Question

someone please help me with precalc ?

Answer 1

only if you promise to type an actual question that can be helped on :/

Answer 2

Simplify the expression: cot x sin x - sin quantity pi divided by two minus x. + cos x

cos x

sin x

2 sin x

2 cos x

Answer 3

@amistre64

Answer 4

\(\large\sf \color{limegreen}{\frac{cot(x)sin(x)-sin(\frac{\pi}{2})}{2-x+cos(x)}}\) ????

Answer 5

oy vey! im pretty sure they invented math notation for a reason.

cot(x)sin(x) - sin(pi/2 - x) + cos(x)

Answer 6

\[\cot x \sin x - \sin (\pi/2-x) + \cos x\]

Answer 7

my strategy would be to get everything in a sin cos term; and the middle can be rewritten by the property:

sin(a+b) = sin(a)cos(b) + sin(b)cos(a)

Answer 8

what is cot in terms of sin and cos?

Answer 9

i have no clue how to do that

Answer 10

\(\sf \color{blue}{cot(x)=\frac{cos(x)}{sin(x)}}\)

Answer 11

you need to be familiar with the basics:

csc = 1/sin
sec = 1/cos
cot = cos/sin
tan = sin/cos

Answer 12

okay

Answer 13

cot(x)sin(x) - sin(pi/2 - x) + cos(x)

cos sin/sin - sin(pi/2 - x) + cos(x)

2cos - sin(pi/2 - x)

can you expand that sin part based on the property i mentioned?

Answer 14

i don't understand the question

Answer 15

i'm sorry , i just don't understand how to do any of this ....... this lesson was very confusing for me .

Answer 16

the lesson most likely is trying to get you to apply the previous lessons content. Trig, or pre calc as they call it, it like 99% memorization.

Answer 17

i've noticed ......

Answer 18

we pretty much have this setup
\[cot~sin-sin(a+b)+cos\]
cot can be defined in terms of sin cosparts
\[\frac{cos}{\cancel{sin}}~\cancel{sin}-sin(a+b)+cos\]
and cos+cos = 2 cos
\[2cos-sin(a+b)\]
now we have to recall the indentity or proterty that I mentioned above

Answer 19

or, there is another property that we can remember ..... but this one seems fine

Answer 20

okay i think i'm starting to get it .

Answer 21

\[2cos-sin(a+b)\]
\[2cos-[sin(a)cos(b)+sin(b)cos(a)]\]
\[2cos-sin(a)cos(b)-sin(b)cos(a)\]
a=pi/2 b = -x

\[2cos-sin(pi/2)cos(-x)-sin(-x)cos(pi/2)\]
sin pi/2 = 1, cos pi/2 = 0

\[2cos-cos(-x)\]

cos -x = cos x

\[2cos-cos(x)\]

etc ...

Answer 22

2 cos x ?