Trig identity help (1 Viewer)

[dumNerd]

Someone please do L (identity)

 

[Randomstudent123]

for L you expand and get sin^2A+cos^2A+sin^2B+cos^2B+2sinAsinB+2cosAcosB then u use sin^2x+cos^2x=1 and
cos(A-B)=cosAcosB+sinAsinB to get 2+2(cos(A-B))then you use cos2x=2cos^2(x)-1to get 2+2(2cos^2((A-B)/2)-1)=4cos^2((A-B)/2)

 

[Trebla]

For future reference, can you please post subject specific content content in subject specific forums?

 

[YonOra]

How can you use the identity of Cos2x here? Or am i just being dumb

 

[dumNerd]

For future reference, can you please post subject specific content content in subject specific forums?

Oh yeh sure sorry

 

[dumNerd]

How can you use the identity of Cos2x here? Or am i just being dumb

?

 

[dumNerd]

How can you use the identity of Cos2x here? Or am i just being dumb

I don't know what the hell to do about the 1/2

 

[dumNerd]

for L you expand and get sin^2A+cos^2A+sin^2B+cos^2B+2sinAsinB+2cosAcosB then u use sin^2x+cos^2x=1 and
cos(A-B)=cosAcosB+sinAsinB to get 2+2(cos(A-B))then you use cos2x=2cos^2(x)-1to get 2+2(2cos^2((A-B)/2)-1)=4cos^2((A-B)/2)

Thank but I'm a bit confused about what happened from here 2+2(2cos^2((A-B)/2)-1) --> 4cos^2((A-B)/2)

 

[Randomstudent123]

u just expand and the +2 and -2 cancel leaving with 4cos^2((A-B)/2)

 

[dumNerd]

Someone please do L (identity)View attachment 29432

Can someone also explain how to do 14 a. Thanks

 

[idkkdi]

Can someone also explain how to do 14 a. Thanks

y = cos2a = cos^a - sin^a = cos^a - (1-cos^a) = 2cos^a -1
express costheta in terms of y, sub into x equation.

 

[dumNerd]

y = cos2a = cos^a - sin^a = cos^a - (1-cos^a) = 2cos^a -1
express costheta in terms of y, sub into x equation.

Ahh Shit I'm stupid

 

[5uckerberg]

y = cos2a = cos^a - sin^a = cos^a - (1-cos^a) = 2cos^a -1
express costheta in terms of y, sub into x equation.

Can you just do .
Next, where


I reckon that through practice will come easier.