• January 5th 2008, 08:56 PM
oceanmd
Please find tan(B+C) given that sin C= $\frac {1}{4}$, with C in quadrant 2, and sin B= $\frac {-1}{2}$, with B in quadrant 4.
The correct answer is $\frac {4 \sqrt{3} + \sqrt{15}}{-11}$

I have fount that tan C= $\frac{1}{- \sqrt{15}}$ and tan B= $\frac {-1}{\sqrt{3}}$
$\frac {-1}{\sqrt{3}}$+ $\frac{1}{-\sqrt{15}}$ over 1- $\frac{-1}{\sqrt{3}}$ $(\frac{1}{-\sqrt{15}})$
Is this correct? What do I do next to get the correct answer?

Thank you
• January 5th 2008, 09:12 PM
Jhevon
Quote:

Originally Posted by oceanmd
Please find tan(B+C) given that sin C= $\frac {1}{4}$, with C in quadrant 2, and sin B= $\frac {-1}{2}$, with B in quadrant 4.
The correct answer is $\frac {4 \sqrt{3} + \sqrt{15}}{-11}$

I have fount that tan C= $\frac{1}{- \sqrt{15}}$ and tan B= $\frac {-1}{\sqrt{3}}$
$\frac {-1}{\sqrt{3}}$+ $\frac{1}{-\sqrt{15}}$ over 1- $\frac{-1}{\sqrt{3}}$ $(\frac{1}{-\sqrt{15}})$
Is this correct? What do I do next to get the correct answer?

Thank you

umm...you don't have a lot of options here. combine the fractions in the numerator and denominator and simplify (note: you may need to conjugate). for the record, i used a different formula and got the answer, it's still a pain to simplify though. i always have trouble remembering the addition formula for tangent, so i used $\tan (B + C) = \frac {\sin (B + C)}{\cos (B + C)}$. but i'm sure you'll get the answer doing it your way. your values for tanB and tanC are correct
• January 6th 2008, 08:34 AM
Soroban
Hello, oceanmd!

You're doing great !!

Quote:

Find $\tan(B+C)$, given that: $\sin C = \frac {1}{4}\text{, with C in quadrant 2}$
and $\sin B = -\frac {1}{2}\text{, with B in quadrant 4.}$

The correct answer is: . $\frac {4 \sqrt{3} + \sqrt{15}}{-11}$
UGH! . . . Who writes fractions that way? . ${\color{blue}-\,\frac{4\sqrt{3}+\sqrt{15}}{11}}$

I have found that: . $\tan C\:=\:-\frac{1}{\sqrt{15}}\:\text{ and }\:\tan B\:=\:-\frac {1}{\sqrt{3}}$

Then: . $\frac{-\frac {1}{\sqrt{3}} -\frac{1}{\sqrt{15}}}{1 - \left(-\frac{1}{\sqrt{3}}\right)\left(-\frac{1}{\sqrt{15}}\right)}$

Is this correct? . . . . Yes!
What do I do next to get the correct answer? . . . . Simplify!

We have: . $-\,\frac{\dfrac{1}{\sqrt{3}} + \dfrac{1}{\sqrt{15}}}{1 - \dfrac{1}{\sqrt{3}\sqrt{15}}}$

Multiply top and bottom by $\sqrt{3}\sqrt{15}$

. . $-\,\frac{\dfrac{1}{\sqrt{3}} + \dfrac{1}{\sqrt{15}}}{1 - \dfrac{1}{\sqrt{3}\sqrt{15}}}\,\cdot\,\frac{\sqrt{ 3}\sqrt{15}}{\sqrt{3}\sqrt{15}} \;=\;-\,\frac{\sqrt{15} + \sqrt{3}} {\sqrt{3}\sqrt{15} - 1} \;=\;-\,\frac{\sqrt{15} + \sqrt{3}}{\sqrt{45} - 1} \;=\;-\,\frac{\sqrt{15}+\sqrt{3}}{3\sqrt{5}-1}$

Rationalize: .multiply top and botttom by $(3\sqrt{5} + 1)$

$-\frac{\sqrt{15}+\sqrt{3}}{3\sqrt{5} - 1}\,\cdot\,\frac{3\sqrt{5} + 1}{3\sqrt{5}+1} \;=\;-\,\frac{(\sqrt{15})(3\sqrt{5}) + (\sqrt{15})(1) + (\sqrt{3})(3\sqrt{5}) + (\sqrt{3})(1)}{(3\sqrt{5})^2 - 1^2}$

. . $= \;-\,\frac{{\color{red}3\sqrt{75}} +\sqrt{15} +3\sqrt{15} + \sqrt{3}}{45 - 1} \;=\;-\,\frac{{\color{red}15\sqrt{3}} + \sqrt{15} + 3\sqrt{15} + \sqrt{3}}{44}\;\;{\color{red}**}$

. . $= \;-\,\frac{16\sqrt{3} + 4\sqrt{15}}{44} \;=\;-\,\frac{4(4\sqrt{3} + \sqrt{15})}{44} \;=\;\boxed{-\,\frac{4\sqrt{3} + \sqrt{15}}{11}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** . $3\!\cdot\!\sqrt{75} \:=\:3\!\cdot\!\sqrt{25\cdot3} \;=\;3\!\cdot\!\sqrt{25}\!\cdot\!\sqrt{3} \;=\;3\!\cdot\!5\!\cdot\!\sqrt{3} \;=\;15\sqrt{3}$
• January 6th 2008, 11:11 AM
oceanmd
Thank you very much.