Asked by Amy
solve cos 5 thetha cos 15 -sin5theta sin 15 =squareroot of 3/2
Thank you
Thank you
Answers
Answered by
drwls
Remember that
cos(a+b) = cosa cosb - sina sinb
Therefore
cos5theta*cos 15 -sin5theta*sin 15
= squareroot of 3/2
cos(5theta + 15)
= cos^-1(sqrt3)/2
Take the arccosines of both sides
5 theta + 15 = 30 degrees5 theta = 15 degrees
theta = 3 degrees
cos(a+b) = cosa cosb - sina sinb
Therefore
cos5theta*cos 15 -sin5theta*sin 15
= squareroot of 3/2
cos(5theta + 15)
= cos^-1(sqrt3)/2
Take the arccosines of both sides
5 theta + 15 = 30 degrees5 theta = 15 degrees
theta = 3 degrees
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