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Hello please help me with these trig identities and double angles as I am not sure where I am going wrong but I keep getting the wrong answer

This is the problem $$ \sin(\theta+30) = 2\cos(\theta) $$ This is my one of my incorrect solutions

$$\sin(\theta +30) = 2\cos(\theta)$$ $$\sin(\theta)\cos(30) + \sin(30)\cos(\theta)=2(1 - \sin(\theta))$$ $$\sin(\theta)(\frac{\sqrt3}{2})+(\frac{1}{2})(1 - \sin(\theta))=2-2\sin(\theta)$$ $$\frac{\sin(\theta)\sqrt3+1-\sin(\theta)}{2}+2 \sin(\theta)=2$$

I get stuck and I am not sure what to do with this problem.

Please help as I am trying to self teach my A -level maths.

Thanks in advance

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Hint: you should eventually arrive at the equation $\tan\theta=\sqrt{3}$. –  icurays1 Nov 12 '12 at 21:32
    
Wimbledon vs. York are going into extra time. FA cup. –  Will Jagy Nov 12 '12 at 21:46
    
Your problem lies with the substitution of $\cos\theta$ by $1-\sin\theta$. You're confusing it with $\cos^2\theta \equiv 1-\sin^2\theta$. –  Fly by Night Nov 12 '12 at 21:47

3 Answers 3

Your addition identity is almost correct. You can't say $\cos(\theta) = 1 - \sin(\theta)$. However, you can say $\cos^2(\theta) = 1 - \sin^2(\theta)$ from the Pythagorean Identity. Just be aware that you CANNOT drop the squares in this equation by taking the square root of both sides. Now, \begin{array}{ccc} \sin(\theta + 30) & = & \sin(\theta)\cos(30) + \cos(\theta)\sin(30) \\ & = & \sin(\theta)\frac{\sqrt{3}}{2} + \cos(\theta)\frac{1}{2} \\ & = & \frac{1}{2}\left[\sqrt{3}\sin(\theta) + \cos(\theta)\right] \end{array}

If this expression was equal to $2\cos(\theta)$, we would have:

\begin{array}{ccc} \frac{1}{2}\left[\sqrt{3}\sin(\theta) + \cos(\theta)\right] & = &2\cos(\theta) \\ \sqrt{3}\sin(\theta) + \cos(\theta)& = & 4\cos(\theta) \\ \sqrt{3}\sin(\theta) & = & 3\cos(\theta) \\ \sin(\theta) & = & \sqrt{3}\cos(\theta)\\ \end{array}

Since this doesn't hold for all $\theta$, this is not an identity. As mentioned in the comments, you can get to $\tan(\theta)$ from what I ended up with. Then, you can solve for $\theta$.

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Thanks for your help.. It seems to make sense now –  user866190 Nov 12 '12 at 21:54

What you have written cannot be an identity. If it were then $\sin(\theta + 30)$ must equal $2\cos\theta$ for all values of $\theta$. However, while $\sin(\theta+30)$ oscillates between $-1$ and $1$, we see that $2\cos\theta$ oscillates between $-2$ and $2$. They have different ranges and so cannot possibly be identical functions. Let's assume you want to find particular values of $\theta$ for which $\sin(\theta + 30) = 2\cos\theta.$

The double angle formula tells us that $\sin(\theta + 30) = \sin\theta\cos(30) + \sin(30)\cos\theta.$ Two well-known values of sine and cosine are $\cos(30^{\circ}) = \sqrt{3}/2$ and $\sin(30^{\circ}) = 1/2.$ Thus:

$$\sin(\theta + 30) = \frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta. $$

It follows that $\sin(\theta+30) = 2\cos\theta$ if and only if

$$\frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta = 2\cos\theta \iff \frac{\sqrt{3}}{2}\sin\theta - \frac{3}{2}\cos\theta = 0 \, . $$

Consider the numerator: $\sqrt{3}\sin\theta - 3\cos\theta = 0 \iff \tan\theta = \sqrt{3} \iff \theta = 60^{\circ} + 180n^{\circ},$ where $n$ is any integer. The multiples of $180^{\circ}$ are added because $\tan\theta$ is periodic with a period of $180^{\circ}$; it repeats itself every $180^{\circ}$. The finally answer is then:

$$\theta = \ldots, -120^{\circ}, \ 60^{\circ}, \ 240^{\circ},\ldots $$ $$ \theta \in \{ (60 + 180n)^{\circ} : n \in \mathbb{Z} \}.$$

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Thanks man, it is starting to sink in –  user866190 Nov 12 '12 at 23:34

As $\cos^2 \theta + \sin^2 \theta = 1,$ I'm afraid $$ \cos \theta = \pm \sqrt{1 - \sin^2 \theta} $$ which is useless for your purposes. So leave the cosine on you right-hand side as it is, your substitution is just wrong.

Alright, if you do it properly, you get a relationship between $\sin \theta$ and $\cos \theta$ which can be rewritten as specifying a value for $\tan \theta.$

Meanwhile, I am a big fan of drawing graphs. I recommend you get some graph paper or quadrille paper and draw $\theta,y$ axes with, say, $0 \leq \theta \leq 360^\circ$ and then draw your two curves, $y = \sin (\theta + 30^\circ)$ and $y = 2 \cos \theta.$ It will be fairly clear when they cross.

Here is a jpeg of empty axes I just made. Since one axis is degrees and one real numbers, I did not worry about relative scale. I also used negative degrees, that aspect does not really matter, but I can understand if degrees up to 360 would be preferred. Anyway, anyone can email me for a jpeg. Believe me, the practice in drawing these things is invaluable. I guess I will draw this specific graph next.

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Alright, I did the graphs. As you can see, (and confirm yourself once you see the likely candidate points), the graphs cross at $60^\circ$ and $-120^\circ.$ By adding $360^\circ,$ we find that $-120^\circ$ is equivalent to $240^\circ.$ And these are the places where $\tan \theta = \sqrt 3.$

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enter image description here

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Doing these graphs yourself really does help. It's true.

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Thanks for the drawing, you guys are really helpful!! Between stack exchange and youtube and MIT open courseware I should smash my A Levels! –  user866190 Nov 12 '12 at 23:36
    
@user866190, I did this graph because a complete solution had been given and to illustrate the general idea. I am quite serious about the importance of you drawing some (many) graphs. We will not be there during the examinations. –  Will Jagy Nov 13 '12 at 1:07

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