y is not in the exponent
2006-12-19 13:17:13 · 6 answers · asked by DadinTrouble 1 in Science & Mathematics ➔ Mathematics
cos^3 y + cos y*sin^2 y = cos y (cos^2y + sin^2y)
From the trigonometric identity (cos^2y + sin^2y) = 1
we get
cos^3 y + cos y*sin^2 y = cos y (1)
therefore,
cos^3 y + cos y*sin^2 y = cos y
2006-12-19 13:21:49 · answer #1 · answered by dkrudge 2 · 0⤊ 0⤋
Cos(y) y is not in the exponent. this is found by factoring out cos(y) and then using a trigonimetric identity cos^2(y)+sin^2(y)=1
2006-12-19 13:20:44 · answer #2 · answered by master_furches 2 · 1⤊ 0⤋
cos^3 y + cos y*sin^2 y
= cos y (cos^2 y + sin^2 y)
= cos y,
since cos^2 y + sin^2 y = 1 for all y.
This reminds me a little of multiple-angle formulas. If it were cos^3 y - cos y* sin^2 y, then that would be:
cos y(cos^2 y - sin^2 y)
= cos y * cos (2y)
2006-12-19 14:29:26 · answer #3 · answered by alnitaka 4 · 0⤊ 0⤋
cos^3 y + cos y*sin^2 y
=cos y ( cos ^2 y +sin^2 y)
=cos y * (1)
=cos y
since, cos^2+sin^2=1
2006-12-19 13:22:49 · answer #4 · answered by angel 2 · 0⤊ 0⤋
cos y
2006-12-19 13:19:41 · answer #5 · answered by Rick 5 · 0⤊ 0⤋
cos(y)^3 + cos(y)sin(y)^2
cos(y)^3 + cos(y)(1 - cos(y)^2)
cos(y)^3 + cos(y) - cos(y)^3
cos(y)
ANS : cos(y)
2006-12-19 13:56:36 · answer #6 · answered by Sherman81 6 · 0⤊ 0⤋