Trigonometric problem: Sin7x= Sin24 * 6.4

[Eobardrush]

Homework Statement

Sin7x= Sin24 * 6.4

Relevant Equations

SinA/A= SinB/B

So I came across a question regarding sine rule. This is to find a missing angle. The question goes as follows

Sin7x= Sin24 * 6.4
I tried two methods to solve this.

Method 1:

x= (Sin24 * 6.4)/ Sin 7
=21.35

I basically divided both sides by Sin 7

Method 2:

Sin x= (Sin 24 * 6.4)/ 7

x= Sin-1 (Ans)

x= 21.83

Note: I used my calculator "Ans" function after dividing by 7 to calculate the Sine inverse to avoid rounding errors and such.

I am confused as to which one is the right method here. Both gives very similar answers but the fact that there is a miniscule difference still makes me worried

 

[anuttarasammyak]

I would like you to do some clarification of the problem.
Do you mean 24 as 24 degree = $\frac{2}{15}\pi$ radian ?
Which do you mean by sin7x as $7 \sin x, x \sin 7$ degree or $\sin(7x)$?
Do these angles belong to a triangle if so what are side length A and B?

 

[Eobardrush]

I would like you to do some clarification of the problem.
Do you mean 24 as 24 degree = $\frac{2}{15}\pi$ radian ?
Which do you mean by sin7x as $7 \sin x, x \sin 7$ degree or $\sin(7x)$?
Do these angles belong to a triangle if so what are side length A and B?

Hello yeah these are in degrees
Here is the question, sorry i did not post it before

Using Sine rule I did:

Sin x/ 6.4= Sin60 * 7
But at the final part I was not sure if I should use inverse sine to solve X or divide both sides by sine 7 to solve X

 

[anuttarasammyak]

Thanks. It seems you can apply

Relevant Equations:: SinA/A= SinB/B

with
sin A / BC = sin x / AB
to know the value of sin x which is used to express y.

 

[Eobardrush]

Thanks. It seems you can apply

with
sin A / BC = sin x / AB
to know the value of sin x.

I understand that yeah. I did apply that equation and arrived at
Sin7x= Sin24 * 6.4

At this point I was having a problem whether to use inverse sine after dividing both sides by 7 to solve for X
OR
To divide both sides by Sine 7 to solve for X
Because both seems to be giving similar answers but are not exactly the same(As shown in my workings above in the original question)

So my question is which method is best to solve for X.

 

[Orodruin]

Sin x/ 6.4= Sin60 * 7

This is not a correct application of the sine theorem. The opposite side of the angle $x$ in the triangle where the opposite side of the $60^\circ$ angle is 7 cm is y cm, not 6.4 cm. You have also multiplied by 7 instead of dividing. Your problems therefore arise earlier than when you have to deduce an angle from the sine of the angle (which is done by applying the inverse of the sine function).

Your formulas could also use some parentheses around the sine arguments for clarity.

Also, please note that the homework guidelines require you to state the full problem formulation exactly as it was given, not just an equation. This is important as we are otherwise stabbing in the dark at a random equation that we don't know where it came from and you may have misinterpreted.

 

[Eobardrush]

This is not a correct application of the sine theorem. The opposite side of the angle $x$ in the triangle where the opposite side of the $60^\circ$ angle is 7 cm is y cm, not 6.4 cm. You have also multiplied by 7 instead of dividing. Your problems therefore arise earlier than when you have to deduce an angle from the sine of the angle (which is done by applying the inverse of the sine function).

Your formulas could also use some parentheses around the sine arguments for clarity.

Also, please note that the homework guidelines require you to state the full problem formulation exactly as it was given, not just an equation. This is important as we are otherwise stabbing in the dark at a random equation that we don't know where it came from and you may have misinterpreted.

I apologize for that. The reason I included just the equation is to know which method is best to solve for trigonometric equations. Whether to use Sine inverse to solve for X, or to divide both sides by Sine 7
I only had problems with solving that equation and wanted to know that and not the whole question. So I thought it would be irrelevant to include the whole question

As for the multiplication by 7 part, I just did cross multiplication of the equation to get to that

 

[Orodruin]

Whether to use Sine inverse to solve for X, or to divide both sides by Sine 7
I only had problems with solving that equation and wanted to know that and not the whole question

No, you had severe problems because your lack of parenthesis confuses you and you wrote down the equation wrong.

As for the multiplication by 7 part, I just did cross multiplication of the equation to get to that

No, you did not. If you did you applied the sine theorem wrong. Also, multiplying by 7 does not get a 7 inside the argument of the sine: $7 \sin(x) \neq \sin(7x)$.

Sin7x= Sin24 * 6.4

The correct application of the sine theorem is
$$
\frac{\sin(x)}{6.4} = \frac{\sin(24^\circ)}{7}
$$
which gives
$$
7\sin(x) = 6.4 \sin(24^\circ).
$$
The 7 is not inside the argument of the sine.

 

[Eobardrush]

The correct application of the sine theorem is
$$
\frac{\sin(x)}{6.4} = \frac{\sin(24^\circ)}{7}
$$
which gives
$$
7\sin(x) = 6.4 \sin(24^\circ).
$$
The 7 is not inside the argument of the sine.

Ahh I see thank you. I am still at very basic level of trigonometry but I understand more now
So in this case I should do sine inverse to solve for 6.4sine(24) right?

Or should I divide both sides by 7 sine ? Both seems to give similar answers which is where I am stuck at, hence my original problem

 

[Delta2]

You should first divide both sides by 7 (not $\sin 7$) and then take the inverse sine of $\frac{6.4\sin 24}{7}$, that is $$x=\arcsin \frac{6.4\sin 24}{7}$$.

The impression you give to me (I might be wrong here), is that you are not fully familiar with solving first order equations of the type $$ax+b=0, a\neq 0$$ that is. Or if you do, the presence of sines confuses you a lot.
In this specific example we have a first order equation, it is $a=7$ and $b=-{6.4\sin 24}$ and here it is what it might confuse you, the role of x is played by $y=\sin x$, so the first order equation is $$7y-{6.4\sin 24}=0$$. Solve this for y and then solve for x, $$y=\sin x\Rightarrow x=\arcsin y$$

 

[Orodruin]

Or should I divide both sides by 7 sine ?

$\sin$ is not a number, it is a function. You cannot treat it as something you can "divide by" and be left with the argument, i.e., "dividing by $7 \sin$" makes no sense - the sine has an argument.

 

[Eobardrush]

Thanks guys I have got it now.

Replying to Delta: Yeah Trigonometry really confuses me a lot. I can solve normal equations but when it comes to Trig I just go blank. Specially when it comes to manipulating equations and all

Replying to Orodruin: Thank you for your explanation. I think I got what u meant by now. I am still getting into trig so hoping I would get used to it more once I do more problems

 

[Orodruin]

Let us take a simpler function: $f(x) = x^2$. If you have an equation on the form $2 f(x) = 32$ you cannot "divide by $2f$" and obtain $x = 16/f$, the $f$ here makes no sense, it has no argument. The only way to solve for $x$ is to apply the inverse of the function $f(x)$, i.e., the square root.