# Trig question I need to understand this tonight!

• May 30th 2008, 02:58 PM
aigiqinf
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! (Bow)
• May 30th 2008, 03:54 PM
Mathstud28
Quote:

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! (Bow)

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
$128=213-\frac{66}{2}-\frac{104}{2}$ 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
• May 30th 2008, 03:57 PM
aigiqinf
Quote:

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
$128=213-\frac{66}{2}-\frac{104}{2}$ 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.
• May 30th 2008, 04:11 PM
sean.1986
Quote:

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! (Bow)

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)
• May 30th 2008, 04:14 PM
Soroban
Hello, aigiqinf!

The hint refers to:

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

Quote:

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

Find $\sin81^o$

Hint: Double, Difference

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

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

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

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

Now substitute the values given in the table . . .

• May 30th 2008, 04:32 PM
aigiqinf
Quote:

Originally Posted by Soroban
Hello, aigiqinf!

The hint refers to:

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

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

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

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

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

Now substitute the values given in the table . . .

Thank you!

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