The solution to the problem is as follows:
<span>cos2A - cos4A = -2 sin(6A/2).sin(-2A/2) = +2 sin(3A).sinA
and
sin4A - sin2A = 2 cos(6A/2).sin(2A/2) = 2 cos(3A).sinA
Hence RHS = (cos(2A)-cos(4A))/ (sin(4A)- Sin(2A)) = sin 3A / cos 3A = tan 3A = LHS
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- The angles are supplementary angles
- The measure of angle m<1 and m<2 are 72 and 108 degrees respectively
Find the diagram attached. From the figure, the sum of the angles m<1 and m<2 are supplementary
Given the following angles
m<1 = 2x + 6
m<2 = 3x + 9
Since both angles are supplementary, hence;
2x + 6 + 3x + 9 = 180
5x + 15 = 180
5x = 180 - 15
5x = 165
x = 165/5
x = 33
Get the measure of m<1;
m<1 = 2x + 6
m<1 = 2(33) + 6
m<1 = 66 + 6
m<1 = 72degrees
Get the measure of m<2:
m<2 = 180 - m<1
m<2 = 180 - 72
m<2 = 108degrees
Hence the measure of angle m<1 and m<2 are 72 and 108 degrees respectively
Learn more here: brainly.com/question/22309882
Period =2r /b
assume frequency factor meaning angular frequency "b"
then
2r= 2r/b => b=1
A. sine function whose period 2pi radians