HW3 Answer Key PDF

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Mat1275 Homework #3 answer key ...

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NYCCT/ MA1275/ SP20 Prof. Nadmi April 29, 2020

Name:__________________________

HW#3, Answer Key Answer all questions and show all work. Clearly indicate the necessary steps. For all questions in this homework, a correct numerical answer with no work shown will receive 0 credits. Each correct answer will receive full credit. Exam will be graded on clarity, neatness, organization and correctness. Simplify if possible and Box your final answer.

1) Simplify: LCD: 𝑥 2 1 1 𝑥 2 (1 − 𝑥 ) 1−𝑥 𝑥 2 − 𝑥 𝑥(𝑥 − 1) = 2 = 2 = 1 1 𝑥 +1 1 + 2 𝑥 2 (1 + 2 ) 𝑥 + 1 𝑥 𝑥 2) Solve the system of equations: 𝑥−𝑦 =2 𝑥 2 − 𝑦 2 = 16

𝑥 =2+𝑦

Let substitute 𝑥 = 2 + 𝑦

into

𝑥 2 − 𝑦 2 = 16

(2 + 𝑦)2 − 𝑦 2 = 16 (2 + 𝑦)(2 + 𝑦) − 𝑦 2 = 16 4 + 4𝑦 + 𝑦 2 − 𝑦 2 = 16 4𝑦 = 16 − 4 4𝑦 12 = 3 4 𝑦=3

Let’s solve for x by substituting 𝑦 = 3

into 𝑥 = 2 + 𝑦

𝑥 = 2 + 𝑦 = 2 + (3) = 5. 𝑆 = {(5 , 3)} 3) Find the reference angle with the angle given. (Draw the graph and state the quadrant). a) 𝜃 = 225°

QIII

𝜃𝑟 = 225° − 180° = 45°. 𝜃𝑟 1

b) 𝜃 = −60°

QIV

𝜃𝑟 = 360° − 60° = 300°.

𝜃𝑟

c) 𝜃 =

11 3

𝜋

QIV

6

5

5

5

𝜃 = 𝜋 + 𝜋 = 2𝜋 + 𝜋 = 𝜋 3 3 3 3 5

𝜃𝑟 = 2𝜋 − 𝜋 = 3

𝜋 3

7

d) 𝜃 = − 𝜋 3

𝜃𝑟

QIV 𝜋

6

𝜋

𝜋

𝜃 = − (3 𝜋 + 3 ) = − (2𝜋 + 3 ) = − 3 𝜋

5𝜋

𝜃𝑟 = 2𝜋 − 3 = 3

𝜃𝑟

4) Without using a calculator, state the exact value of the trig function for the given angle (Draw a graph). 3𝜋

a) sin ( 2 ) =? (𝑐𝑜𝑠𝜃, 𝑠𝑖𝑛𝜃) = (0, -1) 3𝜋

sin ( 2 ) = −1

(0 , -1) 3𝜋

b) tan (

−10𝜋

𝜃=−

6

) =?

3 10𝜋 3

𝜃 = − (3 𝜋 + 4𝜋

2

QII 4𝜋

) = − (2𝜋 + 3

4𝜋

)=− 3

4𝜋 3

𝜋

𝜃𝑟 = 3 − 𝜋 = 3 2

√3 10𝜋 4𝜋 √3 = −√3 tan (− ) = tan (− ) = 21 = −1 3 3 − 2 5) Given P(3,  4) draw  . Deter min e the value of the six trigonometric functions , 0    360 .

To find the value of the six trigonometry functions, we need to solve for the unknown. We know that tanθ =

𝑦 𝑥

(

opposite

𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡

), so we need to find r, the hypotenuse.

By using the Pythagorean theorem, we get:

𝑟 2 = 𝑥 2 + 𝑦 2 , 𝑟 = √𝑥 2 + 𝑦 2 𝑦 −4 tanθ = = (Note that tangent is negative in QIV and cosine is positive in QIV, x is 3 𝑥 positive and y is negative). 𝑟 = √𝑥 2 + 𝑦 2 = √(3)2 + (−4)2 = √9 + 16 = √25 = 5 𝑦 −4 = , 𝑟 5 𝑥 3 cos θ = = , 𝑟 5 𝑦 −4 tan θ = = , 𝑥 3 sinθ =

𝑟 5 csc θ = = 𝑦 −4 𝑟 5 sec θ = = 𝑥 3 𝑦 3 cot θ = = 𝑥 −4

3

6) Without the use of a calculator ,draw and state the value of t in [0, 2 ]of : 1 t in QIV cos t  , 2

(10pts )

The cosine function is positive in QI and QIV, where x > 0 and y < 0. 1 𝜋 From the chart, we saw that is x which adjacent to the reference angle 𝑡𝑟 = , 2 3

therefore 𝑡 = 2𝜋 − 𝑡𝑟 = 2𝜋 −

𝜋

3

=

5𝜋 4

in QIV , This yield to: 𝑡 = 5𝜋 4

12 and cos  0, 5 Draw , indicate the quadrant of the ter min al side of  , and state the value of the six trig functions of  . 7) If tan   

To find the value of the other five trigonometry functions, we need to solve for the unknown. We know that tanθ =

𝑦 𝑥

(

opposite

𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡

), so we need to find r, the hypotenuse.

By using the Pythagorean theorem, we get: 𝑟 2 = 𝑥 2 + 𝑦 2 , 𝑟 = √𝑥 2 + 𝑦 2 𝑦

−12

tanθ = = (Note that tangent is negative in QIV and cosine is positive in QIV, x 5 𝑥 is positive and y is negative). 𝑟 = √𝑥 2 + 𝑦 2 = √(5)2 + (−12)2 = √25 + 144 = √169 = 13

𝑦 −12 = , 𝑟 13 𝑥 5 cos θ = = , 𝑟 13 𝑦 −12 tan θ = = , 𝑥 5 sinθ =

𝑟 13 csc θ = = 𝑦 −12 𝑟 13 sec θ = = 5 𝑥 𝑦 5 cot θ = = 𝑥 −12

4...