Question 392592: please explain or show work. this exactly how the question was written.
using the exact values,find the value of:
a.12 sin 30degrees - 6 tan 45 degress + sec 45 degrees / 52
b.sin 2A + tan 3A/2 - cos A + sec(A+15), when A= 30degrees
There were no parentheses and no exponents. please help by showing the work in details. Thank you!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! using the exact values,find the value of:
a.12 sin 30degrees - 6 tan 45 degress + sec 45 degrees / 52
b.sin 2A + tan 3A/2 - cos A + sec(A+15), when A= 30degrees
All you need to know is the value of the trigonometry functions at the special angles. For this problem, you need to know the following:
sin 30 deg = 1/2
tan 45 deg = 1
sec 45 = sqrt(2)
cos 30 =sqrt(3)/2
sin 60 = sqrt(3)/2
a. 12 sin 30 deg - 6 tan 45 deg +sec 45 deg/52
= 12*1/2-6*1+sqrt(2)/52=6-6+sqrt(2)/52
=sqrt(2)/52
b.sin 2A + tan 3A/2 - cos A + sec(A+15)
=sin (2*30)+tan (3*30/2)-cos 30+sec(30+15)
= sin 60+tan 45-cos 30+sec 45
= sqrt(3)/2+1-sqrt(3)/2+sqrt(2)
= 1+sqrt(2)
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