Question

find sin2x, cos2x ,tan2x if secx=-13/5 pi/2<x<pi​

Answer 1

Answer:

please see the attachment

Answer 2

$\sin 2x=-\frac{120}{169}$

$\cos 2x=\frac{-119}{169}$

$\tan 2x=\frac{120}{119}$

Step-by-step explanation:

Given : If  $\sec x=-\frac{13}{5}$ , $\frac{\pi}{2}<x<\pi$

To find : $\sin 2x,\cos 2x,\tan 2x$ ?

Solution :

We know,  $\frac{\pi}{2}<x<\pi$ x lies in 2nd quadrant.

According to trigonometric properties,

$\sec x=-\frac{13}{5}=\frac{H}{B}$

Applying Pythagorean theorem,

$P=\sqrt{H^2-B^2}$

$P=\sqrt{(13)^2-(5)^2}$

$P=\sqrt{169-25}$

$P=\sqrt{144}$

$P=12$

Now, $\sin x=\frac{P}{H}=\frac{12}{13}$

$\cos x=\frac{B}{H}=-\frac{5}{13}$ (as cos is negative in 2nd quadrant)

Applying trigonometric formulas,

1) $\sin 2x=2\sin x\cos x$

$\sin 2x=2(\frac{12}{13})(\frac{-5}{13})$

$\sin 2x=-\frac{120}{169}$

2) $\cos 2x=2\cos^2 x-1$

$\cos 2x=2(\frac{-5}{13})^2-1$

$\cos 2x=\frac{50}{169}-1$

$\cos 2x=\frac{-119}{169}$

3) $\tan 2x=\frac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{\frac{-120}{169}}{\frac{-119}{169}}$

$\tan 2x=\frac{120}{119}$

#Learn more

If cos x = -3/5 and π < x < 3π/2 , find the value of sin2x ,cos2x ,tan2x

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