Trig functions and Square roots

[rjefferey]

The instructions read:

Sketch a right triangle corresponding to the trigonometric funtion of the acute angle (symbol for theta). Use the Pythagorean Theorem to determine the third side and then find the other five trigonometric functions of (symbol for theta).

The given values:

sin (theta) = 3/4

from the given (opp=3, hyp=4) I got the adjacent to be the sqRt of 7 and angle (theta) to be 36.86deg (37deg)

For secant I would have put 4 over the square root of 7 since secant is just the inverse of sine

and for tangent I would have put 3 over the sqRt of seven.

But the answers in the back of the book indicate that:

secant is 4 times the sqRt of 7 divided by seven, and

tangent is 3 times the sqRt of seven, divided by seven.

Am I overlooking a basic rule?

 

[soroban]

Hello, rjefferey!

Your first step is incorrect . . .

Sketch a right triangle corresponding to the trigonometric funtion of the acute angle \(\displaystyle \theta\).
Use the Pythagorean Theorem to determine the third side and then find the othe five trigonometric functions of \(\displaystyle \theta\)

The given values: \(\displaystyle \,\sin\theta\:=\:\frac{3}{4}\)

From the given I got the hypotenuse to be \(\displaystyle \sqrt{7}\;\) . . . no

\(\displaystyle \sin\theta \:=\:\frac{3}{4}\:=\:\frac{opp}{hyp}\)

So we are given: \(\displaystyle \,opp\,=\,3,\;hyp\,=\,4\)

From Pythagorus: \(\displaystyle \,(opp)^2\,+\,(adj)^2\:=\hyp)^2\)

\(\displaystyle \;\;\)so we have: \(\displaystyle \,3^2\,+\,(adj)^2\:=\:4^2\;\;\Rightarrow\;\; (adj)^2\,=\,7\;\;\Rightarrow\;\;adj\,=\,\sqrt{7}\)

Now write the five trig values . . .

 

[rjefferey]

edited my previous post

Thank you for responding...

I edited my post:

I had put that I determined that the \(\displaystyle \,hyp\,=\,sqrt{7}\), well... I meant that I had determined that the \(\displaystyle \,adj\,=\,sqrt{7}\) since the given values of:

\(\displaystyle \,sin\,\theta\,=\,\frac{3}{4}\)

states that the \(\displaystyle opp\,=\,3\) and \(\displaystyle hyp\,=\,4\)

Since \(\displaystyle sec\,=\,\frac{hyp}{opp}\) I am having difficulty understanding why \(\displaystyle sec\,=\,\frac{4sqrt{7}}{7}\) instead of the inverse of cosine which I thought would be \(\displaystyle \frac{4}{sqrt{7}}\).

and why

\(\displaystyle \,tan\,=\,\frac{3sqrt{7}}{7}\)

instead of: \(\displaystyle \,tan\,=\,\frac{3}{sqrt{7}}\)


I'm looking over the other answers and notice that there are no square roots in any of the denominators. Does this have something to do with my problem?

 

[rjefferey]

Solution

Ok I see it...

I can't have a square root in the denominator so:

\(\displaystyle tan\,\theta\,=\,\frac{opp}{adj}\,=\,\frac{3}{sqrt{7}}\,=\,\frac{3}{sqrt{7}}\cdot\frac{sqrt{7}}{sqrt{7}}\,=\,\frac{3sqrt{7}}{7}\)

and the same would apply to the secant.
Thank you!
RJ