I need some help on solving this equation:
$$\sin x-\cos x=\frac{1}{\sqrt2}$$
If I do $\sin x=\sqrt{1-\cos^2x}$ and then $\cos x=t$ but don't get anything. Or $\sin x-\cos x=1/√2)/√2$ Or $\sin x-\cos x=1/√2)*√2$
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1Related : http://math.stackexchange.com/questions/698964/solve-a-trigonometric-equation-sqrt3-sinx-cosx-sqrt2 – lab bhattacharjee Mar 04 '14 at 17:34
2 Answers
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Use $$\sin(x)=\cos(\tfrac \pi2 -x)$$ and
$$\cos(A)+\cos(B)=2\cos\frac{A+B}2\cos\frac{A-B}2$$
to get
$$ \sin(x)-\cos(x)=\cos(\tfrac \pi2 -x)+\cos(\pi+x)=2\cos(\tfrac{3\pi}4)\cos(\tfrac\pi4+x)=-\sqrt2\cos(\tfrac\pi4+x) $$ Now apply $\cos(\pm\frac{2\pi}3+2k\pi)=-\frac12$.
Lutz Lehmann
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1@Lifeforbetter : Just apply trigonometric theorems, $\cos(90°+x)=\cos(90°)\cos(x)-\sin(90°) \sin(x)=-\sin(x)$. What exactly did you want to express? – Lutz Lehmann Nov 01 '19 at 16:05
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Now apply $\cos(\pm\frac{2\pi}3+2k\pi)=-\frac12$. Why $-2\pi/3$ included? – Dini Nov 01 '19 at 17:28
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@Lifeforbetter : I assumed that the task was to find all solutions, each value inside $(-1,1)$ is taken twice per period, thus $$\frac\pi4+x=\pm\frac{2\pi}3+2k\pi.$$ – Lutz Lehmann Nov 01 '19 at 17:34
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1@Lifeforbetter : The point to the angle $-2\pi/3$ is in the third quadrant, both sine and cosine values are negative there. Also, cosine is an even function, $\cos(-x)=\cos(x)$. – Lutz Lehmann Nov 01 '19 at 17:50
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$$ \sin(x)-\cos(x)=\cos(\tfrac \pi2 - x)+\cos(\pi-x)=2\cos(\tfrac{3\pi}4-x)\cos(-\tfrac\pi4)=\sqrt2\cos(\tfrac{3\pi}4-x) $$ how about this? – Dini Nov 01 '19 at 17:58
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1@Lifeforbetter : Yes, that also works. Might be a bit more straightforward. The solution of course does not change, $$x-\frac{3\pi}4=\pm\frac\pi3+2k\pi,$$ add $\pi$ to both sides to get the previous solution form. – Lutz Lehmann Nov 01 '19 at 18:22
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Hint: Square both sides, then use Pythagorean and double-angle identities. Don't forget to verify potential solutions in the original equation!
Cameron Buie
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You can rearrange that to get $$\sin2x=\frac12.$$ Now, for which angles $\theta$ is $\sin\theta=\frac12?$ – Cameron Buie Mar 04 '14 at 18:08
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30°, but when i come to solutions can i do 2x=π/6+2kπ , 12x=π+12kπ, x=π/12+kπ – dona12 Mar 04 '14 at 18:33
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That doesn't quite tell the whole story. Rather, we can say that $\theta=\frac\pi6+2k\pi$ or $\theta=\frac{5\pi}6+2k\pi$ for some $k\in\Bbb Z.$ Then we can conclude that $x=\frac\pi{12}+k\pi$ or $x=\frac{5\pi}{12}+k\pi$ for some $k\in\Bbb Z$. – Cameron Buie Mar 04 '14 at 21:53
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It would be more beneficial, though, to say that there is some $m\in\Bbb Z$ for which $x=\frac\pi{12}+2m\pi$ or $x={5\pi}{12}+2m\pi$ or $x=\frac{13\pi}{12}+2m\pi$ or $x=\frac{17\pi}{12}+2m\pi.$ (Do you see why we can say this?) However, we still need to check these potential solutions in the original equation, since our squaring may have introduced false solutions. – Cameron Buie Mar 04 '14 at 21:57