Asked by Dora
3 cos^2 𝛼 + 2 cos^2 𝛽 =4
3 sin 2 𝛼 − 2sin 2 𝛽=0
Find the values of cos 2𝛼 and cos 2𝛽.
3 sin 2 𝛼 − 2sin 2 𝛽=0
Find the values of cos 2𝛼 and cos 2𝛽.
Answers
Answered by
oobleck
I already did this for you. Take the trouble to read it this time, ok?
since sin^2 + cos^2 = 1, the 2nd equation becomes
3(1-cos^2 𝛼) − 2(1-cos^2 𝛽) = 0
3 - 3cos^2 𝛼 − 2 + 2cos^2 𝛽 = 0
Now the two equations read
3cos^2 𝛼 + 2cos^2 𝛽 = 4
-3cos^2 𝛼 + 2cos^2 𝛽 = -1
subtract to get
6cos^2𝛼 = 5
cos^2𝛼 = 5/6
Now just finish solving for cos^2𝛼 and cos^2𝛽
Then use your double-angle formula
cos 2x = 2cos^2x - 1
post your work if you get stuck, but don't just post the same problem all over again.
since sin^2 + cos^2 = 1, the 2nd equation becomes
3(1-cos^2 𝛼) − 2(1-cos^2 𝛽) = 0
3 - 3cos^2 𝛼 − 2 + 2cos^2 𝛽 = 0
Now the two equations read
3cos^2 𝛼 + 2cos^2 𝛽 = 4
-3cos^2 𝛼 + 2cos^2 𝛽 = -1
subtract to get
6cos^2𝛼 = 5
cos^2𝛼 = 5/6
Now just finish solving for cos^2𝛼 and cos^2𝛽
Then use your double-angle formula
cos 2x = 2cos^2x - 1
post your work if you get stuck, but don't just post the same problem all over again.
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