Thread: Word problems on trigonometric ratios.

Hi MHF, I need some help with 2 problems that I have been stuck with please

1) If and
0 <= theta <= pi/4 , find the value of .

2) In triangle ABC,the angle B is twice the size as compared to that of angle A.
Show that
a)
b) the area of triangle ABC =
(a is the side opposite angle A and b....angle B .....)

For question 1 , I could not draw in terms of a triangle in the quadrant and hence I tried to manipulate it and got to (cheated a little by changing radians to degrees here) and got stuck. I am totally lost on this from here on and my approach is wrong too i guess,since we need to find the value of .

For question 2, i tried to use the law of cosines (to find length of side) to see if i could derive anything but i am stuck with

Any help would be greatly appreciated.

1) If sin(π/4 - θ) = 5/13, then cos(π/4 - θ) = 12/13.

Again sin(π/4 - θ) = sin[π/2 - (π/4 + θ)] = cos(π/4 + θ)

coa2θ= sin(π/2 - 2θ) = 2sin(π/4 - θ)cos(π/4 - θ)

Substitue hese values and find the result.

2) B = 2A

So sinB = sin2A = 2sinAcosA.

Now proceed.

Originally Posted by sa-ri-ga-ma

1) If sin(π/4 - θ) = 5/13, then cos(π/4 - θ) = 12/13.

Again sin(π/4 - θ) = sin[π/2 - (π/4 + θ)] = cos(π/4 + θ)

coa2θ= sin(π/2 - 2θ) = 2sin(π/4 - θ)cos(π/4 - θ)

Substitue hese values and find the result.

2) B = 2A

So sinB = sin2A = 2sinAcosA.
Now proceed.

why is sin(π/2 - 2θ) = 2sin(π/4 - θ)cos(π/4 - θ) and not cos(2θ)?

Originally Posted by arccos

why is sin(π/2 - 2θ) = 2sin(π/4 - θ)cos(π/4 - θ) and not cos(2θ)?

I didn't understand youe question. In fact sin(π/2 - 2θ) = 2sin(π/4 - θ)cos(π/4 - θ) = cos(2θ)

Originally Posted by sa-ri-ga-ma

I didn't understand youe question. In fact sin(π/2 - 2θ) = 2sin(π/4 - θ)cos(π/4 - θ) = cos(2θ)

Oh yes sorry. My bad. I overlooked sin2A = 2sinAcosA. I will attempt the questions now and let you know in a bit.

Originally Posted by sa-ri-ga-ma

1) If sin(π/4 - θ) = 5/13, then cos(π/4 - θ) = 12/13.

Again sin(π/4 - θ) = sin[π/2 - (π/4 + θ)] = cos(π/4 + θ)

coa2θ= sin(π/2 - 2θ) = 2sin(π/4 - θ)cos(π/4 - θ)

Substitue hese values and find the result.

2) B = 2A

So sinB = sin2A = 2sinAcosA.

Now proceed.

I am still unsure of what to do after we made sinB = sin2A = 2sinAcosA.
Could you please give me more hints?

EDIT: I have solved part 1(b = 2a cosA). Now i'm stuck on part 2.

sinB = 2sinAcosA

So b = 2a*cosA

A + B + C = 180

C = 180 - (A +B) = 180 - 3A

Area od the triangle = 1/2a*b*sinC = 1/2*a*2a*cosA*sin[180 - (A + B)] = a^2*cosA*sin3A.

Originally Posted by sa-ri-ga-ma

sinB = 2sinAcosA

So b = 2a*cosA

A + B + C = 180

C = 180 - (A +B) = 180 - 3A

Area od the triangle = 1/2a*b*sinC = 1/2*a*2a*cosA*sin[180 - (A + B)] = a^2*cosA*sin3A.

Thank you so much! I was trying to figure out angle C. Totally overlooked that.