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Deriving the sum of sin and cos formula

  1. Mar 15, 2017 #1
    1. The problem statement, all variables and given/known data
    Show that ##a \sin x + b\cos x = c \sin (x + \theta)##, where ##c = \sqrt{a^2 + b^2}## and ## \displaystyle \theta = \arctan (\frac{b}{a})##

    2. Relevant equations


    3. The attempt at a solution
    We see that ##c \sin (x + \theta) = c \cos \theta (\sin x) + x \sin \theta (\cos x)##. So we compare coefficients. ##a = c \cos \theta## and ##b = c sin \theta##. We can assume that neither ##a## or ##b## are zero, so we find by division that ##c = \sqrt{a^2 + b^2}##. Next, we see that ##a^2 + b^ = c^2 \cos^2 \theta + c^2 \sin^2 \theta = c^2##, so ##c = \pm \sqrt{a^2+b^2}##. So it seems that I am done, since the we can take the positive root. However, what about the negative root? Would that give a valid answer too? Why isn't the negative root the one that is desired by the question?
     
  2. jcsd
  3. Mar 15, 2017 #2
    I think you mean "so we find by division that ##\theta = \arctan(b/a)##.

    I can think of two ways why ##c >0 ##

    First because the term on LHS is the equation of a straight line in polar coordinates and ##c## corresponds to ##r## in polar. Since ##r## is always +ve and thus so is ##c##.

    Second, for given ##a,b > 0## you can always construct a right triangle with ##h = \sqrt{a^2 + b^2}##. (h is hypotenuse).

    So we can write ##a\sin x + b\cos x = \sqrt{a^2 + b^2}\left({a\over \sqrt{a^2 + b^2}}\sin x + {b\over \sqrt{a^2 + b^2}}\cos x\right)##

    Lets angle between ##h## and ##a## as ##\theta##, So we get ##\cos \theta = {a\over \sqrt{a^2 + b^2}}, \ \ \ \sin \theta = {b\over \sqrt{a^2 + b^2}}## and ##\tan \theta = {b \over a}##

    From here we can get,##\theta = \arctan(b/a)##,

    And ##\sqrt{a^2 + b^2}\left({a\over \sqrt{a^2 + b^2}}\sin x + {b\over \sqrt{a^2 + b^2}}\cos x\right) = \sqrt{a^2 + b^2}\left(\cos \theta\sin x + \sin \theta\cos x\right) = \sqrt{a^2 + b^2}\sin (\theta + x)##

    From here you can see ## c = h = \sqrt{a^2 + b^2}##, and hypotenuses are not negative, Do they ?
     
  4. Mar 15, 2017 #3

    Charles Link

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    Homework Helper

    In the determination of the angle ## \theta ##, ## a ## and ## b ## need to retain their sign, and you draw a diagram to find ## \theta ##. If both ## a ## a ## b ## are negative, then ## \theta ## winds up in the third quadrant, instead of the first. There is ambiguity in the expression ## arctan(\frac{b}{a}) ## if you don't preserve the signs of ## b ## and ## a ##. It is important to preserve the signs of ## b ## and ## a ## in this calculation. The ## \theta ## you compute will be different for ## a=1 ## and ## b=1 ## as compared to ## a=-1 ## and ## b=-1 ##. To preserve the signs, you need to use the positive root for ## c ##.
     
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