Question

simplify the expression...

Answer 1

\[(\sin ^{2}\theta+\cos ^{2}\theta) \div (\sin \theta+\cos \theta)\]

Answer 2

numerator is 1

Answer 3

oh im sorry, the power 2 is supposed to be 3

Answer 4

on account of sine and cosine are points on the unit circle

Answer 5

Trig Cheat Sheet: http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

Answer 6

ah in that case factor as the sum of two cubes. it has nothing to do with trig at all

Answer 7

in that case factor the numerator as sum of two cubes. it has nothing whatsoever to do with trig

Answer 8

\[(\sin ^{3}\theta+\cos ^{3}\theta) \div (\sin \theta+\cos \theta)\]

Answer 9

Sum and difference of two cubes: http://www.purplemath.com/modules/specfact2.htm

Answer 10

in that case factor the numerator as sum of two cubes. it has nothing whatsoever to do with trig

Answer 11

\[\frac{a^3+b^3}{a+b}=\frac{(a+b)(a^2-ab+b^2)}{a+b}=a^2-ab+b^2\]

Answer 12

now comes the trig. replace
\[a=\sin(\theta), b =\cos(\theta)\] get
\[\sin^2(\theta) -2\sin(\theta)\cos(\theta) + \cos^2(\theta)\]
\[1-2\sin(\theta)\cos(\theta)\]

Answer 13

i'm doing my work online, it rejected that answer.

Answer 14

\[2\cos \theta-[1/(\sin \theta+\cos \theta)\]]