find sine and tangent of angle, given cos(angle) = 4/x

[Rictus]

I am taking Algebra and Trig 2. and I ran into a problem that wants me to find the sine and tangent of an angle by giving me:

. . .cos (of the angle) = 4/x

We haven't done varibles and this is throwing me for a loop. The answer in the book for sin is:

. . .sin (angle) =sq.rt.(x^2 - 16)/x

I'm trying to use the identity, like so:

. . .(4/x)^2 + sin^2 = 1

. . .16/x^2 = sin^2 = 1
. . .sin^2= 1 - 16/x^2

But this isn't getting even close to the book. I hope you can understand what I've posted (I'm not sure of some of the syntax).

Thank you for your help!
Rick

 

[tkhunny]

Re: find sin and tan with a variable

cos (of the angle) = 4/x

Why does it matter if it is a number or a representation of a number? Just use it.

Draw a right triangle.
Label an angle "of the angle"
Label the adjacent leg "4"
Label the hypotenuse "x'
Use the Pythagorean Theorem to calculate the length of the opposite leg, \(\displaystyle \sqrt{x^{2} - 16}\)

Now find the sine and tangent of the angle.

Using algebra, you had the right idea.

\(\displaystyle [sin(of\;the\;angle)]^{2} = 1 - \frac{16}{x^{2}} = \frac{x^{2} - 16}{x^{2}}\)

Is it looking any closer?

 

[Guest]

    *
    * *
 y *    *    x
    *       *
    *          *
    * * * * * * *
         4

\(\displaystyle y = \sqrt {x^2 - 4^2 } = \sqrt {x^2 - 16}\)


cos\(\displaystyle \theta\)=4/x
sin\(\displaystyle \theta\) =\(\displaystyle \frac{y}{x}\)=\(\displaystyle \frac{{\sqrt {x^2 - 16} }}{x}\)

tan\(\displaystyle \theta\)=\(\displaystyle \frac{y}{4}\)=\(\displaystyle \frac{{\sqrt {x^2 - 16} }}{4}\)

 

[skeeter]

let A = the angle

cos(A) = 4/x ... cosine is the ratio adjacent/hypotenuse, correct?

4 = adjacent side, x = hypotenuse

using Pythagoras, the opposite side = sqrt(x² - 16)

so ... sin(A) = opposite/hypotenuse = sqrt(x² - 16)/x

tan(A) = ??? ... you finish.

 

[Mrspi]

Re: find sin and tan with a variable

16/x^2 + sin^2 = 1
sin^2= 1 - 16/x^2
Which isn't getting even close to the book.
Hope you can understand evrything, not sure of some of the syntax.

Rick

Oh, but it IS close to what is in the book!!!

sin^2 A = (x^2/x^2) - (16/x^2)

(you need a common denominator to add or subtract fractions, so write that 1 as x^2/x^2)

sin^2 A = (x^2 - 16) / x^2

Take the square root of both sides:

sin A = + sqrt(x^2 - 16) / x

See?

 

[Rictus]

WOW, I wish I would of found this forum awhile ago. Thanks for the help.
It's funny how my mind works, I forget the basics sometimes. Using x^2/X^2 instead of 1 didn't even occur to me.
Thanks, this place is great.