+0

+1
95
1
+5

If cos A = -4/5 for angle A in Quadrant II, find sin 2A

Harmmyy  Jun 6, 2017

#1
+4784
+1

sin( 2A )  =  2 sin( A ) cos( A )

We can find   sin( A )   using the Pythagorean identity.

sin2 A  +  cos2 A   =   1

sin2 A  +  (-4/5)2   =   1

sin2 A  +  (16/25)  =   1

sin2 A   =   1 - 16/25

sin2 A   =   9/25                     Take the square root of both sides.

sin A = 3/5                             Since sin is positive in Quadrant II, we only take the positive root.

Now we can find   sin( 2A )   using the double-angle identity.

sin( 2A )  =  2 sin( A ) cos( A )

sin( 2A )  =  2 (3/5) (-4/5)

sin( 2A )  =  -24/25

hectictar  Jun 6, 2017
Sort:

#1
+4784
+1

sin( 2A )  =  2 sin( A ) cos( A )

We can find   sin( A )   using the Pythagorean identity.

sin2 A  +  cos2 A   =   1

sin2 A  +  (-4/5)2   =   1

sin2 A  +  (16/25)  =   1

sin2 A   =   1 - 16/25

sin2 A   =   9/25                     Take the square root of both sides.

sin A = 3/5                             Since sin is positive in Quadrant II, we only take the positive root.

Now we can find   sin( 2A )   using the double-angle identity.

sin( 2A )  =  2 sin( A ) cos( A )

sin( 2A )  =  2 (3/5) (-4/5)

sin( 2A )  =  -24/25

hectictar  Jun 6, 2017

### 24 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details