A level trignometry?
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Given that sin A= 3/5 and cos B = 12/13 where A is obtuse and B is acute find exact values of cos (A+B) and cot(A-B)
**here is what i got for cos (A+B) **
cos (A+B) = cosAcosB - sinAsinB
= (12/13)(cosA) - (3/5)(sinB)
if (cos A)^2 = 1 - (sinA)^2 = 1 - (3/5)^2 = 16/25
cosA =4/5
if (sinB)^2=1- (cosB)^2 = 1 - (12/13)^2 = 25/169
sinB =5/13
therefor, cos (A+B)= (12/13)(4/5) - (3/5)(5/13) = 33/65
the answer for this at the back of the txtbook is -63/65, why is this so??
**Please explain how to do cot(A-B), for i did not get that**
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i'm using handphone wifi at the moment, so it will be hard to answer. For a hint, A is obtuse, so cos must be negative. Try first.
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Hi,
Since sin A = 3/5 and A is obtuse, you draw a right-angled triangle in the 2nd quadrant and obtain:
cos A = -4/5 and tan A = -3/4.
Since cos B = 12/13 and B is acute, you draw a right-angled triangle in the 1st quadrant and obtain:
sin B = 5/13 and tan B = 5/12.
Consider these results:
cot (A- B) = 1 / tan (A - B)
tan (A - B) = [ tan A - tan B ] / [ 1 + tan A tan B ]
Thanks!
P.S. H2 maths does not explicitly assess this knowledge.Cheers,
Wen Shih -
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Originally posted by limywv:
Oh man how come I have no idea that trigometry is actually part of the A lvl H2 syllabus?
And I think my batch (1991) is pretty screwed up. We were not taught double angle formulas, and we were the last batch to learn A maths vectors at O levels
If you stay near hougang, you can always look to wenshih for A level maths group tuition
Or me if near YCK or Tampines :D
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Hi,
If you read the syllabus document carefully, you will realise that trigo knowledge is embedded in some topics, e.g. calculus. That is the reason I mentioned that trigo is not assessed explicitly.
I did my A-level in 1991 and I was taught double-angle formulae in AJC. I guess schools may have taught knowledge selectively.
Thanks and merry christmas!
Cheers,
Wen Shih