Pre-calculus math help question?

Question

I need help with a few questions,

if you could answer any, it would help a lot!

explanations would also be greatly appreciated!

1. http://i37.tinypic.com/24zlr9w.jpg

2. Verify the identity: sec y - cos y = tan y sin y

3. Find all x satisfying: tan^2(2x) = 3, where 0 less than/equal to x less than/equal to pi

4. If 0 < q < 2p find all values of q for which 2sinq = tanq

Answer 1

1. I&#039;ll just use x instead of theta. We know that tan x = sin x / cos x so tan x cos x is simply sin x. so sin x = s/(((s^2)+(r^2))^(1/2))

2. sec y - cos y = tan y sin y

1/ cos  y - cos y = tan y sin y

(1- cos^2 y)/cos y = tan y sin y

We know that sin^2 y + cos^2 y=1 so:

sin^2 y/cos y= tan y sin y

sin y/ cos y * sin y = tan y sin y

tan y sin y = tan y sin y

3. tan^2(2x)=3

tan(2x)=3^(1/2)

we know that 2x= pi/3 + k* pi for every k integer.

so x= pi/6 +k* pi/2

x=30 , 120 degree

4. 2 sin q = sin q/cos q

first, suppose that sin q =0

it satisfy the equation. So, sin q = 0 is one of the solution.

or q=180 degree = pi radian

second, suppose sin q is not 0, then we can cancel it from both sides.

we get cos q = 1/2. from this we get q= 60/300 degree= (pi/3) or (5pi/3 ) radian

Answer 2

Let&#039;s replace theta with § (just because I don&#039;t know the unicode for theta).

Well we first have to know more about the tangent.

tan(§) = (sin§)/(cos§)

Since tan(§)*(cos§) = (sin§)*cos§)/(cos§), we can cancel out cosine and then we have just sin(§), which is simply your opposite divided by the hypotenuse, or (s/h).

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2. Well sec y is also 1/(cos y)

So now you have 1/(cos y ) - (cos y)

By finding the common denominator, you have [1-cos^2(y)]/(cos y)

now for the right side, (tan y) = (sin y)/(cos y) so from there you have sin^2(y)/(cos y)

now it should look like this:

[1-cos^2(y)] / (cos y) = sin^2(y)/(cos y)

add cos^2(y) to both sides and then multiply cos y to both sides (to cancel denominator), you finally have (rewritten so 1 is on the other side)

sin^2(y) + cos^2(y) = 1   which is the fundamental identity of Trigonometry (say so in my Trig class lol, I dunno if that&#039;s an actual statement though).

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3. First you need to SQRT both sides to get tan(2x) = 3

Now multiply the inverse of tan to both sides to get:

2x=60 (because 60 = tan^ -1(SQRT 3)

divide by 2,

x=30

HOWEVER since pi is in radians, you need to convert 30 into radians by multiplying 30 by π and dividing by 180 (don&#039;t decimal pi).

You have x= (π/6)

Your answer should be something along the lines of {x|x= (π/6)+2πr}

2πr is every co terminal angle that follows π/6

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4. Well I&#039;m not so sure about this one but since tan(q) = sin(q)/cos(q), you can multiply cos(q) to both sides to get a double angle identity of sin which is: sin(2q)=2sin(q)cos(q) however understand the identity for this problem isn&#039;t necessary

you now divide sin to both sides to get

2cos(q)=1

divide 2 to both sides and multiply by the inverse of cosine to get

q = 60 or π/3

Now rewrite in set notation: {q|q=π/3 +2πr}

Hope this helps. ^_^

Answer 3

Only one problem per question please.

1.

Calculate (tanθ)(cosθ) = (sinθ/cosθ)(cosθ) = sinθ = s / √(r² + s²)