$

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Question

A triangle has sides of length 12, 17, and 22. Of the measures of the three interior angles, which is the greatest of the three?

Answer

We can apply the Law of Cosines to find the measure of this angle, which we will call :

$c^{2} = a^{2} + b^{2} -2ab \cos C$

The widest angle will be opposite the side of length 22, so we will set:

, ,

$22^{2} = 12^{2} + 17^{2} -2\cdot 12\cdot 17 \cos C$

$484 = 144 + 289 -408 \cos C$

$51= -408 \cos C$

$\cos C = -\frac{51}{408} = -0.125$

$C = \cos^{-1} 0.125 = 97.2^{\circ }$

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Question

In $\Delta ABC$ , $m \angle A = 66^{\circ }$ , , and . To the nearest tenth, what is ?

Answer

By the Law of Cosines:

$\left ( BC\right )^{2} = \left ( AB\right )^{2} + \left ( AC\right )^{2} - 2 \cdot AB \cdot AC \cdot \cos m \angle A$

or, equivalently,

$BC =\sqrt{ \left ( AB\right )^{2} + \left ( AC\right )^{2} - 2 \cdot AB \cdot AC \cdot \cos m \angle A}$

Substitute:

$BC =\sqrt{ 26^{2} +23^{2} - 2 \cdot 26 \cdot 23 \cdot \cos 66 ^{\circ }}$

$\approx \sqrt{ 676 +529 - 1,196 \cdot 0.4067}$

$\approx \sqrt{ 718.6} \approx 26.8$

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Question

In $\Delta ABC$ , , , and . To the nearest tenth, what is $m \angle A$ ?

Answer

By the Triangle Inequality, this triangle can exist, since .

By the Law of Cosines:

$\left ( BC\right )^{2} = \left ( AB\right )^{2} + \left ( AC\right )^{2} - 2 \cdot AB \cdot AC \cdot \cos m \angle A$

Substitute the sidelengths and solve for $\cos m \angle A$ :

$32^{2} = 26^{2} + 23^{2} - 2\cdot 26 \cdot 23 \cdot \cos m \angle A$

$1,024 =676 + 529 - 1,196 \cdot \cos m \angle A$

$1,024 =1,205 - 1,196 \cdot \cos m \angle A$

$-181 = - 1,196 \cdot \cos m \angle A$

$\cos m \angle A = 181 \div 1,196 \approx 0.1513$

$m \angle A \approx \cos^{-1} 0.1513 \approx 81.3 ^{\circ }$

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Question

$\textup{Which of the following expressions is equal to }\frac{\tan \theta }{\sec \theta }:\textup{?}$

Answer

$\textup{Use identities to solve: }\tan\theta=\frac{\sin \theta }{\cos \theta }:::::\sec\theta=\frac{1}{\cos \theta }$

$\frac{\tan \theta }{\sec \theta }=\frac{\sin \theta }{\cos \theta }\times\frac{\cos \theta }{1}=\sin\theta$

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Question

What is the $\sin$ of $\theta$ ?

Answer

When working with basic trigonometric identities, it's easiest to remember the mnemonic: .

$\sin=\frac{opposite}{hypotenuse}$

$\cos=\frac{adjacent}{hypotenuse}$ ,

$\tan=\frac{opposite}{adjacent}$

When one names the right triangle, the opposite side is opposite to the angle, the adjacent side is next to the angle, and the hypotenuse spans the two legs of the right angle.

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Question

What is the $\cos$ of $\theta$ ?

Answer

When working with basic trigonometric identities, it's easiest to remember the mnemonic: .

$\sin=\frac{opposite}{hypotenuse}$

$\cos=\frac{adjacent}{hypotenuse}$ ,

$\tan=\frac{opposite}{adjacent}$

When one names the right triangle, the opposite side is opposite to the angle, the adjacent side is next to the angle, and the hypotenuse spans the two legs of the right angle.

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Question

What is the $\tan$ of $\theta$ ?

Answer

When working with basic trigonometric identities, it's easiest to remember the mnemonic: .

$\sin=\frac{opposite}{hypotenuse}$

$\cos=\frac{adjacent}{hypotenuse}$

$\tan=\frac{opposite}{adjacent}$

When one names the right triangle, the opposite side is opposite to the angle, the adjacent side is next to the angle, and the hypotenuse spans the two legs of the right angle.

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Question

Simplify $(1-\cos^2\theta )\cdot\cot\theta\cdot\csc^2\theta$ .

Answer

Simplifying trionometric expressions or identities often involves a little trial and error, so it's hard to come up with a strategy that works every time. A lot of times you have to try multiple strategies and see which one helps.

Often, if you have any form of $\tan\theta,$ $\cot\theta,$ $\csc\theta,$ or $\sec\theta,$ in an expression, it helps to rewrite it in terms of sine and cosine. In this problem, we can use the identities $\cot\theta=\frac{\cos\theta }{\sin\theta }$ and $\csc\theta=\frac{1}{\sin\theta }$ .

$(1-\cos^2\theta )\cdot\cot\theta\cdot\csc^2\theta=(1-\cos^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot(\frac{1}{\sin\theta})^2$

$=(1-\cos^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}$ .

This doesn't seem to help a whole lot. However, we should recognize that $(1-\cos^2\theta )=\sin^2\theta$ because of the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$ .

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

We can cancel the $\sin^2\theta$ terms in the numerator and denominator.

$(\sin^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}=\frac{\cos\theta }{\sin\theta }=\cot\theta$ .

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Question

$\textup{Which of the following expressions is equal to }\frac{\tan \theta }{\sec \theta }:\textup{?}$

Answer

$\textup{Use identities to solve: }\tan\theta=\frac{\sin \theta }{\cos \theta }:::::\sec\theta=\frac{1}{\cos \theta }$

$\frac{\tan \theta }{\sec \theta }=\frac{\sin \theta }{\cos \theta }\times\frac{\cos \theta }{1}=\sin\theta$

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Question

What is $\theta$ if and ?

Answer

In order to find $\theta$ we need to utilize the given information in the problem. We are given the opposite and adjacent sides. We can then, by definition, find the $\tan$ of $\theta$ and its measure in degrees by utilizing the $\arctan$ function.

$\tan=\frac{opposite}{adjacent}$

$\tan=\frac{10}{8}$

$\tan=1.25$

Now to find the measure of the angle using the $\arctan$ function.

$\theta=\arctan1.25$

$\rightarrow 51.34^{\circ}$

If you calculated the angle's measure to be $0.90^{\circ}}$ then your calculator was set to radians and needs to be set on degrees.

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Question

$o=7.6, h=15.7, \Theta = ?$

Answer

In order to find $\theta$ we need to utilize the given information in the problem. We are given the opposite and hypotenuse sides. We can then, by definition, find the $\sin$ of $\theta$ and its measure in degrees by utilizing the $\arcsin$ function.

$\sin=\frac{opposite}{hypotenuse}$

$\sin=\frac{7.6}{15.7}$

$\sin=0.4841$

Now to find the measure of the angle using the $\arcsin$ function.

$\theta=\arcsin0.4841$

$\rightarrow 28.95^{\circ}$

If you calculated the angle's measure to be $\dpi{100} 0.50^{\circ}}$ then your calculator was set to radians and needs to be set on degrees.

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Question

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

What is ?

Answer

To get rid of $\sin(x^\circ)$ , we take the $\arcsin$ or $\sin^{-1}$ of both sides.

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

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

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

$x^\circ=30^\circ$

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Question

What is the length of CB?

Answer

$\tan X= \frac{OPPOSITE}{ADJACENT}$

$\tan C=\frac{AB}{CB}$

$CB= \frac{AB}{\tan C}$

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Question

If $\theta$ equals $55^{\circ}$ and is , how long is ?

Answer

This problem can be easily solved using trig identities. We are given the hypotenuse and $\theta (55^{\circ})$ . We can then calculate side using the $\sin$ .

$\sin=\frac{opposite}{hypotenuse}$

$\sin55^{\circ}=\frac{o}{15ft}$

Rearrange to solve for .

$o=\sin55^{\circ}*15ft$

$\dpi{100} \rightarrow 12.3ft$

If you calculated the side to equal then you utilized the $\cos$ function rather than the $\sin$ .

Compare your answer with the correct one above

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