Mathematics, 03.04.2020 00:32 markleal9484

Given 1+cosx\sinx+sinx\1+cosx=4, find a numerical value of one trigonometric function of x.

Answers: 3

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Answer from: vanvalenpeyt

Option d. $sin(x)=\frac{1}{2}$

Step-by-step explanation:

The complete question is

Given (1+cosx)/(sinx) + (sinx)/(1+cosx) =4, find a numerical value of one trigonometric function of x.

a. tanx=2

b. sinx=2

c. tanx=1/2

d. sinx=1/2

we have

$\frac{1+cos(x)}{sin(x)}+\frac{sin(x)}{1+cos(x)}=4$

Find the common denominator and adds the fractions

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

Expanded the numerator

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

Remember that

sin^2(x)+cos^2(x)=1 ----> trigonometric identity

substitute

$\frac{1+2cos(x)+1}{(1+cos(x))sin(x)}=4$

$\frac{2+2cos(x)}{(1+cos(x))sin(x)}=4$

Factor 2 in the numerator

$\frac{2(1+cos(x))}{(1+cos(x))sin(x)}=4$

Simplify

$\frac{2}{sin(x)}=4$

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

Answer from: Quest

I’d say letter a..
Which complex number is represented by the pointe graft on the complex plane below? a. -4+3i b. 3+4

Answer from: Quest

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2:

(2 x^3 + 7 x^2 + (x + 4 x) + 2)/(x (x + 2))

x + 4 x = 5 x:

answer: (2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

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Given 1+cosx\sinx+sinx\1+cosx=4, find a numerical value of one trigonometric function of x....

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