[SPLIT] Find sine and cosine, given that x = 9 tan(a)

[Rictus]

Even though the following exercise has an explanation, I'm afraid I'm still just not seeing it.

. . .Given x = 9 tan(a), find the sine and cosine.

I know tan(a) = x/9, thus sin^2(a) + cos^2(a) = x/9. The book says that sin = x/9 + cos(a). I understand that, and I understand this:

. . .1 = (x^2/81)cos^2(a) + cos^2(a)

But then the book says:

. . .= cos^2(a)(x^2/81+1)

. . .cos^2(a) = 81/(x^2_81)

Then it finishes the exercise, which I get. But how you get a + 1 and get rid of one of the cos^2(a)? Do you multiply both sides by sin^2(a)? I can't seem to get it to work out like it shows.

Thank you.

 

[skeeter]

x = 9tan(a)

tan(a) = x/9 = opposite/adjacent, correct?

hypotenuse will be sqrt(x² + 81)

sin(a) = opposite/hypotenuse = x/sqrt(x² + 81)

cos(a) = adjacent/hypotenuse = 9/sqrt(x² + 81)

the signs of the trig functions depend upon what quadrant angle "a" resides.

 

[stapel]

I know tan(a) = x/9, thus sin^2(a) + cos^2(a) = x/9.

No; sin²(a) + cos²(a) = 1, regardless of the value of the angle "a".

Draw a right triangle, labelling the angle of interest as "a". If tan(a) = x/9, then the "opposite" side is "x", and the "adjacent" side is "9". Label these sides of your triangle. Use the Pythagorean Theorem to find the length of the hypotenuse, and then read the values of sine and cosine from the triangle.

Eliz.

 

[Rictus]

We havent got into triangle and the pythagorian formula yet, I think we are supposed to figure it out by algebra.
I know tan(a) = x/9, thus sin^2(a) + cos^2(a) = x/9. <----this is a typo.
tan(a) = x/9 so sin(a)/cos(a) = x/9 is what I meant to write.
The book does it differently.
Your way makes much more sense.
Thanks for the replies.

 

[soroban]

Hello, Rictus!

I'll do it "the long way" . . . and it's quite long.
I'll assume angle \(\displaystyle a\) is acute (all trig values are positive).

Given: \(\displaystyle \, x \:= \:9\tan(a)\), find \(\displaystyle \sin(a)\) and \(\displaystyle \cos(a)\)

We have: \(\displaystyle \:\tan(a)\:=\:\frac{x}{9}\)

Identity: \(\displaystyle \:\sec^2(a) \:=\:\tan^2(a)\,+\,1\)

So we have: \(\displaystyle \:\sec^2(a)\:=\:\left(\frac{x}{9}\right)^2\,+\,1\:=\:\frac{x^2\,+\,81}{81}\;\;\Rightarrow\;\;\sec(a)\:=\:\frac{\sqrt{x^2\,+\,81}}{9}\)

Then: \(\displaystyle \:\cos(a) \:=\:\frac{1}{\sec(a)} \:=\:\frac{1}{\frac{\sqrt{x^2+81}}{9}}\)

. . Therefore: \(\displaystyle \L\:\fbox{\cos(a)\:=\:\frac{9}{\sqrt{x^2\,+\,81}}}\)


Identity: \(\displaystyle \:\sin^2(a)\,+\,\cos^2(a)\:=\:1\;\;\Rightarrow\;\;\sin(a)\:=\:\sqrt{1\,-\,\cos^2(a)}\)

Then: \(\displaystyle \:\sin(a)\;=\;\sqrt{1\,-\,\left(\frac{9}{\sqrt{x^2+81}}\right)^2} \;=\;\sqrt{\frac{x^2}{x^2+81}}\)

. . Therefore: \(\displaystyle \L\:\fbox{\sin(a)\:=\:\frac{x}{\sqrt{x^2\,+\,81}}}\)

 

[Rictus]

I really appreciate all the trouble everyone is going to to try and help, but what I don't get is how my student solutions manual gets the answer. Here's what's really confusing me. This is what is in my book, word for word:

First notice that tan(a) = x/9 so tan(a) = sin(a) / cos(a) = x / 9, so sin(a) (x / 9) cos(a). Now to find cos(a) by using 1 = sin^2(a) + cos^2(a) = (x^2/81) cos^2(a) + cos^2(a) = cos^2(a)(x^2/81+1), so cos^2(a) = 81 / (x^2 + 81) and cos(a) = 9 / sqrt(x^2 + 81). Thus, sin(a) = (x / 9)(9 / sqrt(x^2 + 81)) = x / sqrt(x^2 + 81).

What I'm missing here is how do we get from:

. . .(x^2/81) cos^2(a) + cos^2(a)

...to this:

. . .cos^2(a)(x^2/81 + 1)

...and from that to this:

. . .cos^2(a) = 81 / (x^2 + 81)

Or is there a mistake in my book?

Thank you for all your help!

 

[skeeter]

First notice that tan(a) = x/9 so tan(a) = sin(a)/cos(a) = x/9, so sin(a) (x / 9) cos(a).

Now what I'm missing here is how do we get from
(X^2/81) cos^2(a)+cos^2(a)

to this
cos^2(a)(x^2/81+1)

and from that, to this
cos^2(a) = 81/(x^2+81)

the bolded part should say sin(a) = (x/9)*cos(a)

from this point, square both sides of the equation ...

sin^2(a) = (x^2/81)cos^2(a)

now they use the Pythagorean identity 1 = sin^2(a) + cos^2(a) ... substitute (x^2/81)cos^2(a) for sin^2(a) ...

1 = (x^2/81)cos^2(a) + cos^2(a)

note that the two terms on the right have cos^2(a) in common ... factor out cos^2(a) ...

1 = cos^2(a)[(x^2/81) + 1]

get a common denominator for the [(x^2/81) + 1] and add ...

1 = cos^2(a)[(x^2/81) + (81/81)]

1 = cos^2(a)[(x^2 + 81)/81]

multiply both sides by the reciprocal of [(x^2 + 81)/81] ...

1*[81/(x^2 + 81)] = cos^2(a)[(x^2 + 81)/81]*[81/(x^2 + 81)]

81/(x^2 + 81) = cos^2(a)

hope this clears up the "muddled" method they used.

 

[Rictus]

Again it's seems to be the basics I get lost on. Factoring out the cos^2(a) is the key here to my confusion.

This site is invaluable!

Thanks again.