All categories

3 years ago

Using basic trigonometric identities simplify the

Mathematics

1 answer:

katrin2010 [14]3 years ago

6 0

Answer:

Step-by-step explanation:

Using the trigonometric identities

cot x = $\frac{cosx}{sinx}$ and tan x = $\frac{sinx}{cosx}$ thus

$\frac{cos0}{cot0}$ + tanΘ × cosΘ

= $\frac{cos0}{\frac{cos0}{sin0} }$ + $\frac{sin0}{cos0}$ × cosΘ

= cosΘ × $\frac{sin0}{cos0}$ + sinΘ

= sinΘ + sinΘ

= 2sinΘ → b

You might be interested in

Subtract 0000 1000 whats the answer

Finger [1]

If u subtract 1,000 - 0000 it will equal 1,000

8 0

3 years ago

Nicole has 17 nickels n and dimes d. If the value of her coins is $1.30, how many of each coin does she have?

denis-greek [22]

Answer:

8 nickels and 9 dimes

Step-by-step explanation:

If all were dimes, the value would be $1.70. It is $0.40 less than that. Changing a dime for a nickel reduces the value by $0.05, so there must have been $0.40/$0.05 = 8 such changes.

There are 8 nickels and 9 dimes.

_____

Check

8 · 0.05 + 9 · 0.10 = 0.40 + 0.90 = 1.30 . . . the answer checks OK

7 0

3 years ago

Keisha throws a rock down an old well the distance D in Feet The Rock Falls after T seconds can be represented by D =16t ^ 2 + 6

sesenic [268]

Answer:

1 second 

Step-by-step explanation:

16t^2+64t=80

16t^2+64t-80=0

64^2-4(16)(-80)

9216

-64-96

______ = -5

 32

-64+96

_______ = 1 sec

 32

7 0

3 years ago

What is the completely factored form of f(x)=x^3-7x^2+2x+4

xxMikexx [17]

$Solution, \mathrm{Factor}\:x^3-7x^2+2x+4:\quad \left(x-1\right)\left(x^2-6x-4\right)$

$Steps:$

$x^3-7x^2+2x+4$

$Use\:the\:rational\:root\:theorem,\\a_0=4,\:\quad a_n=1,\\\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:4,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\\\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:4}{1},\\\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x-1,\\\left(x-1\right)\frac{x^3-7x^2+2x+4}{x-1}$

$\frac{x^3-7x^2+2x+4}{x-1}$

$\mathrm{Divide}\:\frac{x^3-7x^2+2x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^3-7x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{x^3}{x},\\\mathrm{Quotient}=x^2,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}x^2:\:x^3-x^2,\\\mathrm{Subtract\:}x^3-x^2\mathrm{\:from\:}x^3-7x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-6x^2+2x+4,\\Therefore,\\\frac{x^3-7x^2+2x+4}{x-1}=x^2+\frac{-6x^2+2x+4}{x-1}$

$\mathrm{Divide}\:\frac{-6x^2+2x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-6x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{-6x^2}{x}=-6x,\\\mathrm{Quotient}=-6x,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-6x:\:-6x^2+6x,\\\mathrm{Subtract\:}-6x^2+6x\mathrm{\:from\:}-6x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-4x+4,\\\mathrm{Therefore},\\\frac{-6x^2+2x+4}{x-1}=-6x+\frac{-4x+4}{x-1}$

$\mathrm{Divide}\:\frac{-4x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-4x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{-4x}{x}=-4,\\\mathrm{Quotient}=-4,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-4:\:-4x+4,\\\mathrm{Subtract\:}-4x+4\mathrm{\:from\:}-4x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=0,\\\mathrm{Therefore},\\\frac{-4x+4}{x-1}=-4$