Question

Simplify this expression

Answer 1

\[ \huge \ln e^{\sin^2x} + \ln e^{\cos^2x} \]

Answer 2

hah.,
a

Answer 3

the answer is one.. but how ?!

Answer 4

its a nice problem

Answer 5

its 0

Answer 6

no my answerkey says 1

Answer 7

:'( walks in shame

Answer 8

\[\cos^2x \ln e+\sin^2x \ln e\]
but \[\ln e\] is 1

Answer 9

since \[\ln e = \log_{e} ^e\]

Answer 10

\[\sin^2x+\cos^2x\]

Answer 11

we know that \[\cos^2x+\sin^2x=1\]
in trigonometry

Answer 12

thast why the answer is one

Answer 13

log e is = to lin

Answer 14

thank you :D

Answer 15

so log e to e to 1 is 1

Answer 16

you're welcome..