# Thread: Trig question I need to understand this tonight!

1. ## Trig question I need to understand this tonight!

Okay...

Given that:
sin cos
66 degrees 0.9135 0.4067

104 degrees 0.9703 -0.2419

213 degrees -0.5446 -0.8387

Find sin 81 degrees

Hint: Double, Difference

Apparently I don't understand the concept. Thanks!

2. Originally Posted by aigiqinf
Okay...

Given that:
sin cos
66 degrees 0.9135 0.4067

104 degrees 0.9703 -0.2419

213 degrees -0.5446 -0.8387

Find sin 81 degrees

Hint: Double, Difference

Apparently I don't understand the concept. Thanks!
Find a way to get 81 to be a combination of 213,104,66 for example this wouldnt work but lets say you wanted 128 degrees
$\displaystyle 128=213-\frac{66}{2}-\frac{104}{2}$ and once you find a way of doing that to get 81, use your trig identities to get them into a form you can use the given information with

3. Originally Posted by Mathstud28
Find a way to get 81 to be a combination of 213,104,66 for example this wouldnt work but lets say you wanted 128 degrees
$\displaystyle 128=213-\frac{66}{2}-\frac{104}{2}$ and once you find a way of doing that to get 81, use your trig identities to get them into a form you can use the given information with
I've been trying to do that... I'll continue to do so. I'm pretty sure that that's not the way I'm supposed to do it.

4. Originally Posted by aigiqinf
Okay...

Given that:
sin cos
66 degrees 0.9135 0.4067

104 degrees 0.9703 -0.2419

213 degrees -0.5446 -0.8387

Find sin 81 degrees

Hint: Double, Difference

Apparently I don't understand the concept. Thanks!
Sin(81) = Sin(213-66-66) = Sin((213-66)-66)

Sin(213-66) = Sin213Cos66 - Cos213Cos66

Cos(213-66) = Cos213Cos66 + Sin213Sin66

Sin((213-66) - 66) = (Sin213Cos66 - Cos213Cos66)(Cos66) - (Cos213Cos66 + Sin213Sin66)(Sin66)

5. Hello, aigiqinf!

The hint refers to:

. . a Difference identity: .$\displaystyle \sin(A-B) \:=\:\sin(A)\cos(B) - \cos(A)\sin(B)$
. . two Double-Angle identities: .$\displaystyle \begin{array}{ccc}\sin(2A) &=& 2\sin(A)\cos(A) \\ \cos(2A) &=&2\cos^2(A) - 1 \end{array}$

Given that: .$\displaystyle \begin{array}{c|cc} & \text{sin} & \text{cos} \\ \hline 66^o & 0.9135 & 0.4067 \\ 104^o & 0.9703 & \text{-}0.2419 \\ 213^o & \text{-}0.5446 & \text{-}0.8387 \end{array}$

Find $\displaystyle \sin81^o$

Hint: Double, Difference
We note that: .$\displaystyle 81 \:=\:213 - 132$

So: .$\displaystyle \sin81^o \;=\;\sin(213^o - 132^o) \;=\;\sin(213^o)\cos(132^o) - \cos(213^o)\sin(132^o)$

. . Since $\displaystyle 132 \:=\:2(66)$, we have: .$\displaystyle \sin(213^o)\overbrace{\cos(2\!\cdot\!66^o)} - \cos(213^o)\overbrace{\sin(2\!\cdot\!66^o)}$

. . . . . . . . . . . . . . . . $\displaystyle = \;\sin(213^o)\overbrace{\bigg[2\cos^2(66^o) - 1\bigg]} - \cos(213^o)\overbrace{\bigg[2\!\cdot\!\sin(66^o)\!\cdot\!\cos(66^o)\bigg]}$

Now substitute the values given in the table . . .

6. Originally Posted by Soroban
Hello, aigiqinf!

The hint refers to:

. . a Difference identity: .$\displaystyle \sin(A-B) \:=\:\sin(A)\cos(B) - \cos(A)\sin(B)$
. . two Double-Angle identities: .$\displaystyle \begin{array}{ccc}\sin(2A) &=& 2\sin(A)\cos(A) \\ \cos(2A) &=&2\cos^2(A) - 1 \end{array}$

We note that: .$\displaystyle 81 \:=\:213 - 132$

So: .$\displaystyle \sin81^o \;=\;\sin(213^o - 132^o) \;=\;\sin(213^o)\cos(132^o) - \cos(213^o)\sin(132^o)$

. . Since $\displaystyle 132 \:=\:2(66)$, we have: .$\displaystyle \sin(213^o)\overbrace{\cos(2\!\cdot\!66^o)} - \cos(213^o)\overbrace{\sin(2\!\cdot\!66^o)}$

. . . . . . . . . . . . . . . . $\displaystyle = \;\sin(213^o)\overbrace{\bigg[2\cos^2(66^o) - 1\bigg]} - \cos(213^o)\overbrace{\bigg[2\!\cdot\!\sin(66^o)\!\cdot\!\cos(66^o)\bigg]}$

Now substitute the values given in the table . . .

Thank you!

-.5446*(2(.4067^2-1)--.8387(2*.9135*.4067)=.9876 degrees