Sin, Cos, Tan - Trigonometry

Card 0 of 60

Question

If cos x = 0.2 $\small \cosx$ and sin x = 0.4, what is the value of tan x?

Answer

$\small \tan{x}=\frac{\sin{x}}{\cos{x}}$

$\small \small \tan{x}=\frac{0.4}{0.2}$

$\small \tan{x}=0.035$

$\small x=\tan^{-1}0.035$

$\small x=2$

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Question

Find the value of the trigonometric function in fraction form for triangle .

What is the cosine of $\angle B$ ?

Answer

The cosine of an angle is the value of the adjacent side over the hypotenuse.

Therefore:

$cos \angle B = \frac{adjacent}{hypotenuse} = \frac{7}{25}$

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Question

Which of the following describes the ratio of sine?

Answer

Sine is by definition of sides in a right triangle is opposite side over hypotenuse.

To remember this, use SOH CAH TOA.

SOH: Sine=Opposite/Hypotenuse

CAH: Cosine=Adjacent/Hypotenuse

TOA: Tangent=Opposite/Adjacent

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Question

What is the value of ?

Answer

Solve each term separately.

$cos(30)= \frac{\sqrt3}{2}$

$cos(60)=\frac{1}{2}$

Add both terms.

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

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Question

Determine the value of .

Answer

Rewrite in terms of sines and cosines.

$2tan(120)=2(\frac{sin(120)}{cos(120)})=2(\frac{\frac{\sqrt3}{2}}{-\frac{1}{2}})$

Simplify the complex fraction.

$2(\frac{\frac{\sqrt3}{2}}{-\frac{1}{2}})= 2\times\frac{\sqrt3}{2}\times-2 = -2\sqrt3$

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Question

Find the value of $\frac{1}{2}sin(45)+ tan(60)$ .

Answer

To find the value of $\frac{1}{2}sin(45)+ tan(60)$ , solve each term separately.

$\frac{1}{2}sin(45)=\frac{1}{2} \cdot \frac{\sqrt2}{2} = \frac{\sqrt2}{4}$

$tan(60) = \sqrt3$

Sum the two terms.

$\frac{\sqrt2}{4}+\sqrt3 = \frac{\sqrt2}{4}+\frac{4\sqrt3}{4} = \frac{\sqrt2+4\sqrt3}{4}$

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Question

Select the ratio that would give Tan B.

Answer

We need the Tan B. Which side lengths correspond to this ratio?

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

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Question

Calculate $tan(\frac{4\pi}{3})$ .

Answer

The tangent function has a period of $\pi$ units. That is,

$tan(x+n\pi)=tanx$

for all $n\in\mathbb{Z}$ .

Since $\frac{4\pi}{3}=\frac{\pi}{3}+\pi$ , we can rewrite the original expression $tan(\frac{4\pi}{3})$ as follows:

$tan(\frac{4\pi}{3})=tan(\frac{\pi}{3}+\pi)$

$=tan(\frac{\pi}{3})$

$=\frac{sin(\frac{\pi}{3})}{cos(\frac{\pi}{3})}$

$=\frac{(\frac{\sqrt{3}}{2})}{(\frac{1}{2})}$

$=\sqrt{3}$

Hence,

$tan(\frac{4\pi}{3})=\sqrt{3}$

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Question

Calculate $cos(\frac{5\pi}{3})$ .

Answer

First, convert the given angle measure from radians to degrees:

$cos(\frac{5\pi}{3})=cos(300^{\circ})$

Next, recall that $300^{\circ}$ lies in the fourth quadrant of the unit circle, wherein the cosine is positive. Furthermore, the reference angle of $300^{\circ}$ is

$360^{\circ}-300^{\circ}=60^{\circ}$

Hence, all that is required is to recognize from these observations that

$cos(\frac{5\pi}{3})=cos(300^{\circ}) =cos(60^{\circ})$ ,

which is $\frac{1}{2}$ .

Therefore,

$cos(\frac{5\pi}{3})=\frac{1}{2}$

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Question

What is the result when the following expression is simplified as much as possible?

$\sin(2h)\sec(h)+2\sin(-h)$

Answer

Because $\sin x$ is an odd function, we can rewrite the second term in the expression.

$2\sin(-h)=-2\sin h$ .

We now use a double-angle formula to expand the first term.

$\sin(2h)\sec h=2\sin h\cos h\sec h$ .

Because they are reciprocals, $\cos h\sec h=1$ .

$2\sin h\cos h\sec h-2\sin h=2\sin h-2\sin h=0.$

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Question

Round to the nearest hundredth.

Use your calculator to find:

$\sin72^{\circ}$

Answer

Before plugging the function into the calculator make sure the mode of the calculator is set to degrees,

Plug in $\sin72^{\circ}$ which equals to .

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Question

Round to the nearest hundredth.

Use your calculator to find:

$\cos \frac{3\pi}{8}$

Answer

Before plugging the function into the calculator make sure the mode of the calculator is set to radians,

plug in

$\cos \frac{3\pi}{8}$

which is equal to .

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Question

$\left ( \sin15 + \cos\15 \right )^{2}$ What is the closest value to this expression?

Answer

First you have to know the equation for $\left ( a+b\right )^{2} = a^{2}+2ab+b^{2}$ and apply it to our equation. Second the formula of $\sin^{2}+\cos^{2}=1$ will allow you to get rid of the square sin and cos. Third the equation $\sin2x = 2\sin \cos$ will allow you to get rid of the . Therefore, we will have a $\sin(30)$ , and that will be equal to . Sum and and you will get

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Question

if $\cos x = \frac{4}{5}$ What is $\sin2x$ ?

Answer

Remember two things. First if $\cos x=\frac{4}{5}$ , find the $\sin(x)$ by using the Pythagoras Theorem. If one side is and the hypotenuse is , then the other side is . $\sin x$ will be $\frac{3}{5}$ . Finally remember the formula for $\sin2x = 2\sin x *\cos x$ . And just place the things we found to the equation.

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Question

What is the value of cos30o . sin30o . tan60o

Answer

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Question

If cos x = 0.2 $\small \cosx$ and sin x = 0.4, what is the value of tan x?

Answer

$\small \tan{x}=\frac{\sin{x}}{\cos{x}}$

$\small \small \tan{x}=\frac{0.4}{0.2}$

$\small \tan{x}=0.035$

$\small x=\tan^{-1}0.035$

$\small x=2$

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Question

Find the value of the trigonometric function in fraction form for triangle .

What is the cosine of $\angle B$ ?

Answer

The cosine of an angle is the value of the adjacent side over the hypotenuse.

Therefore:

$cos \angle B = \frac{adjacent}{hypotenuse} = \frac{7}{25}$

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Question

Which of the following describes the ratio of sine?

Answer

Sine is by definition of sides in a right triangle is opposite side over hypotenuse.

To remember this, use SOH CAH TOA.

SOH: Sine=Opposite/Hypotenuse

CAH: Cosine=Adjacent/Hypotenuse

TOA: Tangent=Opposite/Adjacent

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Question

What is the value of ?

Answer

Solve each term separately.

$cos(30)= \frac{\sqrt3}{2}$

$cos(60)=\frac{1}{2}$

Add both terms.

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

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Question

Determine the value of .

Answer

Rewrite in terms of sines and cosines.

$2tan(120)=2(\frac{sin(120)}{cos(120)})=2(\frac{\frac{\sqrt3}{2}}{-\frac{1}{2}})$

Simplify the complex fraction.

$2(\frac{\frac{\sqrt3}{2}}{-\frac{1}{2}})= 2\times\frac{\sqrt3}{2}\times-2 = -2\sqrt3$

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