1. find cot θ if csc θ = square root of five divided by two and tan θ > 0. (6 points) a) square root of five b) 2 c) square root of five divided by five d) one half 2. find cos θ if sin θ = negative twelve divided by thirteen and tan θ > 0. (6 points) a) negative five divided by twelve b) negative five divided by thirteen c) twelve divided by five d) negative thirteen divided by twelve 3. use basic identities to simplify the expression. (6 points) tangent of theta divided by secant of theta. a) cos3θ b) tan2θ c) sec2θ d) sin θ 4. simplify the expression. (6 points) sine of x to the second power minus one divided by cosine of negative x a) -sin x b) cos x c) sin x d) -cos x 5. find all solutions in the interval [0, 2π). (6 points) sin2x + sin x = 0 a) x = 0, π, four pi divided by three , five pi divided by three b) x = 0, π, pi divided by three , two pi divided by three c) x = 0, π, pi divided by three , five pi divided by three d) x = 0, π, three pi divided by two 6. find all solutions to the equation. cos x = sin x (6 points) a) pi divided by four plus n pi such that n equals zero, plus or minus one, plus or minus two to infinity b) pi divided by four plus two n pi comma seven pi divided by four plus two n pi such that n equals zero, plus or minus one, plus or minus two to infinity c) pi divided by two plus n pi such that n equals zero, plus or minus one, plus or minus two to infinity d) three pi divided by four plus two n pi comma five pi divided by four plus two n pi such that n equals zero, plus or minus one, plus or minus two to infinity 7. find an exact value. sin 105° (6 points) a) quantity square root of six plus square root of two divided by four b) quantity negative square root of six plus square root of two divided by four c) quantity square root of six minus square root of two divided by four d) quantity negative square root of six minus square root of two divided by four 8. find an exact value. (6 points) tangent of seven pi divided by twelve a) quantity two minus square root of three divided by four b) -2 - square root of three c) quantity two plus square root of three divided by four d) 2 + square root of three 9. write the expression as the sine, cosine, or tangent of an angle. (6 points) sin 57° cos 13° - cos 57° sin 13° a) cos 70° b) cos 44° c) sin 44° d) sin 70° 10. write the expression as the sine, cosine, or tangent of an angle. (6 points) cos 94° cos 37° + sin 94° sin 37° a) sin 131° b) sin 57° c) cos 131° d) cos 57° 11. write the expression as the sine, cosine, or tangent of an angle. (6 points) sin 5x cos x - cos 5x sin x a) cos 6x b) cos 4x c) sin 6x d) sin 4x 12. rewrite with only sin x and cos x. (6 points) sin 2x - cos 2x a) 2 sin2x - 2 sin x cos x + 1 b) 2 sin x c) 2 sin2x + 2 sin x cos x - 1 d) 2 sin2x - 2 sin x cos x - 1 13. find the exact value by using a half-angle identity. (6 points) cosine of pi divided by twelve a) one half times the square root of quantity one plus square root of three b) one half times the square root of quantity one minus square root of three c) one half times the square root of quantity two plus square root of three d) one half times the square root of quantity two minus square root of three 14. verify the identity. (7 points) 4 csc 2x = 2 csc2x tan x 15. verify the identity. (8 points) cotangent of x divided by quantity one plus cosecant of x equals quantity cosecant of x minus one divided by cotangent of x 16. verify the identity. (7 points) sin quantity x plus pi divided by two = cos x i would really appreciate priority for this! you!

1. find cot θ if csc θ = square root of five divided by two and tan θ > 0. (6 points) a) square root of five b) 2 c) square

Answers

rowdycar313p0ao5k

Step-by-step explanation:

See figure 1 attached

Radius of circle equal 1. This radius is at the same time the hypotenuse of triangle OMP . You can see:

sin∠POM = opposite leg/hypotenuse given that hypotenuse is 1

sin∠POM = opposite leg = PM Note PM never change sign when

rotating from 0 up to π/2 (quadrant one). Its value will be

0 ≤ sin∠POM ≤ 1

cos∠POM = adjacent leg/hypotenuse /hypotenuse given that hypotenuse is 1 then for the same reason

cos∠POM = adjacent leg = OM

OM never change sign in the first quadrant, and can tak vals beteen 1 for 0° up to 1 for π/2

Tan∠POM = sin∠POM /cos∠POM

The last relation is always positive (in the first quadrant) and

tan∠POM = opposite leg/adjacent leg

Determine the value of the signs of the functional values of sine, cosine and tangent in quadrant on

lzrd3rik

STEP 3

Step-by-step explanation:

Francesca drew point (–2, –10) on the terminal ray of angle $\Theta$ , which is in standard position. She found values for the six trigonometric functions using the steps below.

Step 1

A unit circle is shown. A ray intersects point (negative 2, negative 10) in quadrant 3. Theta is the angle formed by the ray and the x-axis in quadrant 1.

Step 2

$r = (\sqrt{(-2)^2+(-10)^2}=\sqrt{104}=2\sqrt{26$

Step 3

$Sin \theta = \frac{-2}{2\sqrt{26} }=-\frac{1}{\sqrt{26} }=-\frac{\sqrt{26}}{26 }$

$cos \theta = \frac{-10}{2\sqrt{26} }=-\frac{5}{\sqrt{26} }=-\frac{5\sqrt{26}}{26 }$

$tan \theta = \frac{2}{-10 }=-\frac{1}{5}\\$

$cosec \theta = \frac{1}{sin \theta}=\frac{1}{-\frac{\sqrt{26}}{26 }}=-\frac{\sqrt{26}}{5}$

$Secant \theta = 1/cos \theta =\frac{1}{-\frac{5\sqrt{26} }{26} } =-\frac{\sqrt{26}}{5} \\cotangent \theta=1/tan \theta=\frac{1}{1/5} =5$

Francesca made her first error in step 3 because the sine, cosine, and tangent ratios are incorrect, which also resulted in incorrect cosecant, secant, and tangent functions.

The correct values are:

$Sin \theta = \frac{-10}{2\sqrt{26} }=-\frac{5}{\sqrt{26} }=-\frac{5\sqrt{26} }{26}\\cos \theta = \frac{-2}{2\sqrt{26} }=-\frac{1}{\sqrt{26} }=-\frac{\sqrt{26} }{26}\\tan \theta = \frac{-10}{-2}=5$

ValerieRubyPena

Section 1.

sin∠X =5/13

cos∠X =12/13

tan∠X = 5/12

Step-by-step explanation:

The sine of an angle is defined as; Opposite side/the Hypotenuse. From the right angle triangle given; the opposite side of angle X is 5 while the hypotenuse is 13.

The cosine of an angle is defined as; Adjacent side/Hypotenuse. From the right angle triangle given; the adjacent side of angle X is 12.

The tangent of an angle is defined as; Opposite side/Adjacent side.

Section 2

sin∠Y = 12/13

cos∠Y = 5/13

tan∠Y = 12/5

Step-by-step explanation:

The sine of an angle is defined as; Opposite side/the Hypotenuse. From the right angle triangle given; the opposite side of angle Y is 12 while the hypotenuse is 13.

The cosine of an angle is defined as; Adjacent side/Hypotenuse. From the right angle triangle given; the adjacent side of angle Y is 5.

The tangent of an angle is defined as; Opposite side/Adjacent side.

Section 3

The sin∠X and the cos∠Y are equal, their value is 5/13.

Step-by-step explanation:

The sine of angle is always equal to the cosine of its complement. Complement angles add up to 90 degrees. In this case, ∠X+∠Y =90 hence ∠X is a complement of ∠Y.

Section 4

The tangents of ∠X and ∠Y are reciprocals of each other. That is;

tan∠X = 5/12 and tan∠Y = 12/5. Clearly; tan∠Y = 1/tan∠X .

Step-by-step explanation:

The tangent of an angle will always be equal to the reciprocal of the tangent of its complement. In this case, ∠X+∠Y =90 hence ∠X is a complement of ∠Y.

zara76

Option C is correct.

Step-by-step explanation:

$\theta=\frac{5\pi }{6}$

We need to find reference angle and signs of sinФ, cosФ and tanФ

We know that $\theta=\frac{5\pi }{6}radians$ is equal to 150°

and 150° is in 2nd quadrant.

So, Ф is in 2nd quadrant.

And In 2nd quadrant sine is positive, while cos and tan are negative

The reference angle Ф' is found by: π - Ф

=> Ф = 5π/6

so, Reference angle Ф' = π - 5π/6

Ф' = 6π - 5π/6

Ф' = π/6

So, Option C Θ' = pi over 6; sine is positive, cosine and tangent are negative is correct.

cjulius9437

D.She made her first error in step 3 because the sine, cosine, and tangent ratios are incorrect, which also results in incorrect cosecant, secant, and tangent functions.

Step-by-step explanation:

on e 2020

rrrrainy

A) sine: positive cosine: positive tangent: positive

Step-by-step explanation:

Consider the first quadrant in the coordinate diagram below:

x and y are positive

$Sin \theta = \dfrac{Opposite}{Hypotenuse} =\dfrac{y}{\sqrt{x^2+y^2} } \\Cos \theta = \dfrac{Adjacent}{Hypotenuse} =\dfrac{x}{\sqrt{x^2+y^2} } \\Tan \theta = \dfrac{Opposite}{Adjacent} =\dfrac{y}{x }$

For positive x and y, $\sqrt{x^2+y^2}$ is also positive. Therefore:

$Sin \theta$ is positive

$Cos \theta$ is positive

$Tan \theta$ is positive

mremoney530

6.d.Quantity square root of six plus square root of two divided by four.

7.:Quantity negative square root 2 minus square root three divided by two.

8.C.Sin 8x

9.d. $Cos 67^{\circ}$

10.b. 2sin x cos x-1 +2 sin^2 x

Step-by-step explanation:

6. $Sin75^{\circ}$ =Sin(45+30)

Sin(A+B)=Sin A Cos B+Sin B Cos A

Using identity

Sin(45+30)= Sin 45 Cos 30+ Cos 45 Sin 30= $\frac{1}{\sqrt2}\cdot \frac{\sqrt3}{2}+\frac{1}{\sqrt2}\cdot\frac{1}{2}$

$Sin 45^{\circ}=Cos 45^{\circ}=\frac{1}{\sqrt2}$

$Sin 30=\frac{1}{2},Cos 30=\frac{\sqrt3}{2}$

$Sin(45+30)=\frac{\sqrt3}{2\sqr2}+\frac{1}[2\sqt2}$

$Sin(45+30)=\frac{\sqrt3\times \sqrt2}{2\sqrt2\times\sqrt2}+\frac{\sqrt2}{2\sqrt\times \sqrt2}$

$Sin(45+30)=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}=\frac{\sqrt6+\sqrt2}{4}$

d.Quantity square root of six plus square root of two divided by four.

7. $Sin(-\frac{11\pi}{12})$

$Sin(-\frac{11\pi}{12})=-Sin\frac{11\pi}{12}$

Sin (-x)=-Sin x

$Sin (-\frac{11\pi}{12})=-Sin(\pi-\frac{\pi}{12})=-Sin\frac{\pi}{12}$

$-Sin\frac{\frac{\pi}{6}}{2}=-\sqrt{\frac{1-cos\frac{\pi}{6}}{2}}$

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

$Sin(-\frac{11\pi}{12})=-\frac{\sqrt{2-\sqrt3}}{2}$

Answer :Quantity negative square root 2 minus square root three divided by two.

8. sin 9x Cos x-Cos 9x sin x

sin (A-B)=Sin A Cos B- Sin B Cos A

Using this identity

Then we get

Sin 9x Cos x- Sin x Cos 9x= Sin (9x-x)=Sin 8x

C.Sin 8x

9 $.Cos 112^{\circ}Cos 45^{\circ}+ sin 112^{\circ} sin 45 ^{\circ}$

Cos (A-B)=Cos A Cos B+ Sin A Sin B

Using this identity then we get

$Cos (112-45)=Cos 67^{\circ}$

d. $Cos 67^{\circ}$

10. sin 2x -cos 2x

Sin 2 x=2 sin x cos x

Cos 2x =1- 2 sin^2 x

Using above identities

Therefore, sin 2x- cos 2x=2sin xcos x-1+2 sin^2x

b. 2sin x cos x-1 +2 sin^2 x

ziyonsolange

5π/6 is in quadrant 2.

The reference angle, θ' = π/6.

sin(5π/6) is positive, cosine & tangent are both negative

emilaw3233

1.sin x = \frac{5}{13}

cos x = \frac{12}{13}

tan x = \frac{5}{12}

sin y = \frac{12}{13}

cos y = \frac{5}{13}

tan y = \frac{12}{5}

2. ditance travelled = hypotenuse = \frac{200}{sin 40} = 311.14yards

Step-by-step explanation:

sin x = $\frac{height}{hypotenuse}$

cos x = $\frac{base}{hypotenuse}$

tan x = $\frac{height}{base}$

therefore from the figure sin x = \frac{5}{13}

cos x = \frac{12}{13}

tan x = \frac{5}{12}

according to angle y height will be 12 and base will be 5

therefore sin y = \frac{12}{13}

cos y = \frac{5}{13}

tan y = \frac{12}{5}

2. given that height = 200 yards and angle x= 40

we know that tan x = \frac{height}{base}

therefore tan 40 = \frac{200}{base}

therefore base = \frac{200}{tan 40}

they have asked us to find hypotenuse therefore

sin x = \frac{height}{hypotenuse}

hypotenuse = \frac{200}{sin 40} = 311.14yards

7letters22

6 Find an exact value.
sin 75°
sin(A+B)=sin(A)cos(B)+cos(A)sin(B)
sin(45)=cos(45)=(2^0.5)/2 sin(30)=0.5 cos(30)=(3^0.5)/2
sin(45+30)=sin(45)cos(30)+cos(45)sin(30)=(6^0.5+2^0.5)/4
the answer is the letter d) quantity square root of six plus square root of two divided by four.

7. Find an exact value.
sine of negative eleven pi divided by twelve.

sin(-11pi/12) = -sin(11pi/12) = -sin(pi - pi/12) = -sin(pi/12) = -sin( (pi/6) / 2)

= - sqrt( (1-cos(pi/6) ) / 2) = -sqrt( (1-√3/2) / 2 ) = -(√3-1) / 2√2=(√2-√6)/4
the answer is the letter c) quantity square root of two minus square root of six divided by four.

8. Write the expression as the sine, cosine, or tangent of an angle.
sin 9x cos x - cos 9x sin x

sin(A−B)=sinAcosB−cosAsinB

sin(9x−x)= sin9xcosx−cos9xsinx= sin(8x)

the answer is the letter c) sin 8x

9. Write the expression as the sine, cosine, or tangent of an angle.
cos 112° cos 45° + sin 112° sin 45°

cos(A−B)=cosAcosB+sinAsinB

cos(112−45)=cos112cos45+sin112sin45=cos(67)

the answer is the letter d) cos 67°

10. Rewrite with only sin x and cos x.

sin 2x - cos 2x

sin2x = 2sinxcosx
cos2x = (cosx)^2 - (sinx)^2 = 2(cosx)^2 -1 = 1- 2(sinx)^2

sin2x- cos2x=2sinxcosx-(1- 2(sinx)^2=2sinxcosx-1+2(sinx)^2

sin2x- cos2x=2sinxcosx-1+2(sinx)^2

the answer is the letter b) 2 sin x cos2x - 1 + 2 sin2x

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