# Math Help - Problems From Trig Test

1. ## Problems From Trig Test

Could someone please post a solution to these 4 questions:

2. Originally Posted by SportfreundeKeaneKent
Could someone please post a solution to these 4 questions

Hello,

to #3.a)

$\cos^3(2x)=\cos(2x)$
$\cos^3(2x)-\cos(2x)=0$
$\cos(2x)(\cos^2(2x)-1)=0$
$\cos(2x)=0 \vee \cos^2(2x)-1=0$
$\cos(2x) = 0$ $\vee \cos(2x) = -1 \vee \cos(2x) = 1$

(For some reason the Latex doesn't work properly anymore. I'll repeat the equations in plain text)

cos(2x) = 0 or cos(2x) = -1 or cos(2x) = 1

This character π means pi
From the 1rst equ.:

2x = -π/2 --> x = -π/4 or
2x = -3π/2 ---> x = -3π/4

from the 2nd equ.:

2x = -π ---> x = -π/2

from the 3rd equ.:

2x = 0 ---> x = 0 or
2x = -2π ---> x = -π

EB

3. Hello, SportfreundeKeaneKent!

Here's the very first one . . .

1. If cos(x) = -12/13, where π < x < 3π/2

then find the exact value of sin(2x)

We'll use the identity: .sin(2x) .= .2·sin(x)·cos(x)

We know that: cos(x) = -12/13 . . . we need sin(x).

Angle x is in Quadrant 3.
Since cos(x) = -12/13 = adj/hyp,
. . we use Pythagorus and find that: opp = -5
Hence: .sin(x) = -5/13

Therefore: .sin(2x) .= .2(-5/13)(-12/13) .= .120/169

4. Hello again, SportfreundeKeaneKent!

Here's #4(b) . . .

Prove:
. . . . . . . . . . . . . .tan(x) - tan(y)
. . tan(x - y) . = . --------------------
. . . . . . . . . . . . .1 + tan(x)·tan(y)
We will use these identities:

. . sin(x - y) .= .sin(x)·cos(y) - sin(y)·cos(x)
. . cos(x - y) .= .cos(x)·cos(y) + sin(x)·sin(y)

. . . . . . . . . . . sin(x - y) . . . . . sin(x)·cos(y) - sin(y)·cos(x)
tan(x - y) . = . ------------ . = . -----------------------------------
. . . . . . . . . . . cos(x - y) . . . . .cos(x)·cos(y) + sin(x)·sin(y)

Divide top and bottom by cos(x)·cos(y):

. . .sin(x)·cos(y) . .sin(y)·cos(x) . . . . . . sin(x) . sin(y)
. . .--------------- - --------------- . . . . . . ------- - -------
. . .cos(x)·cos(y) . cos(x)·cos(y) . . . . . . cos(x) . cos(y)
. . ------------------------------------ . = . ---------------------
. . .cos(x)·cos(y) . .sin(x)·sin(y) . . . . . . . . sin(x) sin(y)
. . .--------------- + --------------- . . . . .1 + -------·--------
. . .cos(x)·cos(y) . .cos(x)·cos(y) . . . . . . . .cos(x) cos(y)

. . . . . . . tan(x) - tan(y)
. . . = . ---------------------
. . . . . . 1 + tan(x)·tan(y)

5. Originally Posted by SportfreundeKeaneKent
Could someone please post a solution to these 4 questions[/IMG]
Hello, Sportsfreund!

#3.b)

$\sec^4(x)-4\sec^2(x)+4=0,\ 0 < x < 2\pi$

Use substitution: $y=\sec^2(x)$. Then you'll get the equation:

$y^2-4y+4=0 \Longleftrightarrow (y-2)^2=0 \Longleftrightarrow y=2$

Now re-substitute:

$\sec^2(x)=2\Longleftrightarrow \sec(x)=\sqrt{2}\ \vee \ \sec(x)=-\sqrt{2}$

From the 1rst equation you get:

$\sec(x)=\sqrt{2} \Longleftrightarrow x=\frac{1}{4} \cdot \pi\ \vee \ x=\frac{7}{4} \cdot \pi$

From the 2nd equation you get:

$\sec(x)=-\sqrt{2} \Longleftrightarrow x=\frac{3}{4} \cdot \pi\ \vee \ x=\frac{5}{4} \cdot \pi$

EB

6. Originally Posted by SportfreundeKeaneKent
Could someone please post a solution to these 4 questions
Hello, Sportsfreund,

#2

Use the property:

$\cos(w-x)=\cos(w) \cos(x)+\sin(w) \sin(x)$

As Soroban has demonstrated use the Pythagoran theorem:

$\sin^2(x)+\cos^2(x)=1 \Longrightarrow \sin(x)=\sqrt{1-\left(\frac{4}{5} \right)^2}=-\frac{3}{4}$. Because x is in the 4. quadrant the sine-value must be negative.

$\sin^2(w)+\cos^2(w)=1 \Longrightarrow \cos(w)=\sqrt{1-\left(\frac{7}{25} \right)^2}=-\frac{24}{25}$. Because w is in the 2. quadrant the cosine-value must be negative.

Now you know all values necessary:

$\cos(w-x)=\left(-\frac{24}{25}\right)\cdot \left(\frac{4}{5}\right)+\left(\frac{7}{25}\right) \cdot \left(-\frac{3}{4}\right)=\left(-\frac{489}{500}\right)$

EB

7. Hello, SportfreundeKeaneKent!

There's something wrong with 4(a).
. . Is there a typo?

$4a)\;\sec2x(1 - \sin2x) \:=\:\sec^2x - \tan^2x$

The right side is always equal to 1.
. . So it's a silly way to write a problem.

Besides, the statement is not an identity.

Try x = π/3
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._
The left side is: .sec(π/3)[1 - sin(π/3)] .= .2(1 - √3/2) . . 1