+0  
 
+1
315
1
avatar+5 

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

 Jun 6, 2017

Best Answer 

 #1
avatar+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
avatar+7347 
+1
Best Answer

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

18 Online Users

avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.