# Simple(?) trig problem

**URL:** <https://boards.straightdope.com/t/simple-trig-problem/685868>\
**Category:** Factual Questions\
**Created:** [April 11, 2014, 12:42pm UTC](https://boards.straightdope.com/t/simple-trig-problem/685868 "2014-04-11T12:42:27Z")\
**Posts on this page:** 3\
**Page:** 2

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**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\
**Post date:** [April 15, 2014, 12:16am UTC](https://boards.straightdope.com/t/simple-trig-problem/685868/21 "2014-04-15T00:16:16Z")

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Well, no, I suppose that’s not fair… there’s still the danger that students may inappropriately forget that x is determined (modulo complete revolutions) by cos(x) and sin(x) together and not by either individually even with **Hogarth** ’s approach. It seems for some reason less tempting to me on that approach (where one explicitly, correctly solves for the one in terms of the other at the beginning, rather than waiting till the end), but I suppose students are tempted by everything all the time anyway.

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**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\
**Post date:** [April 15, 2014, 12:59am UTC](https://boards.straightdope.com/t/simple-trig-problem/685868/22 "2014-04-15T00:59:59Z")

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In fact, since you get the same quadratic equation whether using the professor’s approach or **Hogarth** ’s, it was a very silly thing to say, and I wish I hadn’t said it.

I’ll redeem myself by pointing out:

> [@sbunny8](#):
>
> sqrt(1-sin2(x)) = 1.2 - .8sin(x)  
> Now square both sides.  
> 1-sin2(x) = 1.44 - 1.92sin(x) - .64sin2(x)

You mean “+ .64sin2(x)”.

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**Author:** ![Manlob](https://avatars.discourse-cdn.com/v4/letter/m/96bed5/32.png) [@Manlob](https://boards.straightdope.com/u/Manlob)\
**Post date:** [April 15, 2014, 5:22am UTC](https://boards.straightdope.com/t/simple-trig-problem/685868/23 "2014-04-15T05:22:11Z")

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If you have a 0.8 by 1 rectangle inclined by angle x relative to the ground, the length of its projection onto the ground is .8_sin(x)+cos(x) = 1.2 (from projecting each of 2 sides to the ground and adding). This equals the length of the projection of its diagonal onto the ground. The diagonal has length sqrt(1+.8_.8), and is inclined angle x-atan(.8) to the ground:  
sqrt(1+.8\*.8)\*cos(x-atan(.8))=1.2.

x=atan(.8)+acos(1.2/sqrt(1+.8\*.8)) and  
x=atan(.8)-acos(1.2/sqrt(1+.8\*.8))

You could also use complex numbers. The equation is the real part of  
exp(ix)_(1-0.8i), with 1-.8i=sqrt(1+.8_.8)_exp(-i_atan(.8)), giving  
1.2=Real part of sqrt(1+.8\*.8)\*exp( i(x-atan(.8) )… same equation as above.

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