Finding the exact value!! help!

[mathfun]

how do u do this?

find the exact value of sec195 degrees.

thanks in advance

 

[Unco]

G'day, Mathfun.

One way would be to use 195 = 90 + 45 + 60 and apply the compound angle identities a couple of times. sec(195) = 1/cos(195).

 

[mathfun]

G'day, Unco

So then the equation would be
1/(cos(90+45+60))
right? and then i use compound trig identities

cos(a + b) = cosacosb - sinasinb

right?

what is the identities for three variables?

 

[Gene]

You have to apply sum of two variables twice.
(a+(b+c)) then
(b+c)
to that answer.

 

[Unco]

Nice one.

cos(a + b + c) = cos((a + b) + c) = cos(a + b).cos(c) - sin(a + b).sin(c)

cos(a + b) and sin(a + b) are found using compound angle identities as well.

 

[mathfun]

oh!! thanks so much.
i'm gonna try that out tomorrow...its too late now ...*yawnz*

happy boxing day!

 

[mathfun]

after i worked it out, i got

is that correct? because according to the answers (that my teacher gave me) it is not.

 

[Unco]

.\(\displaystyle \L \cos{(135 + 60) = \cos{135} \cdot \cos{60} \, - \, \sin{135} \cdot \sin{60}\)

. Where

. . .\(\displaystyle \L \begin{align*}
\cos{135} &= \cos{(90 \, + \, 45)} \\
\\
\\
&= \cos{90} \cdot \cos{45} \, - \, \sin{90} \cdot \sin{45} \\
\\
\\
&= 0 \cdot \frac{1}{\sqrt{2}} \, - \, 1 \cdot \frac{1}{\sqrt{2}} \\
\\
\\
&= -\frac{1}{\sqrt{2}} \\
\end{align*}\)

. . .\(\displaystyle \L \cos{60} = \frac{1}{2}\)

. . .\(\displaystyle \L \begin{align*} \sin{135} &= \sin{(90 + 45)} \\
\\
\\
&= \sin{90} \cdot \cos{45} \, + \, \cos{90} \cdot \sin{45} \\
\\
\\
&= 1 \cdot \frac{1}{\sqrt{2}} \, + \, 0 \cdot \frac{1}{\sqrt{2}} \\
\\
\\
&= \frac{1}{\sqrt{2}} \\
\end{align*}\)

. . .\(\displaystyle \L \sin{60} = \frac{\sqrt{3}}{2}\)

So we have
.\(\displaystyle \L \begin{align*}\cos{195} &= \left(-\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\right) - \left(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}\right) \\
\\
\\
&= -\frac{1}{2\sqrt{2}} \, - \, \frac{\sqrt{3}}{2\sqrt{2}} \\
\\
\\
&= -\frac{1 + \sqrt{3}}{2\sqrt{2}} \\

\end{align*}\)

So
. . .\(\displaystyle \L \begin{align*}
\sec{195} &= -\frac{2\sqrt{2}}{1 + \sqrt{3}} \\
\\
\\
&= -\frac{2\sqrt{2}(1 - \sqrt{3})}{1 - 3} \\
\\
\\
&= -\frac{2\sqrt{2} - 2\sqrt{6}}{-2} \\
\\
\\
&= \sqrt{2} - \sqrt{6} \\
\end{align*}\)

 

[mathfun]

wow! thanx so much! thank youuuuuuu!