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The questions asks to find the intersections of

$$f(x) = 2 \sin(x-7) + 6$$ and $$g(x) = \cos(2x-10) + 8$$

within the interval $[6,14]$.

So my general strategy was, 1) equate the functions, 2) get all the $X$s on one side and 3) convert to the same trig function.

So

$$2 \sin(x-7) + 6 = \cos(2x-10) + 8$$

I recognized the double angle in the cosine function, so

$$2 \sin(x-7) + 6 = \cos[ 2 (x-5) ] + 8$$

then

$$2 \sin(x-7) + 6 = \cos^2(x-5) - \sin^2(x-5) + 8$$

$\cos^2$ can be replaced with an identity, so

$$2 \sin(x-7) + 6 = 1 - \sin^2(x-5) - \sin^2(x-5) + 8$$

Group like terms and move then around,

$$2 \sin(x-7) + 2 \sin^2(x-5) = 3$$

Extracting the $2$ from the left side.

$$\sin(x-7) + \sin^2(x-5) = \frac 3 2$$

So here is where I hit a mental wall.

I could use the sine addition formula, but that would reintroduce cosine.

I can't simplify the terms any further since the angles are different.

Where would I go from here? Or is my approach off completely?

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    $\begingroup$ Is it $x-7$ or $x+7$? $\endgroup$
    – Amzoti
    Commented Dec 22, 2013 at 22:09
  • $\begingroup$ Maybe consider plotting, including a parametric plot, see both using WA. Look at the point $6$ in the parametric plot, which is a hint. $\endgroup$
    – Amzoti
    Commented Dec 22, 2013 at 22:22
  • $\begingroup$ Excuse my typos. I corrected it to x-7. $\endgroup$
    – Marty B.
    Commented Dec 22, 2013 at 22:23
  • $\begingroup$ I graphed the functions on Desmos and I can see that there are two solutions in the given interval. However, I still don't know how to derive the answer. $\endgroup$
    – Marty B.
    Commented Dec 22, 2013 at 22:31

1 Answer 1

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$$\sin(x-7)+\sin^2(x-5)=\frac32\iff\sin(t-2)+\sin^2t=\frac32$$

$$\sin t\cdot\cos(-2)+\cos t\cdot\sin(-2)+\sin^2t=\frac32$$

$$\sin^2t+\cos2\cdot\sin t-\sin2\cdot\cos t=\frac32\iff y^2+\cos2\cdot y-\sin2\cdot\bigg(\!\!\pm\sqrt{1-y^2}\bigg)=\frac32$$

$$y^2+\cos2\cdot y-\frac32=\sin2\cdot\bigg(\!\!\pm\sqrt{1-y^2}\bigg)\iff\bigg(y^2+\cos2\cdot y-\frac32\bigg)^2=\sin^22\cdot(1-y^2)$$ Now you're left with solving a quartic equation in $y=\sin t=\sin(x-5)\iff x=5+\arcsin y$

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  • $\begingroup$ Thank you @Lucian. Trying it out but it's not working. Substituting on the right, I get cos [ 2 (t+2) ] + 8. Then, cos^2 (t+2) - sin^2 (t+2) + 8, which becomes 9 - 2sin^2 (t+2), followed by 9 - 2 ( sin t cos 2 + cos t sin 2 )^2, and 9 - 2 ( sin^2 t cos^2 2 + 2 sin t cos t sin 2 cos 2 + cos^2 t sin^2 2 ). Replacing cos^2 with 1-sin^2 I get, 9 - 2 ( sin^2 t + sin^2 2 - 2sin^2 t sin^2 2 + 2 sin t cost t sin 2 cos 2 ). It just gets messier. It appears that I am misunderstanding your solution. Please advise. $\endgroup$
    – Marty B.
    Commented Dec 25, 2013 at 12:52
  • $\begingroup$ @MartyB.: Done! $\endgroup$
    – Lucian
    Commented Dec 25, 2013 at 16:37
  • $\begingroup$ Thank you @Lucian. This was an involved problem. $\endgroup$
    – Marty B.
    Commented Dec 28, 2013 at 20:45

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