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I am currently taking a Precalc II (Trig) course in college. There is a question in the book that I can't figure out how to complete it. The question follows:

Write the expression in terms of sine only:

$\sin(4x)-\cos(4x)$

So far I have $A\sin(x)+B\cos(x)=k\cdot\sin(x+\theta)$

I believe I have found k: $k=\sqrt{A^2+B^2}=\sqrt{2}$

So I think it would be $\sqrt{2}\cdot\sin(4x+\theta)$ but I do not know how I would find $\theta$.

Thanks in advance for all of your help. You have no idea how much I appreciate it!

CiaPan
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BlazeRyder
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3 Answers3

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$$\sin(4x)-\cos(4x)=\sqrt{2}\left(\frac{\sin(4x)}{\sqrt{2}}-\frac{\cos(4x)}{\sqrt{2}}\right)=$$

$$\sqrt{2}\sin(4x-\tfrac{\pi}{4})$$

so your $ \theta=\frac{-\pi}{4}$.

CiaPan
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hamam_Abdallah
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  • This is similar to what I think the answer is supposed to be. The question before the one I asked had the answer in the back of the book and it looks exactly like this but different values. I was confused about how to actually find the last value, but now I clearly see how to do so. Thanks very much! – BlazeRyder Oct 18 '19 at 21:44
  • @BlazeRyder Thanks for your question . Thanks for your comment. But some unknowns will make me loosing points for No reason. – hamam_Abdallah Oct 18 '19 at 21:53
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Use that

$$\cos(4x)=1-2\sin^2(2x)=1-2(2\sin x \cos x)^2=1-8\sin^2 x(1-\sin^2x)$$

user
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Use the two double angle identities that you know: \begin{align} \sin2\theta&=2\sin\theta\cos\theta\\ \cos2\theta&=1-2\sin^2\theta \end{align} You will also need to know $$\cos x=\sqrt{1-\sin^2x}$$

As such, we get \begin{align} \sin4x-\cos4x&=\sin2(2x)-\cos2(2x)\\ &=2\sin2x\cos2x-(1-2\sin^22x)\\ &=2(2\sin x\cos x)(1-2\sin^2x)+1+2(2\sin x\cos x)^2\\ &=2\sin x\sqrt{1-\sin^2x}(1-2\sin^2x)+1+8\sin^2x(1-\sin^2x)\\ &=2\sin x\sqrt{1-\sin^2x}-4\sin^3x\sqrt{1-\sin^2x}+1+8\sin^2x-8\sin^4x \end{align}

Andrew Chin
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  • I'd like to point out: if the question then asked you to evaluate the expression given a particular value for $\sin x$, then instead of writing $\cos x=\sqrt{1-\sin^2x}$, we can simply solve for $\cos x$ using the Pythagorean theorem. – Andrew Chin Oct 18 '19 at 21:06