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1)sin127 cos82- cos127 sin82

2)sin220+ cos220

3) sin4alpha+ sin2alpha• cos2alpha+ cos2alpha=1

4) sin2x-sinx=0

 

Thanks for the help!!!

Guest May 30, 2017
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1)sin127 cos82- cos127 sin82

 

Notice  that  this actually simplifies to  sin (127 - 82)  =  sin (45) =  1 / √2

 

 

2)sin^2(20)+ cos^2(20)   =   1         [ using the identity  sin^2(theta) + cos^2(theta)  = 1 ]

 

 

 

3) sin^4alpha+ sin^2alpha• cos^2alpha+ cos^2alpha=1

 

sin^4alpha  + sin^2theta  (1 - sin^2theta)  + cos^2alpha

 

sin^4alpha  + sin^2alpha  - sin^4alpha  + cos^2 alpha =

 

sin^2alpha   +  cos^2alpha   = 1          and this is true

 

 

4) sin2x-sinx=0    

 

2sinxcosx - sinx  = 0      factor

 

sinx ( 2cosx - 1)  = 0         set each factor to 0 and solve

 

sin x  = 0   at   0 + pi *n      where n is an integer

 

And for the other factor

 

2cosx  - 1  = 0       add 1 to both sides

 

2cosx  = 1            divide both sides by 2

 

oos x   = 1/2      

 

And this happens  at  pi/3 + 2pi*n    and   at    5pi/3 + 2pi*n    for some integer, n

 

 

 

cool cool cool

CPhill  May 30, 2017

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