More Trig questions (1 Viewer)

[M@ster P]

Hello more Trig questions that I could not do.


1. If cosA = (p^2 - m^2 / p^2 + m^2) and 0 < A < 90 express tanA and sin2A in terms of p and m.


2. If A, B and C are in arithmetic sequence, prove that sinA + sinC = 2sinBcos(B - A)


3. If 4tan(A - B) = 3tanA, prove that tanB = (sin2A / 7 + cos2A)


4. If tanA = t, express sin2A and cos2A in terms of t. Hence find the values of t for which (k+1)sin2A + (k-1)cos2A = k + 1

thanks

 

[lolokay]

1. draw a triangle and use pythagoras to find the other side

2. express A and C in terms of B and d (the common difference)

3. from another thread:
[t = tanx, c = cosx]
if you rearrange to find tanb, you get
tanb = t/(4+3t^2)
now multiply top and bottom by 2c^2
2sc/(8c^2+6s^2)
= sin2x/(2c^2 + 6(c^2+s^2))
where c^2 + s^2 = 1 and 2c^2 = cos2x + 1
= sin2x/(cos2x + 7)

with this one you'll want to expand (sin2A / 7 + cos2A) and then it should become clear how to convert tanB into it (once you have the term for tanB)


4. sin2A = 2sc/(s^2+c^2)
dividing by c^2/c^2:
2t/(t^2+1)

cos2a = (c^2 - s^2)/(c^2 + s^2)
dividing c^2/c^2
= (1 - t^2)/(1 + t^2)

then just expand out and solve the quadratic

 

[M@ster P]

ok ty

 

[M@ster P]

Sorry but i got 2 more questions:

5. Use the factors of x^3 - y^3 to show that cos^6A - sin^6A = [1 - 1/4sin^2(2A)] cos2A

and finally

6. If tanA = ktanB, show that (k - 1)sin(A+B) = (k + 1)sin(A - B)

 

[undalay]

Sorry but i got 2 more questions:

5. Use the factors of x^3 - y^3 to show that cos^6A - sin^6A = [1 - 1/4sin^2(2A)] cos2A

x^3 - y^3 = (x-y)(x^2 +xy+y)
sub x= cos^2 A
y = sin^2 A

simplify and u'll get it

 

[M@ster P]

ty question 6 left

 

[lolokay]

Sorry but i got 2 more questions:

5. Use the factors of x^3 - y^3 to show that cos^6A - sin^6A = [1 - 1/4sin^2(2A)] cos2A

and finally

6. If tanA = ktanB, show that (k - 1)sin(A+B) = (k + 1)sin(A - B)

k = tanA/tanB
= sinA/cosA * cosB/sinB
= sinAcosB/sinBcosA
(k-1)sin(A+B) =[sinAcosB/sinBcosA -1] [sinAcosB + sinBcosA]

[sinAcosB - sinBcosA] [sinAcosB/sinBcosA + 1]
= (k+1) sin(A-B)