Simple Trig??

Question

Ok, I'm a little confused. I'm working with angles in trig and need some help.

The acute angle given is defined as:
a=adjacent=2
b=opposite=1
c=hypotenuse=√5

Now, it says 'find the six functions' (no problems, until I see this).

For SIN they say;

SIN = b/c = 1/√5 = √5/5

How did it jump from 1/√5 to √5/5... What powers or force made the change happen? (I'm assuming that I am forgetting a rule in regards to sqrt's or something).


Thanks!

Answer 1

They multiplied both the numerator and denominator by sqrt(5).

1/√5 = 1/√5 * √5/√5 = √5/5

The powers that be like to see the radical in the numerator rather than the denominator.

Answer 2

Hi,

When you get a fraction like 1/sqrt(5), it's not considered simplified with a radical in the denominator. To eliminate this, multiply top and bottom both by sqrt(5). This gives sqrt(5)/sqrt(25), but you can simplify the sqrt(25) to the number 5. So this becomes sqrt(5)/5

6 trig functions can be remembered by any of these "patterns".

"soh-cah-toa" is sin = opp/hyp, cos = adj/hyp, tan = opp/adj

Tom's old aunt sat on her coffin and howled.

Tan = opp/adj Sin = opp/hyp Cos = adj/hyp

Some old hippie caught another hippie tripping on acid.

Sin = opp/hyp cos = adj/hyp Tan = opp/adj


Sometimes the sillier, the better.

The other trig functions are csc = hyp/opp, sec = hyp/adj and cot = adj/opp.


Anyhow for your triangle, the 6 functions would be:
Sin = sqrt(5)/5 Csc = sqrt(5) <==These are reciprocals.
Cos = 2sqrt(5)/5 Sec = sqrt(5)/2 <==These are reciprocals.
Tan = 1/2 Cot = 2/1 = 2 <==These are reciprocals.

I hope that helps!! :-)

Answer 3

Note that √5 /√5 = 1 and √5 x √5 = 5
1/√5 = (1/√5) x (√5/√5) = √5 / 5
Pleased to hear that you can now find sin, cos and tan of angles as I didn`t fancy typing them out!
Hope this is OK.

Answer 4

You have a right angled triangle having sides O =1 (opposite side), A= 2 (adjacet side) and H = Sq.rt of 5 (hypotanuse)

Memorize it as...

O, A, H
1, 2, sq. rt 5

Rememember six related trignometrical functions in 3 pairs as...

(1) O/H =Sine...(1/ sq.rt 5)_______ H/O= Cosec...(sq rt 5) /1

(2) A/H =Cosine...(2/ sq.rt 5)_____ H/A= Sec...(sq rt 5) / 2

(3) O/A =Tan...(1/ 2)____________A/O= Cot...( 2 /1)

Said "pair memorising" involves mere " interchanging of numerator and dinominator" (in each said pairs). By that we can just memorize three functions O/H= sine , A/H= Cos and O/A= Tanjant and remaining three by said interchanging of numerator and dinominator!

Once you have firmly memorised "O/H, A/H, O/A" in relation to an "angle theta" ( "adjacent" and "opposite" sides are fixed in relation to theta), your confusion will disappear!

Remember that hypotonuse (H) is always the "longest side of a right angle triangle" and "a side adjacent to to theta" is adjacent side(A). Naturally remaining "opposite side of theta" is regarded opposite side (O)!


Regards!

Answer 5

in algebra in order to simply the irrational denomenator in a fraction.the fraction should be rationalized.mean multiply both numerator and denomenator by sqrt of 5 in order for as to take out the irrational sign.because sqrt5(sqrt5)=5

therefore 1/sqrt5(sqrt5/sqrt5)=sqrt5/5

hope this help.