Question

use the sum/difference identities to evaluate exactly
sin pi/12

Answer 1

\[\pi/12*180/\pi\]=15
sin (45-30 )

Answer 2

Use the difference formula:

sin(x - y) = sin x cos y - sin y cos x

x = pi/3 and y = pi/4; so x - y = pi/3 - pi/4 = pi/12

Answer 3

first, write down the rule for
sin(A-B)
see http://www.intmath.com/analytic-trigonometry/2-sum-difference-angles.php

Answer 4

sin45cos30-cos45sin30

Answer 5

now replace with numbers

Answer 6

\[\sqrt{4-4}/4\]

Answer 7

right

Answer 8

sin(15) won't be 0
did you use these values:
sin 45 = sqr(2)/2
cos 30= sqr(3)/2
cos 45= sqr(2)/2
sin 30= ½

Answer 9

\[\sqrt{2}-\sqrt{2}/4\]

Answer 10

now it is right

Answer 11

you should write down the expression first. You are going wrong somewheres, but I don't know where.

Answer 12

\[\sqrt{6}-\sqrt{2}/4\]

Answer 13

right

Answer 14

sin45cos30-cos45sin30
sin 45 = sqr(2)/2
cos 30= sqr(3)/2
so the first term is
sqr(2)/2 * sqr(3)/2

the 2nd term uses
cos 45= sqr(2)/2
sin 30= ½

it is sqr(2)/2 * ½

and you now have
sqr(2)/2 * sqr(3)/2 - sqr(2)/2 * ½

written more nicely:
\[ \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \]

multiply top times top and bottom times bottom

you will have 4 as a common denominator, so you can put the top all over 4

Answer 15

\[\sqrt{6}/4-\sqrt{2}/4\]

Answer 16

right

Answer 17

or \[\sqrt{6}-\sqrt{2}/4\]

Answer 18

ok, but only if you put parens around the top
\[ (\sqrt{6}-\sqrt{2})/4 = \frac{\sqrt{6}-\sqrt{2}}{4}\]

if you leave out the parens, it means
\[ \sqrt{6} - \frac{\sqrt{2}}{4} \]

but yes, you have the answer

Answer 19

thank you :)