Solving trigonometric equation of a sum of unknowns

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Homework Statement


[itex]\sin (x) = \frac{2}{3}[/itex] and [itex]\sec (y) = \frac{5}{4}[/itex], where [itex]x[/itex] and [itex]y[/itex] lie between 0 and [itex]\frac{\pi}{2}[/itex] evaluate [itex]\sin (x + y)[/itex]

Homework Equations


Looked over some trig laws, don't think I saw anything that's too relevant. There [itex]\sec (x) = \frac{1}{\sin (x)}[/itex]

The Attempt at a Solution


I can't think of anything. Assuming I'm not an idiot, I can simply re-write the secant equation as another sine equation:
[tex]\sec (y) = \frac{5}{4} \longrightarrow \sin (y) = \frac{4}{5}[/tex]
So then we know [itex]\sin (x) = \frac{2}{3}[/itex] and [itex]\sin (y) = \frac{4}{5}[/itex]. From here we can do some inverse sines and substitute in [itex]x[/itex] and [itex]y[/itex] in [itex] \sin (x+y)[/itex], but that looks awful and worse to solve ([itex]\sin (\sin ^{-1} (\frac{2}{3}) + \sin ^{-1} (\frac{4}{5})[/itex]).

Can anyone give me a hint as to the proper next step I should take to evaluate this equation?

Thanks for any help!
 

Answers and Replies

  • #2
Simon Bridge
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You can gain a better understanding by constructing the relevant triangles. sec(y) = 1/cos(y) and cos(y)=4/5 suggests a 3-4-5 triangle... so sin(y)=3/5 and so on. Maybe you can relate sin(A+B) to combinations of sinAsinB, sinAcosB etc?
 
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  • #3
Math_QED
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There is a formula ##\sin(\alpha + \beta) =## that will lead to the solution.

I don't know whether you have covered complex numbers yet but the formula is quite trivial to derive then. Start with:

##\cos(\alpha + \beta) + i\sin(\alpha + \beta) = e^{i(\alpha + \beta)} = \dots##

Now, using the rules of complex numbers and exponentiation, and after comparing the real and imaginary parts, you should be able to find the formula that leads you to the answer, assuming you have already seen complex numbers.
 
  • #4
Ray Vickson
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Homework Statement


[itex]\sin (x) = \frac{2}{3}[/itex] and [itex]\sec (y) = \frac{5}{4}[/itex], where [itex]x[/itex] and [itex]y[/itex] lie between 0 and [itex]\frac{\pi}{2}[/itex] evaluate [itex]\sin (x + y)[/itex]

Homework Equations


Looked over some trig laws, don't think I saw anything that's too relevant. There [itex]\sec (x) = \frac{1}{\sin (x)}[/itex]

The Attempt at a Solution


I can't think of anything. Assuming I'm not an idiot, I can simply re-write the secant equation as another sine equation:
[tex]\sec (y) = \frac{5}{4} \longrightarrow \sin (y) = \frac{4}{5}[/tex]
So then we know [itex]\sin (x) = \frac{2}{3}[/itex] and [itex]\sin (y) = \frac{4}{5}[/itex]. From here we can do some inverse sines and substitute in [itex]x[/itex] and [itex]y[/itex] in [itex] \sin (x+y)[/itex], but that looks awful and worse to solve ([itex]\sin (\sin ^{-1} (\frac{2}{3}) + \sin ^{-1} (\frac{4}{5})[/itex]).

Can anyone give me a hint as to the proper next step I should take to evaluate this equation?

Thanks for any help!
You have ##\sec(x)## given incorrectly: ##\sec(x) = \frac{1}{\cos(x)}##, NOT ##\frac{1}{\sin(x)}##.

See, eg., https://en.wikipedia.org/wiki/Trigonometric_functions .
 
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  • #5
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You can gain a better understanding by constructing the relevant triangles. sec(y) = 1/cos(y) and cos(y)=4/5 suggests a 3-4-5 triangle... so sin(y)=3/5 and so on. Maybe you can relate sin(A+B) to combinations of sinAsinB, sinAcosB etc?
Thanks a bunch for your response! First issue, I was so tired last night that I mixed up secant and cosecant. Correctly transforming secant into cosine (instead of sine) would let me use the sum-difference formula: [itex]\sin (x+y) = \sin (x)\cos (y) + \cos (x)\sin (y)[/itex]. I have values for [itex]\sin (x)[/itex] and [itex]\cos (y)[/itex], but not for [itex]\sin (y)[/itex] and [itex]\sin (y)[/itex].

Using your insights on solving this more visually, I deduced that [itex]\cos (x) = \frac{\sqrt{5}}{3}[/itex] and, as you stated, [itex]\sin (y) = \frac{3}{5}[/itex]. I can then fully solve the equation:
[tex]\sin (x+y) = \sin (x)\cos (y) + \cos (x)\sin (y) = \frac{2}{3} \times \frac{4}{5} + \frac{\sqrt{5}}{3} \times \frac{3}{5} = \frac{8}{15} = \frac{\sqrt{5}}{5} = \frac{8 + 3\sqrt{5}}{15}[/tex]

This turned out to be the correct answer, however I'm confused as to how it was possible. The assumption allowing us to get values for [itex]\sin (y)[/itex] and [itex]\sin (y)[/itex] would be that the triangle in question is a right triangle. How is this assumption possible? Am I just missing something blatantly obvious as to why this is a right triangle?
 
  • #6
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teetar said:
This turned out to be the correct answer, however I'm confused as to how it was possible. The assumption allowing us to get values for [itex]\sin (y)[/itex] and [itex]\sin (y)[/itex] would be that the triangle in question is a right triangle. How is this assumption possible? Am I just missing something blatantly obvious as to why this is a right triangle?
From your OP:
[itex]\sin (x) = \frac{2}{3}[/itex] and [itex]\sec (y) = \frac{5}{4}[/itex], where [itex]x[/itex] and [itex]y[/itex] lie between 0 and [itex]\frac{\pi}{2}[/itex]
Does this answer your question?

Also, from the first equation above, since sin(x) > 0, x must be in the first or second quadrant. From the second equation, since sec(y) > 0, y has to be in the first or fourth quadrant. For both to be true, x and y have to be in the first quadrant, which is also stated explicitly in the last part of what I quoted.
 
  • #7
haruspex
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The assumption allowing us to get values for sin(y)sin⁡(y)\sin (y) and sin(y)sin⁡(y)\sin (y) would be that the triangle in question is a right triangle
It is not so much that there is a triangle "in question". It is the basic definition of sine and cosine (before being extended to arguments outside the range 0 to π/2). If you construct a right triangle in which one angle is theta then sine theta is opposite / hypotenuse and cosine is adjacent/hypotenuse.
Or you can use sin2+cos2=1, and use the given ranges to know to take the positive square root.
 
  • #8
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It is not so much that there is a triangle "in question". It is the basic definition of sine and cosine (before being extended to arguments outside the range 0 to π/2). If you construct a right triangle in which one angle is theta then sine theta is opposite / hypotenuse and cosine is adjacent/hypotenuse.
Or you can use sin2+cos2=1, and use the given ranges to know to take the positive square root.
OHH okay, I was skipping over a pretty fundamental idea there, but I think I get it now. Makes my question, about why we know it's a right triangle, seem pretty silly now.

Thanks for all the input in helping me find the answer, everyone!
 

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