Here's the question you clicked on:
mathslover
Find the value of : sin 47 (degrees) + sin 61 (degrees) - sin 11 (degrees) - sin 25 (degrees)
set the calculator to degrees and start pressing buttons
lol @1on1 , that's a shortcut but I do want to do that without *calculator*
Sorry, that will require much more thought from me !
No problem! @sasogeek any suggestions, if no then I have only 1 .
Not suggestion but an idea infact.
use the calculator to find their actual values and do normal addition... that's if u want to do some work... if not, u may use the calculator to find the whole answer.
hmn quite hard if calculator is not allowed.
earlier i thought it was sin(A+B)-sin(A-B) sort of question but it's not, direct solution with the calculator is the way forward :) unless u know those values off the top of ur head xD
i do not wish to confuse you, but you may want to read this, http://www.trans4mind.com/personal_development/mathematics/trigonometry/sumProductCosSin.htm#Sum_of_Sine_and_Cosine
@mathslover Do u knw the Value of Sin18 and cos36
SinA + SinB = 2Sin (A+B)/2 . Cos (A-B)/2...Use this
@Yahoo! and @sasogeek this was my main idea.
Let me try again to find out the answer through the formula of sina + sinb
OK so if I remember the values of : cos 36 degrees and sin 18 degrees then I am getting cos 7 degrees.
cos Here is it how I did that : \[\large{\sin 47 + \sin 61 = 2 \sin 54 \cos 7 }\] ^that theta(s) are in degrees. similarly : \[\large{-(\sin 11 + \sin 25) = -( 2\sin 18 \cos 7)}\] from the above we get : \(2\cos 7 ( \sin 54 - \sin 18 )\) \(2\cos 7 ( 2\cos 36 \sin 18)\) \(4\cos 7 (\frac{\sqrt{5}+1}{4} \times \frac{\sqrt{5}-1}{4})\) \(\cos 7\) That is it! Is it correct? @sasogeek @Yahoo!
Yup...thats wat i also Got
2Sin54.cos7 - 2sin18cos7 2cos7 ( sin54 - sin18) sin54 = cos 36 2cos7(cos36 - sin18) 2cos7( (sqrt5 + 1)/2 - (sqrt5-1)/2) = cos7
Sorry 2cos7( (sqrt5 + 1)/4 - (sqrt5-1)/4)
np, thanks a lot for your help yahoo.
To..Be honest..U Did All the Hard Work..)