Mathematics, 30.09.2019 16:30 annalaurie7

Which of the following is not an identity? a: (cosx+sinx)^2=1+2cosx sinx b: 1+cos^2x/sin^2=1 c: tan^2x=sec^2x-sin^2x-cos^2x d: (tanx+cotx)^2=csc^2x+sec^2x

Answers: 2

Show answers

Answers

Part A
(cosx+sinx)² = 1 + 2cosxsinx
Foil the left side.
cos²x + 2cosxsinx + sin²x = 1 + 2cosxsinx
Subtract 2cosxsinx from each side.
cos²x + sin²x = 1
And we're left with the Pythagorean identity!
Looks like it's not A.

Part B
( 1 + cos²x ) / sin²x = 1
Multiply each side by sin²x.
1 + cos²x = sin²x
Immediately, you should see a problem with this, and here's why:
Our Pythagorean identity from earlier can be put like this:
sin²x = 1 - cos²x
Substitute 1 - cos²x for sin²x in our earlier equation, and you get
1 + cos²x = 1 - cos²x
Which is clearly incorrect!
(If you really needed to verify this...subtract 1, divide by cos²x to get 1 = -1)
B is not an identity.

Part C
tan²x = sec²x - sin²x - cos²x
You might be able to recognize something from our Pythagorean identity.
Factor out a -1 from -sin²x - cos²x...
tan²x = sec²x -1(sin²x + cos²x)
Using our Pythagorean identity, 1 = sin²x + cos²x...
tan²x = sec²x - 1
Hey, look! Another trig identity! Don't recognize it? here...
sin²x/cos²x = 1/cos²x - 1
Multiply by cos²x...
sin²x = 1 - cos²x
There's also a great diagram for remembering the basic trig identities (besides double angle, half angle, angle addition and subtraction) called the magic hexagon...look it up, it's a great tool!

Part D
(tanx+cotx)² = csc²x + sec²x
This one sucks to prove but we can always use our last resort: expressing in terms of sine and cosine.
(sinx/cosx + cosx/sinx)² = (1/sinx)² + (1/cosx)²
Get the fractions on the left into a common denominator sinxcosx.
(sin²x/sinxcosx + cos²x/sinxcosx)² = (1/sinx)² + (1/cosx)²
(sin²x+cos²x/sinxcosx)² = (1/sinx)² + (1/cosx)²
(1/sinxcosx)² = (1/sinx)² + (1/cosx)²
Get the fractions on the right into a common denominator.
(1/sinxcosx)² = 1/sin²x + 1/cos²x
(1/sinxcosx)² = cos²x/sin²xcos²x + sin²x/sin²xcos²x
(1/sinxcosx)² = (sin²x+cos²x)/sin²xcos²x
(1/sinxcosx)² = 1/sin²xcos²x
(1/sinxcosx)² = (1/sinxcosx)²

Answer from: Quest

Jjutsclopyrsvnkiu it’s c

Answer from: Quest

$2416.6

step-by-step explanation:

per hour = $ 25

total work time = 35

commission = 15%

total she has to make = $ 4,500

let she sold x dollar products

so her commission will be 15 % of x

now

total = weekly basic pay + commission

4500= (25*35) + 15 % of x

4500=875 + $\frac{15}{100}$ * x

4500-875 = $\frac{15}{100}$ * x

3625= $\frac{15}{100}$ * x

multiplying wwith 100 on both sides gives

36250 = 15 x

now

x = $\frac{36250}{15}$

x= $2416.6

Another question on Mathematics

Mathematics, 21.06.2019 13:00

Me what is the value of x? showing work would be appreciated - ! : )

Answers: 1

Answer

Mathematics, 21.06.2019 18:00

Asquare has a perimeter of sixteen centimeters. what is the length of each of its sides? a. 8 cm c. 9 cm b. 4 cm d. 10 cm select the best answer from the choices

Answers: 2

Answer

Mathematics, 21.06.2019 20:00

Aclothing store has the sign shown in the shop window. pani sees the sign and wants to buy 3 shirts and 2 pairs of jeans. the cost of each shirt before the discount is $12, and the cost of each pair of jeans is $19 before the discount. write and simplify an expression to find the amount pani pays if a $3 discount is applied to her total

Answers: 2

Answer

Mathematics, 21.06.2019 23:30

The area (a) of a circle with a radius of r is given by the formula and its diameter (d) is given by d=2r. arrange the equations in the correct sequence to rewrite the formula for diameter in terms of the area of the circle.

Answers: 1

Answer

You know the right answer?

Which of the following is not an identity? a: (cosx+sinx)^2=1+2cosx sinx b: 1+cos^2x/sin^2=1 c: ta...

Questions

Mathematics, 16.07.2019 07:30

We are given ab // de. because the lines are parallel and segment cb crosses both lines someone plz...

Mathematics, 16.07.2019 07:30

Felicia found an apartment that rents for 470 a month the landlord requires one and half months rent for the security deposit and the first month rent...