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I've been asked to prove $$y=\frac{\sqrt{3}} 2 x+\frac 1 2 \sqrt{1-x^2}$$

given $x=\sin(t)$ & $y=\sin(t+\frac \pi 6)$

I did $t=\arcsin(x)$ and plugged that into the $y$ equation. Used the $\sin(a+b)$ identity to get: $$y=x\cos\left(\frac \pi 6\right)+\frac{\cos(\arcsin(x))}2 = \frac{\sqrt3} 2 x+\frac{\cos(\arcsin(x))} 2$$

Now I'm sure there must be an identity for $cos(arcsin(x))$ however I'm unaware of it. I'm also unaware of how to prove it.

I did a quick google and I found this page which says:

enter image description here

which is seemingly exactly what I need to complete the question; however, it wouldn't be proving that $y = \text{answer}$ if I didn't show how to get to this result.

Is there a "more correct" way to complete this question without having to fiddle with this formula / arcsins etc.

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The usual proof involves drawing a triangle. The opposite side will be called $\sin(y)=x$, the hypotenuse will be $1$, and the angle will be $\arcsin x$. Can you use the pythagorean theorem to find $\cos(\arcsin x)$?

enter image description here

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  • $\begingroup$ This makes a lot of sense, I guess the - relative - complication of the question intimidated me... $\endgroup$ – Tobi Dec 11 '16 at 18:46
  • $\begingroup$ @Tobi As is the nature of trigonometry sometimes, but often it ends up beautiful :) $\endgroup$ – Simply Beautiful Art Dec 11 '16 at 22:17
  • $\begingroup$ I find there to be so many directions in which I can take a question, it's often a matter of chance. Maybe with experience I'll be able to tell, faster, which identities to use. $\endgroup$ – Tobi Dec 11 '16 at 23:54
  • $\begingroup$ @Tobi Yes, definitely practice your trig identities. Personally, the most important ones are the Pythagorean identities (what we have here) and the sum of angles formula. Combined, you can get most of the other formulas. Learning how to use the formulas will become especially useful in some parts of calculus btw. $\endgroup$ – Simply Beautiful Art Dec 12 '16 at 0:35
  • $\begingroup$ The solutions can be so illusive sometimes, simple rearrangement / substitution can often take me forever to spot. It's like there's no instant way to do them, just brute force. $\endgroup$ – Tobi Dec 12 '16 at 0:42
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$$\cos^2(\arcsin(x))+\sin^2(\arcsin(x))=1\\\cos^2(\arcsin (x))+x^2=1\\\cos^2(\arcsin(x))=1-x^2\\\cos(\arcsin(x))=\sqrt{1-x^2}$$ The last step is okay because $-1\leq\arcsin x\leq 1$ and $\cos$ is positive there.

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  • $\begingroup$ Nice, algebraic, solution! $\endgroup$ – Tobi Dec 11 '16 at 18:48
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$$y=\sin\left(t+{\pi\over 6}\right)=\cos{\pi\over 6}\sin t+\sin{\pi\over 6}\cos t\\={\sqrt3\over 2}x+{1\over 2}\sqrt{1-x^2}$$

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  • $\begingroup$ What is this showing? $\endgroup$ – Tobi Dec 11 '16 at 18:53
  • $\begingroup$ @Tobi Can't this be a more correct way to do this question rather than to fiddle with the arcsins? $\endgroup$ – Qwerty Dec 11 '16 at 18:55
  • $\begingroup$ How do you go from $\cos{t}$ to $\sqrt{1-x^2}$ $\endgroup$ – Tobi Dec 11 '16 at 18:58
  • $\begingroup$ @Tobi $$\cos t=\sqrt {1-\sin^2 t}=\sqrt{1-x^2}$$ $\endgroup$ – Qwerty Dec 11 '16 at 19:00
  • $\begingroup$ Oh I see, very illusive. $\endgroup$ – Tobi Dec 11 '16 at 19:00
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Draw a right triangle in which the "opposite" side has length $x$ and the hypotenuse has length $1$. Then the sine of the angle having that "opposite" side is $\sin=\dfrac{\text{opposite}}{\text{hypotenuse}} =\dfrac x 1 = x.$ So that angle is $\arcsin x$.

Now use the Pythagorean theorem to show that the "adjacent" side has length $\sqrt{1-x^2}$. Then we have $$ \cos\arcsin x = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{1-x^2}} 1 = \sqrt{1-x^2}. $$

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