C3 TRICKY TRIG EQ: Where have I gone wrong? *Maths HELP*
sin(x+π/6) = 3cos(x-π/6)
sinxcosπ/6 + sinπ/6cosx = 3cosxcosπ/6 + 3sinxsinπ/6
√3/2sinx + 1/2cosx = 3√3/2cosx + 3/2sinx
1-3√3/2cos = 3-√3/2sin
1-3√3/2 = 3-√3/2tan
3-√ 3/2 = arctan(1-3√ 3/2)
-1.78..
Answer should be -1.28 and 1.86
Original post by creativebuzz
sin(x+π/6) = 3cos(x-π/6)
sinxcosπ/6 + sinπ/6cosx = 3cosxcosπ/6 + 3sinxsinπ/6
√3/2sinx + 1/2cosx = 3√3/2cosx + 3/2sinx
1-3√3/2cos = 3-√3/2sin
1-3√3/2 = 3-√3/2tan
3-√ 3/2 = arctan(1-3√ 3/2)
-1.78..
Answer should be -1.28 and 1.86
sin(x+π/6) = 3cos(x-π/6)
sinxcosπ/6 + sinπ/6cosx = 3cosxcosπ/6 + 3sinxsinπ/6
√3/2sinx + 1/2cosx = 3√3/2cosx + 3/2sinx
1-3√3/2cos = 3-√3/2sin
1-3√3/2 = 3-√3/2tan
3-√ 3/2 = arctan(1-3√ 3/2)
-1.78..
Answer should be -1.28 and 1.86
Your 6th line is wrong. . I don't understand what you've done going from your 5th to 6th line?
Original post by Chlorophile
Your 6th line is wrong. . I don't understand what you've done going from your 5th to 6th line?
Ah I've got the answer now, thank you!
But the reason I did that was because I thought that usually, when dealing with trig eq, e.g:
tan(x+50)=30 you do the inverse
(x + 50) = arctan30 and then get x on it's own
x = arctan30 - 50
hence i did the inverse and then divided! so what's wrong with that (I'm not arguing I just want to see where my thought process went wrong)?
Original post by Gaiaphage
It would be much easier to help you if you LaTeX'd this, there might be some brackets missed out because it's not laid out properly
What's LaTeX'd?
Original post by creativebuzz
What's LaTeX'd?
Original post by Chlorophile
.
This is LaTeX, it's really helpful for layout. I'll try and do it for you (not a pro myself though!)
Had a go and I'm no good, perhaps someone else can try :P
(edited 9 years ago)
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