trig identities: find exact values for sin15deg, cos105deg,
- Thread starter dave199
- Start date
pka
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- Jan 29, 2005
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First you need to know the following.
\(\displaystyle \begin{array}{l} \sin \left( {A \pm B} \right) = \sin (A)\cos (B) \pm \sin (B)\cos (A) \\ \cos \left( {A \pm B} \right) = \cos (A)\cos (B) \mp \sin (B)\sin (A) \\ \end{array}\).
The is is matter of making good guesses.
I will a) for you.
\(\displaystyle \sin \left( {15^ \circ } \right) = \sin \left( {45^ \circ - 30^ \circ } \right) = \sin (45^ \circ )\cos (30^ \circ ) - \sin (30^ \circ )\cos (45^ \circ )\)
You should know all the "well-known" angles.
\(\displaystyle \begin{array}{l} \sin \left( {A \pm B} \right) = \sin (A)\cos (B) \pm \sin (B)\cos (A) \\ \cos \left( {A \pm B} \right) = \cos (A)\cos (B) \mp \sin (B)\sin (A) \\ \end{array}\).
The is is matter of making good guesses.
I will a) for you.
\(\displaystyle \sin \left( {15^ \circ } \right) = \sin \left( {45^ \circ - 30^ \circ } \right) = \sin (45^ \circ )\cos (30^ \circ ) - \sin (30^ \circ )\cos (45^ \circ )\)
You should know all the "well-known" angles.
I assume you're not supposed to use a calculator. I will do the first one and give you an idea of one way to tackle the others.
The idea is to write them in the form of already well-known values.
\(\displaystyle sin(15)=sin(45-30)\)
The subtraction formula is \(\displaystyle sin(u-v)=sin(u)cos(v)-cos(u)sin(v)\)
Using the subtraction formula, We can write this one as \(\displaystyle sin(45)cos(30)-cos(45)sin(30)\)
Now, \(\displaystyle cos(45)=sin(45)=\frac{1}{\sqrt{2}}, \;\ sin(30)=\frac{1}{2}, \;\ cos(30)=\frac{\sqrt{3}}{2}\)
You can use the unit circle to find these values or look them up in your book or use a calculator.
You may also be able to use the addition formulae for some of them.
EDIT: Oops, looks like pka beat me and I was thinking along his lines.
