Question

Tan15, I have to find the exact value using sum and difference formulas

Answer 1

@Michele_Laino sorry to bother you but can you help with me with this problem?

Answer 2

we can write this:
\[\Large \begin{gathered}
\tan 15 = \tan \left( {45 - 30} \right) = \hfill \\
= \frac{{\tan 45 - \tan 30}}{{1 + \tan 45 \times \tan 30}} = ...? \hfill \\
\end{gathered} \]

Answer 3

Yes then you get 1-(1/sqrt3)/1+(1)(1/sqrt3)

Answer 4

@Michele_Laino

Answer 5

yes! that's right!

Answer 6

Can you show me how to do the math for the next part

Answer 7

here are my steps:
\[\Large \begin{gathered}
\tan 15 = \tan \left( {45 - 30} \right) = \hfill \\
= \frac{{\tan 45 - \tan 30}}{{1 + \tan 45 \times \tan 30}} = \frac{{1 - \frac{1}{{\sqrt 3 }}}}{{1 + \frac{1}{{\sqrt 3 }}}} \hfill \\
\end{gathered} \]

Answer 8

now the least common multiple at the denominator is sqrt(3), and the least common multiple at the denominator is also sqrt(39, so we can write:
\[\Large \frac{{1 - \frac{1}{{\sqrt 3 }}}}{{1 + \frac{1}{{\sqrt 3 }}}} = \frac{{\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}}}{{\frac{{\sqrt 3 + 1}}{{\sqrt 3 }}}} = \frac{{\sqrt 3 - 1}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 + 1}} = ...?\]

Answer 9

Why did the 3rd piece come out as just sqrt 3-1? @Michele_Laino

Answer 10

What step

Answer 11

the sqrt 3-1/sqrt3

Answer 12

since I computed this:
\[1 - \frac{1}{{\sqrt 3 }} = \frac{1}{1} - \frac{1}{{\sqrt 3 }} = \frac{{1 \times \sqrt 3 - 1}}{{\sqrt 3 }} = \frac{{\sqrt 3 - 1}}{{\sqrt 3 }}\]

Answer 13

namely I have computed the least common multiple between 1 and sqrt(3)

Answer 14

similarly it goes for denominator:
\[1 + \frac{1}{{\sqrt 3 }} = \frac{1}{1} + \frac{1}{{\sqrt 3 }} = \frac{{1 \times \sqrt 3 + 1}}{{\sqrt 3 }} = \frac{{\sqrt 3 + 1}}{{\sqrt 3 }}\]

Answer 15

the last step is since we have to multiply the first fraction by the inverse of the second fraction