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Trigonometry (1 Viewer)
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I have a few more questions i'm having trouble with and i cant believe that these are the easiest of questions
can anyone tell me what textbook has a kinda heartbeat kinda thing??
Ok these are teh questions
1.If cos@ = -4/5 and @ is a positive obtuse angle, find the values of sin @ and tan @.
2. Prove the identities
a) sec2@ - cosec2@ = tan2@ - cot2
b) (cotA+tanB)/(cotB+tanA) = cotAtanB
can anyone tell me what textbook has a kinda heartbeat kinda thing??
Ok these are teh questions
1.If cos@ = -4/5 and @ is a positive obtuse angle, find the values of sin @ and tan @.
2. Prove the identities
a) sec2@ - cosec2@ = tan2@ - cot2
b) (cotA+tanB)/(cotB+tanA) = cotAtanB
tommykins
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sin(pi - @) = sin@
cos(pi-@) = -cos@
tan(pi-@) = -tan@
1. Draw a triangle with adjacent 4, hypotenuse 5, opposite 3. sin@ = o/h = 3/5 (2nd quad as @ is obtuse) and tan@ = -3/4 (tan@<0 in2nd quad)
2. LHS = (1+tan²@) - (1+cot²) = 1+tan²@ - 1 - cot²@ = tan²@ - cot²@ = RHS
3.
LHS = (1/tanA + tanB)/(1/tanB + tanA) = (1+tanAtanB/tanA)/(1+tanAtanB/tanB)
= (1+tanAtanB)/tanA * tanB/*(1+tanAtanB)
= tanB/tanA
= tanB*1/tanA
= tanBcotA
= RHS.
cos(pi-@) = -cos@
tan(pi-@) = -tan@
1. Draw a triangle with adjacent 4, hypotenuse 5, opposite 3. sin@ = o/h = 3/5 (2nd quad as @ is obtuse) and tan@ = -3/4 (tan@<0 in2nd quad)
2. LHS = (1+tan²@) - (1+cot²) = 1+tan²@ - 1 - cot²@ = tan²@ - cot²@ = RHS
3.
LHS = (1/tanA + tanB)/(1/tanB + tanA) = (1+tanAtanB/tanA)/(1+tanAtanB/tanB)
= (1+tanAtanB)/tanA * tanB/*(1+tanAtanB)
= tanB/tanA
= tanB*1/tanA
= tanBcotA
= RHS.
Last edited:
lyounamu
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1.Fortian09 said:
Draw a diagram in the second quadrant where the adacent side is 4 units with the hypotenuse being 5.
Therefore, sin@ = 3/5 and tan@ = -3/4
2 a). sec^2(@) - cosex^2(2) = 1+tan^2(@) - (1+cot^2(@)) = tan^2(@) - cot^2(@)
2 b). LHS = (cotA+tanB)/(cotB+tanA) = (1/tanA + tanB) / (1/tanB + tanA)
= (1+tanBtanA/tanA)/(1+tanBtanA/tanB)
= (tanB+tan^2(B)tanA/tanAtanB)/(tanA + tanBtan^2(A)/tanAtanB)
= (tanB+tan^2(B)tanA)/(tanA+tanBtan^2(A))
= (tanB ( 1+tanBtanA))/(tanA (1+tanBtanA))
= tanB/tanA = tanBcotA = RHS
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Lol I actually gave 2b another go an got it... that qn was actually easy...
ahh well
can anyone answer that textbook quesiton?
oh man I'm gonna be here for ages asking hw questions...
1. If tan2A - 2tan2B = 1. Find the possible values of cosA/cosB
2. If tan@ = (2u)/(1-u2), find sec@ and cosec@ in terms of u.
3. If cosecA = (1/2)(a+(1/a)), find cotA in terms of a.
4. If cot2@+3cosec2@ = 7, show that tan2@ =1.
5. If asin@+bcos@=p nd bsin@-acos@ = q, show that a2+b2 = p2+q2
6. Given 5cos2@+sin2@=4...........(*)
show that tan2@ = 1/3 by:
a) Finding sec2@ first;
b) writing RHS of (*) as a sum of sin2@ and cos2@.
7. If cosec@-cot@ = y, find cos@ in terms of y.
8. Given that sin@+cosec@ = RT(k2-4).
9. If cot@ + cosec@ = 1/4, find a quadratic equation whose roots are cot@ and cosec@
ahh well
can anyone answer that textbook quesiton?
oh man I'm gonna be here for ages asking hw questions...
1. If tan2A - 2tan2B = 1. Find the possible values of cosA/cosB
2. If tan@ = (2u)/(1-u2), find sec@ and cosec@ in terms of u.
3. If cosecA = (1/2)(a+(1/a)), find cotA in terms of a.
4. If cot2@+3cosec2@ = 7, show that tan2@ =1.
5. If asin@+bcos@=p nd bsin@-acos@ = q, show that a2+b2 = p2+q2
6. Given 5cos2@+sin2@=4...........(*)
show that tan2@ = 1/3 by:
a) Finding sec2@ first;
b) writing RHS of (*) as a sum of sin2@ and cos2@.
7. If cosec@-cot@ = y, find cos@ in terms of y.
8. Given that sin@+cosec@ = RT(k2-4).
9. If cot@ + cosec@ = 1/4, find a quadratic equation whose roots are cot@ and cosec@
common typing error
lyounamu
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Ok, despite my laziness I will tackle few of them - not at all once! (I am doing past paper at the moment so I only have few minutes to spare, sorry!)Fortian09 said:
Here we go (wait till I actually write):
cosa/cosb
1. tan^2(A) - 2tan^2(B) = 1
sin^2(A) - 2sin^2(B)cos^2(A)/cos^2(B) = cos^2(A)
-2sin^2(B)cos^2(A)/cos^2(B) = cos^2(A) - sin^2(A)
cos^2(A)/cos^2(B) = (cos^2(A) - sin^2(A))/-2sin^2(B)
cosA/cosB = SQRT ((cos^2(A) - sin^2A)/-SQRT(2)sinB
2. tan@ = (2u/1-u^2)
Here, draw a right-angled triangle.
adjacent = 1-u^2, opposite = 2u and hypotenuse = SQRT (1+2u^2 + u^4)
So sec@ = 1/cos@ = hypotenuse/adjacent = SQRT(1+2u^2+u^4)/(1-u^2)
cosec@ = 1/sin@ = hypotenuse/opposite = SQRT(1+2u^2+u^4)/2u
3. same story for Q 3.
4. cot^2(@) + 3cosex^2(@) = cot^2(@) + 3 + 3cot^2(@) = 7
4cot^2(@) = 4
cot^2(@) = 1
so tan^2(@) = 1
will come back after dinner. sorry~
I just found that I couldn't be bothered to continue...I need to freshen up. I will finish them off tomorrow unless someone else doesn't do any from here.
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lyounamu
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To be honest, the questions are quite tedious in nature. They aren't really difficult at all but they do take tremendous amount of time to complete (all of them together). Why don't you ask for tutors to go with you one by one? I think that seems like a way to go. That way, you learn while you do the questions.
lyounamu
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What??????????!!!!Fortian09 said:
That's not really the best study method if you keep working under the pressure of getting your work done...
Homework is basically "homework", it really should be a light exercise to test your knowledge and trust me: a lot of questions you posted up are pretty damn difficult for HSC standards. The greatest chance is that: you won't really get them for HSC. It's great that your teacher is trying to teach you some advanced mathematics but it's too much for your level (or commitment), it will do more harms than goods.