# Homework Help: Q6 - Some Trigonometry questions

1. Aug 26, 2010

### ExamFever

1. The problem statement, all variables and given/known data

P1
sin 2x + sqrt(2) sin (x - (pi/4)) = 1

P2
Two sides of a triangle are 5cm and 9cm long. The ratio of the opposite angles is 1:3. Determine the angles and the length of the third side of the triangle!

2. Relevant equations

sin(2&) = 2 sin&cos&
sin(& - §) = sin& cos§ - cos& sin §

3. The attempt at a solution

P1
With this problem I'm stuck at:

sin 2x + sqrt(2) sin (x - (pi/4)) = 1
sin 2x + sqrt(2) (sinx (1/2)sqrt(2) - cosx (1/2)sqrt(2)) = 1
sin 2x + sin x - cos x = 1
2 sinx cosx + sin x - cos x = 1

Here I have the equation in one angle, but not in one function. I guess I need to do some changing etc. But I really can't figure out where to start.

P2
Two sides are given and a ratio of opposite angles. I do not even know where to start on this one...help is greatly appreciated!

2. Aug 26, 2010

### hunt_mat

for P1 use
$$a\sin A+c\cos B=r\sin (A+B)$$

3. Aug 26, 2010

### ehild

You can replace 1 by sin2x+cos2x in P1.

ehild

Last edited: Aug 27, 2010
4. Aug 27, 2010

### eumyang

If I'm reading this right, the angles that are opposite the known sides (5 cm and 9cm ) have the ratio 1:3. If so, then you'll need to use the Law of Sines. If you let the smaller of the two angles be θ, then the larger one would be 3θ.
$$\frac{\sin \theta}{5} = \frac{\sin 3\theta}{9}$$

There exists a triple-angle identity for sin 3θ. You can derive it yourself or look it up. I found two versions -- use the one that has just sin, not the one that has sin and cos. See if you can take it from here.

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5. Aug 28, 2010

### ExamFever

Hmm ok, I seem to really struggle with this. I just do not see how any of the identities you gave me helps me further with the problem.

Instead of giving straight out answers could you perhaps explain to me HOW I should look at a trigonometry equation and how to continue further than changing into basic identities?

I am familiar with the reciprocal, pythagorean, sum and product, double angle and half angle identities but I am having trouble looking past those are adapting those identities into something suitable to solve an equation (obviously the basic form can only be used for certain problems and an alternative form will need to be used for others).

Could you give me any good advice on this that will make it possible for me to quickly grasp trig equations?

6. Aug 28, 2010

### ehild

There is only one way: practice. Try to apply what you know.

Here you have a problem which is solved by applying sine law.

$$\frac{\sin \theta}{5} = \frac{\sin 3\theta}{9}$$

To transform sin(3θ) to an expression containing sinθ and cosθ looks a bit tedious, but it can be done. But it is not necessary to start with it. If you have sharp eyes, you notice that 1=2-1 and 3 =2+1. And you know cosine and sine of sum and difference of angles-write both sinθ and sin(3θ) in terms of θ and (2θ).

$$\frac{\sin{(2 \theta-\theta)}}{5} = \frac{\sin {(2\theta+\theta)}}{9}$$

Proceed. It is fun.

ehild

7. Aug 28, 2010

### eumyang

ExamFever, the reason you need to use a trig identity in the equation
$$\frac{\sin \theta}{5} = \frac{\sin 3\theta}{9}$$
is because I want to have an equation in terms of sin θ, and/or powers thereof.

As an example (not related to this problem), look at this:
$$\cos \,2\theta = \cos \theta$$

I use one of the double-angle identities for cosine:
\begin{aligned} 2\cos^2 \theta - 1 &= \cos \theta \\ 2\cos^2 \theta - \cos \theta - 1 &= 0 \end{aligned}

I've now rewritten the equation in terms of a single trig function - cos θ. Look what happens if I let cos θ = y:
\begin{aligned} 2\cos^2 \theta - \cos \theta - 1 &= 0 \\ 2y^2 - y - 1 &= 0 \end{aligned}

Looks like a simple quadratic, doesn't it? Once you solve for y, you'll have to substitute back cos θ for y, and then solve for θ. Don't forget this last part.

My point is this: in this equation:
$$\frac{\sin \theta}{5} = \frac{\sin 3\theta}{9}$$

Rewrite $$\sin 3\theta$$ into an expression that contains only sin θ (and/or powers thereof). There is a triple-angle identity that gives you just that. Cross multiply, move all terms to one side, and you'll have an equation in terms of sin θ that you'll have to solve. If you want, you could substitute sin θ = y like I did in the example above.

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