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Prove that the value of sin3x/cos2x is positive in (13pi/48,14pi/48)?

Prove that the value of sin3x/cos2x is positive in (13pi/48,14pi/48)?

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1 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
 First lets prove that sin3x/cos2x=(2 sin(3 x) cos(2 x))/(cos(4 x) +1)
Cross multiply:
sin(3 x) (cos(4 x) +1)=2 cos(2 x)^2 sin(3 x)
Dividing both sides by sin(3 x):
cos(4 x) +1=2 cos(2 x)^2
cos(2 x)^2 =1/2 (cos(4 x) +1):
cos(4 x) +1=2 (cos(4 x) +1)/2
(2 (cos(4 x) +1))/2 =cos(4 x) +1:
cos(4 x) +1=cos(4 x) +1
The left hand side and right hand side are equal.
So as in (2 sin(3 x) cos(2 x))/(cos(4 x) +1): sin3x,cos2x,cos4x are positive(as all angles lie in first quadrant as it is asked between 13pi/48 to 14pi/48)
Therefore sin3x/cos2x is always positive in (13pi/48,14pi/48).
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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