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I have an exam tomorrow and it is highly likely that there will be a trig identity on it. To practice I tried this identity: $$2 \sin 5x\cos 4x-\sin x = \sin9x$$ We solved the identity but we had to move terms from one side to another. My question is: what are the things that you can and cannot do with trig identities? And what things are must not when doing trig identities? Thank you |
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Those are very general questions: what "can you do" and what "can't you do"...in terms of using trig identities: What you can do is use an identity to replace one expression with its strict equivalent, as determined by the identity in question. It's simply grounded in "substituting one expression for its equivalent", which is legitimate beyond just its handiness with utilizing trig identities to transform one expression to an equivalent expression. You can algebraically manipulate equations involving trigonometric expressions in any way that is "legal" for manipulating any equation. So for a simple example, ${1-\sin^2\theta} = \cos^2 \theta$ because we know the identity: $$\sin^2\theta + \cos^2\theta = 1.$$ We do not change this equivalence when we manipulate it to get: $$\sin^2\theta + \cos^2\theta = 1 \iff \cos^2\theta = 1 - \sin^2\theta$$ In the identity you solved, you can use: $$\sin c+\sin d=2\sin\frac{c+d}2\cos\frac{c-d}2$$ Applied here: $$\sin9x+\sin x=2\sin\frac{9x+x}2\cos\frac{9x-x}2=2\sin5x\cos4x$$ $$2\sin a\cos b=\sin(a+b)+\sin(a-b)$$ Applied here: $$2\sin5x\cos4x=\sin(5x+4x)+\sin(5x-4x)=\sin9x+\sin x$$ |
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$$\sin C+\sin D=2\sin\frac{C+D}2\cos\frac{C-D}2$$ So, $$\sin9x+\sin x=2\sin\frac{9x+x}2\cos\frac{9x-x}2=2\sin5x\cos4x$$ $$2\sin A\cos B=\sin(A+B)+\sin(A-B)$$ So, $$2\sin5x\cos4x=\sin(5x+4x)+\sin(5x-4x)=\sin9x+\sin x$$ |
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Hint: We have $\sin 5x\cos 4x+\cos 5x\sin 4x=\sin 9x$. Replace one of the $\sin 5x\cos 4x$ by $\sin 9x-\cos 5x\sin 4x$. Leave the other one alone. Something nice will happen. You will end up having used the basic addition and subtraction laws $$\sin(a\pm b)=\sin a\cos b\pm \cos a\sin b.$$ |
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The intended techniques all follow from using the sine and cosine addition formulas and normalization, which you should have seen before. However, I wanted to point out that a more unified approach to the general problem of proving trig. identities is to work in the complex plane. For instance $e^{ix}=\cos(x)+i\sin(x)$ allows you to derive identities such as $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ and $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$. Let $w=e^{ix}$. Then you can turn your identity into an equivalent "factorization problem" for a sparse polynomial in $w$, of degree $18$. Try it out! |
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