Trigonometric Functions - Trigonometry

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Question

In this figure, if angle $c=90^\circ$ , side , and side , what is the measure of angle ?

Answer

Since $c=90^\circ$ , we know we are working with a right triangle.

That means that $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$ .

In this problem, that would be:

$\sin(a)=\frac{\text{Z}}{\text{X}}$

Plug in our given values:

$\sin(a)=\frac{8}{12}$

$\sin(a)=\frac{2}{3}$

$a=\sin^{-1}(\frac{2}{3})$

$a=41.8^\circ$

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Question

If $cos\ x=A$ , give $cos\ 2x$ in terms of .

Answer

We need to use the identity .

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Question

If sin 26o = t, find the value of cos 52o in terms of t.

Answer

Since $52^{\circ}=2\times 26^{\circ}$ , we can use a double angle formula:

$cos\ 2x=1-2sin^2x\Rightarrow cos\ 52^{\circ}=1-2sin^2\ (26)^{\circ}$

Substituting $sin \26^{\circ}=t$ , we get $cos\ 52^{\circ}=1-2sin^2\ (26)^{\circ}=1-2t^2$ .

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Question

If $cos\ x=\frac{\sqrt{2}}{2}$ , find $cos\ 2x$ .

Answer

We need to use the identity .

$cos2x=2cos^2x-1=2(\frac{\sqrt{2}}{2})^2-1=2(\frac{1}{2})-1=1-1=0$

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Question

If $sin\ x=A$ , express $sin\ 2x$ in terms of . ( $0<x<30^{\circ}$ )

Answer

We need to use the double angle formula:

$sin\ 2x=2(sin\ x)(cos\ x)$

$sin\ x$ is known, but we need to find $cos \ x$ :

$sin^2x+cos^2x=1\Rightarrow cos\ x=\pm \sqrt{1-sin^2x}$

In this problem is a first quadrant angle ( $0<x<30^{\circ}$ ), so we can only use the positive value for $cos\ x$ .

$cos\ x=\sqrt{1-sin^2x}=\sqrt{1-A^2}$

$sin\ 2x=2(sin\ x)(cos\ x)=2A\sqrt{1-A^2}$

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Question

If $tan\ x=\sqrt{3}$ , what is the value of $cos\ x$ ? ( $0^{\circ}<x<90^{\circ}$ )

Answer

We need to use a Pythagorean identity:

$sec^2x=1+tan^2x=1+(\sqrt{3})^2=1+3=4\Rightarrow sec\ x=\pm 2$

Since $0^{\circ}<x<90^{\circ}$ , we can only use the positive value of $sec\ x$ . That means $sec\ x=2$ .

$cos\ x=\frac{1}{sec\ x}=\frac{1}{2}=0.5$

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Question

If $cot\ x=\sqrt{3}$ , what is the value of $cos\ x$ ? ( $0^{\circ}<x<90^{\circ}$ )

Answer

We need to use a Pythagorean identity:

$csc^2x=1+cot^2x=1+(\sqrt{3})^2=1+3=4\Rightarrow csc\ x=\pm 2$

Since $0^{\circ}<x<90^{\circ}$ , we can only use the positive value for $csc\ x$ . That means $csc\ x=2$ .

$sin\ x=\frac{1}{csc\ x}=\frac{1}{2}$

Based on another Pythagorean identity we have:

$cos^2x+sin^2x=1\Rightarrow cos^2x=1-sin^2x=1-(\frac{1}{2})^2=1-\frac{1}{4}=\frac{3}{4}$

$\Rightarrow cos\ x=\pm \sqrt{(\frac{3}{4})}=\pm \frac{\sqrt{3}}{2}$

Since $0^{\circ}<x<90^{\circ}$ , we can again only use the positive value for $cos\ x$ .

$cos\ x=\frac{\sqrt{3}}{2}$

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Question

$sin (cot^{-1}(4/3)) = ?$

Answer

To solve this, start by setting up a triangle that has an angle with a cotangent of 4/3. This triangle would therefore have two legs of lengths 4 and 3, with the side of length 3 being opposite the angle in question. (Cotangent is adjacent over hypotenuse.) Note that the angle itself does not have to be solved for; we just need to find its sine. To do that, we first need to find the length of the hypotenuse using the Pythagorean Theorem:

$3^{2} + 4^{2} = c^{2}$

Solving this gives a hypotenuse of length 5. Now, the sine of this angle is opposite (i.e. the side of length 3) over hypotenuse (length 5), which gives an answer of $\frac{3}{5}$ .

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Question

If $sin^2x+2cos^2x+3(sin\ x)(cos\ x)=3$ , give the value of $cot\ x$ .

Answer

$sin^2x+2cos^2x+3(sin\ x)(cos\ x)=3$

$\Rightarrow sin^2x+(cos^2x+cos^2x)+3(sin\ x)(cos\ x)=3$

Use the identity .

$\Rightarrow 1+cos^2x+3(sin\ x)(cos\ x)=3\Rightarrow cos^2x+3(sin\ x)(cos\ x)=3-1=2$

Now we should divide both sides by :

$\Rightarrow \frac{cos^2x}{sin^2x}+\frac{3(sin\ x)(cos\ x)}{sin^2x}=\frac{2}{sin^2x}$

We can use the identity $\frac{1}{sin^2x}=1+cot^2x$ .

$\Rightarrow cot^2x+3cot\ x=2(1+cot^2x)$

$\Rightarrow cot^2x+3cot\ x=2+2cot^2x$

$\Rightarrow 2cot^2x-cot^2x-3cot\ x+2=0$

$\Rightarrow cot^2x-3cot\ x+2=0$

$\Rightarrow (cot\ x-1)(cot\ x-2)=0$

$cot\ x-1=0\Rightarrow cot\ x=1$

or

$cot\ x-2=0\Rightarrow cot\ x=2$

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Question

If $sin\ x+\frac{1}{sin\ x}=2$ , find the value of .

Answer

$sin\ x+\frac{1}{sin\ x}=2\Rightarrow \frac{sin^2x+1}{sin\ x}=2$

$\Rightarrow sin^2x+1=2sin\ x$

$\Rightarrow sin^2x-2sin\ x+1$

$\Rightarrow (sinx-1)^2=0$

$\Rightarrow sin\ x-1=0$

$\Rightarrow sin\ x=1$

Therefore, .

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Question

If $x=\frac{\pi}{6}$ , find the value of $\frac{1+cos\ x}{sin^3x}$ .

Answer

$x=\frac{\pi}{6}\Rightarrow \frac{{1+cos\ x}}{}{sin^3x}=\frac{{1+cos\ (\pi/6)}}{sin^3(\pi/6)}=\frac{1+\frac{\sqrt{3}}{2}}{(\frac{1}{2})^3}=4(2+\sqrt{3})=8+4\sqrt{3}$

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Question

Find the value of the following expression:

$sin^2 (\frac{\pi}{15})+\frac{1}{sec^2(\frac{\pi}{15})}$

Answer

We know that $sec\ x= \frac{1}{cos\ x}$ .

$sin^2 (\frac{\pi}{15})+\frac{1}{sec^2(\frac{\pi}{15})}=sin^2 (\frac{\pi}{15})+\frac{1}{\frac{1}{cos^2(\frac{\pi}{15})}}=sin^2 (\frac{\pi}{15})+cos^2 (\frac{\pi}{15})=1$

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Question

If $sec\ x=\frac{\sqrt{5}}{2}$ , give $tan\ x$ :

Answer

We need to use the identity :

$sec^2x=1+tan^2x\Rightarrow (\frac{\sqrt{5}}2)^2=1+tan^2x$

$\Rightarrow tan^2x=\frac{5}{4}-1\Rightarrow tan^2x=\frac{1}{4}$

$\Rightarrow tan\ x=\pm \frac{1}{2}$

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Question

Give the value of $sec\ 26^{\circ} - csc\ 64^{\circ}$ .

Answer

We need to use the identity $csc(90^{\circ}-x)=sec\ x$ :

$sec\ 26^{\circ} - csc\ 64^{\circ}=sec\ 26^{\circ} - csc (90-26)^{\circ}=sec\ 26^{\circ} -sec\ 26^{\circ}=0$

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Question

Find $\frac{cos\ 65^{\circ}}{sin\ 25^{\circ}}$ .

Answer

We need to use the identity $cos(90^{\circ}-x)=sin\ x$ :

$\frac{cos\ 65^{\circ}}{sin\ 25^{\circ}}=\frac{cos (90-25)^{\circ}}{sin\ 25^{\circ}}=\frac{sin\ 25^{\circ}}{sin\ 25^{\circ}}=1$

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Question

Which is not true about the following function?

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

Answer

Breaking down the equation piece by piece, if we look at it as :

. This means both that we do have an amplitude of and a minimum of because .

(because it is not present), meaning that we do have no phase shift.

$b = \pi$ . This means that our period is $\frac{2\pi}{\pi} = 2$ , which means that we do not have a period of 1 and that we therefore have our answer.

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Question

What is the phase shift of $f(x) = sin(2\pi x - \pi)$ ?

Answer

Remember that when a trigonometric equation is written as...

...then the phase shift is (instead of simply .) In this problem, $c = \pi$ (and take careful that you do not set $c = -\pi$ by mistake) while $b = 2\pi$ . Therefore, our phase shift is $\frac{c}{b} = \frac{\pi}{2\pi} = \frac{1}{2}$ .

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Question

In this cosine function, is time measured in seconds:

$f(t) = 3cos(16\pi x - \pi) + 3$

Which is not true about this function?

Answer

Examining the equation based on this form:

. This means that we do have both an amplitude of since and a minimum of since .

$c = \pi$ . This is not relevant until we examine .

$b = 16\pi$ . Let's examine what this means step by step:

Phase shift of $\frac{1}{4}$ is confirmed because our phase shift is equal to $\frac{c}{b} = \frac{1\pi}{16\pi} = \frac{1}{16}$ .

Frequency of Hertz is confirmed. Here is why: Our period is equal to $\frac{2\pi}{b} = \frac{2\pi}{16\pi} = \frac{1}{8}$ seconds, which means that the frequency is $(\frac{1}{8})^{-1} = 8$ cycles per second.

The phase shift and period are the important points here. If our phase shift were a multiple of our period, then our -intercept would be our maximum, which is , because all we are doing in that case is shifting by an amount of periods or cycles. However, $\frac{1}{16}$ is not a multiple of $\frac{1}{8}$ , which means this is not the case and that we do not have a -intercept of (and it can absolutely help if you graph the function to check this.)

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Question

If you want to roughly approximate an EKG (pulse/heart beat diagram) of a person with a pulse of beats per minute using a sine function, what would be equal to in the following equation?

, where is measured in seconds

Answer

Firstly, we need to realize that because time in this function is measured in seconds and we need to produce a function that approximates heartbeats in seconds (or periods in seconds), our frequency is...

$\frac{90}{60} = \frac{3}{2}$ beats per second.

We can take the reciprocal from here to get our period for a single heartbeat:

$(\frac{3}{2})^{-1} = \frac{2}{3}$ seconds.

Finally, since we know that $\frac{2\pi}{b}$ must be our period, we can solve for using algebra.

$\frac{2\pi}{b} = \frac{2}{3}$

$2b = 6\pi$

$b = \frac{6\pi}{2} = 3\pi$

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Question

What is the y-intercept of ?

Answer

Step 1: Find the value of cos(0)

,

The graph of starts from

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