express 7sintheta+24 in the form Rsin(theta+a) where a is between 0

Question

express 7sintheta+24 in the form Rsin(theta+a) where a is between 0<a<90 stating the values of a and R
solve the equation 7sintheta+24costheta=-3 in the interval -90<theta<270
using your solution from part i) or otherwise solve a)the equations 7sin(x-10)+24cos(x-10)=-3 for x in the interval 0<x<180
b) 7cosy+24siny=-3 for y in the interval 0<y<180

abmon98 · Answer

$$7\sin \theta+24\cos \theta=Rsin \theta cosa+Rcos \theta sina$$
$$Rcosa=7$$
$$Rsina=24$$
$$tana=Sina/cosa, tana=24/7, a=73.739$$
$$R=\sqrt{(24)^2+(7)^2}=25$$

abmon98 · Answer

$$25\sin(\theta+24/7)=-3$$
$$\sin(\theta+24/7)=-3/25$$

abmon98 · Answer

@ganeshie8 Can you please help me in this question.

abmon98 · Answer

Can we use 25sin$$25\sin(\theta-\phi)=-3$$

abmon98 · Answer

$$\theta+\tan^{-1}( 24/7)=\sin^{-1}( -3/25)$$

abmon98 · Answer

x=-6.89210258-73.73979529
x=-80.63189787
I am stuck on the part of finding the solution from the interval -90<theta<270

ganeshie8 · Answer

So for part i) you got  $25\sin \left(\theta + \tan^{-1}(\frac{24}{7})\right)$   ?

abmon98 · Answer

yes i do

ganeshie8 · Answer

$25\sin \left(\theta + \tan^{-1}(\frac{24}{7})\right) = -3$   

 $\theta + \tan^{-1}(\frac{24}{7}) = \sin^{-1}(\frac{-3}{25})$   

 $\theta = \sin^{-1}(\frac{-3}{25}) -  \tan^{-1}(\frac{24}{7}) $   

 $\theta = \sin^{-1}(\frac{-3}{25}) -  \tan^{-1}(\frac{24}{7}) $   

 $\theta = -6.89 - 73.74 $   

 $\theta = -80.63 $

ganeshie8 · Answer

good, we both are on same page now :) lets see...

ganeshie8 · Answer

the sin function in question has a period of 360, and since sin(180-x) = sin(x)  :

the general solutions are :
1) $\large \theta =   -80.63  \pm   360n$,
2)  $\large \theta =   180 - (-80.63)  \pm   360n$

ganeshie8 · Answer

simplifying the second general solution gives :

1) $\large \theta =   -80.63  \pm   360n$,
2)  $\large \theta =   260.63  \pm   360n$

abmon98 · Answer

is it because the sine is negative so its used in the 3rd and 4th quadrant according to ASTEC Rule

ganeshie8 · Answer

-90<theta<270

add 90 through out

0 <theta+90<360

ganeshie8 · Answer

so you will get two solutions cuz the given window forms a full period of 360

ganeshie8 · Answer

`is it because the sine is negative so its used in the 3rd and 4th quadrant according to ASTEC Rule`

because, for what ?

abmon98 · Answer

sorry ASTC sine in only -ve in the 3rd quadrant and 4th quadrant
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ganeshie8 · Answer

okay :) whats the question ? 
are you asking why sin(x) = sin(180-x) ?

abmon98 · Answer

yes

ganeshie8 · Answer

btw, $-80.63$ and $260.63$ are the solutions for this problem
notice that they're between (-90, 270)

abmon98 · Answer

yah they are thank you @ganeshie8 for your help. :)

ganeshie8 · Answer

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ganeshie8 · Answer

suppose,  thats the angle $ x$

ganeshie8 · Answer

lets draw the angle $180-x$ , and see whats going on

ganeshie8 · Answer

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