Trig identity help (1 Viewer)

Randomstudent123

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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

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

YonOra

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

dumNerd

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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)
 

5uckerberg

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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.
 
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