5 -13 2) Find sin 2x, cos 2x, tan 2x if secx = -13/5 pi/2 < x < pi
Answers
Step-by-step explanation:
sin2x=−169120
\cos 2x=\frac{-119}{169}cos2x=169−119
\tan 2x=\frac{120}{119}tan2x=119120
Step-by-step explanation:
Given : If \sec x=-\frac{13}{5}secx=−513 , \frac{\pi}{2} < x < \pi2π<x<π
To find : \sin 2x,\cos 2x,\tan 2xsin2x,cos2x,tan2x ?
Solution :
We know, \frac{\pi}{2} < x < \pi2π<x<π x lies in 2nd quadrant.
According to trigonometric properties,
\sec x=-\frac{13}{5}=\frac{H}{B}secx=−513=BH
Applying Pythagorean theorem,
P=\sqrt{H^2-B^2}P=H2−B2
P=\sqrt{(13)^2-(5)^2}P=(13)2−(5)2
P=\sqrt{169-25}P=169−25
P=\sqrt{144}P=144
P=12P=12
Now, \sin x=\frac{P}{H}=\frac{12}{13}sinx=HP=1312
\cos x=\frac{B}{H}=-\frac{5}{13}cosx=HB=−135 (as cos is negative in 2nd quadrant)
Applying trigonometric formulas,
1) \sin 2x=2\sin x\cos xsin2x=2sinxcosx
\sin 2x=2(\frac{12}{13})(\frac{-5}{13})sin2x=2(1312)(13−5)
\sin 2x=-\frac{120}{169}sin2x=−169120
2) \cos 2x=2\cos^2 x-1cos2x=2cos2x−1
\cos 2x=2(\frac{-5}{13})^2-1cos2x=2(13−5)2−1
\cos 2x=\frac{50}{169}-1cos2x=16950−1
\cos 2x=\frac{-119}{169}cos2x=169−119
3) \tan 2x=\frac{\sin 2x}{\cos 2x}tan2x=cos2xsin2x
\tan 2x=\dfrac{\frac{-120}{169}}{\frac{-119}{169}}tan2x=169−119169−120
\tan 2x=\frac{120}{119}tan2x=119120
#Learn more
If cos x = -3/5 and π < x < 3π/2 , find the value of sin2x ,cos2x ,tan2x