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My niece asked for help with an SAT prep question. We are given that

$$\sin a = \cos b$$

where the angles are both acute and $a=4k-22$ and $b=6k-13$.

The only way we could think to solve it is by plotting and using fzero. But since it's an SAT problem, I assume there should be an approach that doesn't require a calculator.

Is there some trig identity I'm overlooking?

Blue
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Fractal20
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  • "Is there some trig identity I'm overlooking?" yes, $\cos b = \sin (90 - b)$ so $\sin a = \sin (90-b)$. But DO be careful. $\sin N =\sin M$ doesn't always mean $N=M$. Does it in this case (when $a,b$ are both acute)? – fleablood May 18 '20 at 15:45
  • Whenever you see cosine, think complimentary sine (that is, the sine of an angle is the complimentary sine of the complimentary angle). – Andrew Chin May 18 '20 at 15:51

3 Answers3

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You are overlooking some identities. $$ \bbox[yellow,5pt, border:2px solid red]{ \sin A = \cos B \iff A+B = 90^\circ }\quad \textrm{(for A and B between 0 and 90 degrees)} \\ \bbox[yellow,5pt, border:2px solid red]{ \sin A = \cos B \iff A-B = 90^\circ }\quad \textrm{(for A between 0 and 90 degrees,B between 0,-90)} \\ \bbox[yellow,5pt, border:2px solid red]{ \sin A = \cos B \iff B-A = 90^\circ }\quad \textrm{(for B between 0 and 90 degrees,A between 0,-90)} $$

From here, of course, you get $4k-22 = 90-(6k-13)$ for the first case, $(6k-13) - (4k-22) = 90$ for the second case, and $(4k-22) - (6k-13) = 90$ for the third case.

Sarvesh Ravichandran Iyer
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Rewrite $\sin(4k-22) = \cos(6k-13)$ as

$$\cos(90-4k+22) - \cos(6k-13)=2\sin\frac{99-2k}2 \sin\frac{115-10k}2 =0$$

which leads to $\frac{99+2k}2 =n\pi,\>\>\>\>\> \frac{134-10k}2 =n\pi$ and the solutions $k= n\pi-\frac{99}2,\> \frac{n\pi}5-\frac{67}5$.

Quanto
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Draw a right triangle with acute angle $a$. The soh-cah-toa rule implies the other angle is $b$. Triangle angles add up to $180^\circ$.

Harish Chandra Rajpoot
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dshin
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