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

Jun 6, 2017

#1
+7347
+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

Jun 6, 2017

#1
+7347
+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