Question

1 .If tanA=1/3 and tanB=1/7 (both A and B are acute),calculate 50sin(2A +B)
2.1Prove that sin2A+2cosA-2cos^3A/1+sinA =sin2A
2.2 for which values of A in the interval[-360;360] is the identity in 2.1 undefined?
3.If tanB=3/4 , 0<B<90 , prove that 4cos2B+3sin2B=4
4.Prove that (1/cos^2B)cos2B=2-1/cos^2B)
5. Write sin2A and cos2A in terms of sinA and cosA

Answer 1

1.
tanA = 1/3, and tanB = 1/7
I usually construct my triangles,
(since the fraction gives you 2 of the sides of right-angled triangles, the third can always be found using Pythagoras)

for tanA = 1/3, sinA = 1/√10 , cosA = 3/√10
for tanB = 1/7, sinB = 1/√50, cosB = 7/√50
we will need sin2A and cos2A,
sin2A = 2sinAcosA = 2(1/√10)(3/√10) = 3/50
cos2A = cos^2 A - sin^2 A
= 9/10 - 1/10 = 4/5
50sin(2A+B)
= 50(sin2AcosB + cos2AsinB)
= 50( (3/50)(7/√50) + (4/5)(1/√50) )
= 50(21/(50√50) + 4/(50√50)
= 25/√50
= 1/√2

hey , I bet 2A +B = 45

2. LS = (2sinAcosA + 2cosA - 2cos^3 A)/ (1+sinA)
= cosA(2sinA + 2 - 2cos^2 A)/(1+sinA)
= 2cosA (1 + sinA - (1 - sin^2 A) /(1+sinA)
= 2cosA (sin^2 A + sinA) /(1+sinA)
= 2cosA sinA (sinA + 1)/(1+sinA)
= 2sinAcosA
= sin 2A
= RS


3 make your sketch, and proceed like in #1

4. Convert to single angle trig ratios, using your formulas

5. All you have to do is look them up and memorize them.

Answer 2

Thank u so much !!
But I still have a question in#1 where did the 50 go?

Answer 3

the denominator is 50√50

Answer 4

but oops: 25/√50 = 5/√2, not 1/√2

Answer 5

tanA=1/7 sin2A cos2A