Expert: Paul Klarreich - 7/26/2009

Question
I'm taking CIE A LEVEL MATHEMATICS and this question is in the pure mathematics trigonometry topic. I know part i) but for part ii) I dont know how to find a. i) express 5sinx+ 12cosx in the form Rsin(x + y), where R >0 & 0degree< y <90degree,giving the value of y correct 2 d.p. ii) hence solve the equation 5sin 2a+ 12cos 2a=11, giving all the solutions in the interval 0degree< a <180degree. Please help me with part ii). For part ii) I only found out (sin^-1 (11/13) )= 57.796degree. I dont know how to continue the rest in finding the value of a. Thanks a million

Answer
Questioner: Warm
Country: China
Category: Calculus
Private: No
Subject: Double angle
Question: I'm taking CIE A LEVEL MATHEMATICS and this question is in the pure mathematics trigonometry topic. I know part i) but for part ii) I dont know how to find a. i) express 5sinx+ 12cosx in the form Rsin(x + y), where R >0 & 0degree< y <90degree,giving the value of y correct 2 d.p. ii) hence solve the equation 5sin 2a+ 12cos 2a=11, giving all the solutions in the interval 0degree< a <180degree. Please help me with part ii). For part ii) I only found out (sin^-1 (11/13) )= 57.796degree. I dont know how to continue the rest in finding the value of a.

Thanks a million
........................................
(a thousand will be quite sufficient)

OK, you have:

5 sin 2a + 12 cos 2a = 11

This is a standard 'phase angle' calculation:

5 sin x + 12 cos x = R sin(x + t)  ["t" means theta]

R sin x cos t + R cos x sin t

R cos t = 5
R sin t = 12

Square and add, we get:

R^2 = 169
R = 13

Divide, we get
tan t = 12/5 = 67.38 degrees

t = arctan 12/5, for which you use your calculator.
.............................
Now for your ii):

5sin 2a+ 12cos 2a=11

Rewrite:

13 sin(2a + t) = 11,  where t is as above.

2a + t = arcsin(11/13)

Use the calculator now.

2a + 67.38 = 57.79

2a = - -9.584

a = -4.79

But if you want the solution in the interval 0 < a < 180, then do this:

2a = - -9.584 + 360 = 350.416

a = 175.208