Basic Trigonometric identities

[evolchild9]

I really have no idea how to do anything, and i have a final in a few days, with my teacher trying to teach us three chapters in one day... Needless to say she isn't very good for a math teacher.. Anyway, if anyone can go through these problems and describe how you did them... i'd be very appreciative.

Solve for values of theta, between 0 degrees and 90 degrees.

If cot theta=2, find tan theta.

if tan theta=sqrt{11}/2, find sec theta.

if cos theta=3/10, find cot theta.

Simplify

tanxcscx/secx

tanAcos(squared)A

tan Beta/cot Beta

1/sin(squared) theta - cos(squared) theta/sin(squared) theta

csc Beta/1+cot(squared) bet(1-sin x)(1+sin x)

 

[soroban]

Hello, evolchild9!

A quick review . . .

You're expected to know the definitions of the six functions
. . and some of their relationships.. .

. . . . . . . . . opp . . . . . . . . . . . . . . . . . hyp
. . . sin θ = ------ . . . . . . . . . . csc θ = ------
. . . . . . . . . hyp. . . . . . . . . . . . . . . . . opp

. . . . . . . . . adj . . . . . . . . . . . . . . . . . hyp
. . . cos θ = ----- . . . . . . . . . . sec θ = ------
. . . . . . . . . hyp . . . . . . . . . . . . . . . . . adj

. . . . . . . . . opp . . . . . . . . . . . . . . . . . adj
. . . tan θ = ------ . . . . . . . . . . cot θ = ------
. . . . . . . . . adj . . . . . . . . . . . . . . . . . .opp

Pythagorus: . opp² + adj² .= .hyp²

Solve for values of θ between 0 degrees and 90 degrees.

If cot θ = 2, find tan θ.

This one is easy . . .
Check the list of defintions . . . the tangent and cotangent are reciprocals.

. . . . . . . . . . . . . .2 . . . . . . . . . . . . 1
So, since cot θ = --, .then .tan θ = --
. . . . . . . . . . . . . .1 . . . . . . . . . . . . 2

If tan θ = sqrt{11}/2, find sec θ.

This one takes more work . . .

For sec θ, we want: hyp ÷ adj

. . . . . . . . . . . . . . . . . . . . . sqrt{11} . . . . opp
We know that: . tan θ . = . ----------- . = . ------
. . . . . . . . . . . . . . . . . . . . . . . .2 . . . . . . . . adj

. . . So we know that: . opp = sqrt{11}, .adj = 2

. . . But we need "hyp".

We use Pythagorus: . hyp² .= .opp² + adj²
. . . . . . so we have: . hyp² .= .(sqrt{11})² + 2² .= .11 + 4 .= .15
. . . . . . . . . . .Hence: . hyp .= .sqrt{15}

. . . . . . . . . . . . . . . . . . hyp . . . . sqrt{15}
Therefore: . sec θ . = . ----- . = . ------------
. . . . . . . . . . . . . . . . . . adj . . . . . . . 2

If cos θ = 3/10, find cot θ.

We want: . cot θ = adj ÷ opp

. . . . . . . . . . . . . . . . . . .3 . . . . . adj
We know: . cos θ . = . ---- . = . -----
. . . . . . . . . . . . . . . . . . 10 . . . . hyp

We want "opp" . . . so: . opp² + adj² .= .hyp²
. . . Then: . opp² + 3² .= .10² . . ---> . . opp = sqrt{91}

. . . . . . . . . . . . . . . . . . adj . . . . . . . 3
Therefore: . cot θ . = . ------ . = . -----------
. . . . . . . . . . . . . . . . . . opp . . . . sqrt{91}