How Does the Rule a Cos Theta + b Sin Theta Simplify Using Trigonometry?

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The discussion centers on the simplification of the expression a cos θ + b sin θ using trigonometric identities. The key formula presented is that a cos θ + b sin θ can be rewritten as √(a² + b²) cos(θ - α), where α is an angle derived from the sides of a right triangle formed by a and b. The user seeks clarification on how this rule is derived, particularly the geometric interpretation involving a right triangle. The textbook explanation involves using the triangle to define the hypotenuse as √(a² + b²) and the angles accordingly. The conversation highlights the need for a clearer understanding of the relationship between the trigonometric functions and their geometric representations.
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Hey,
I’ve got a test in one week’s time and was studying through my textbook of geometry and trigonometry. I came across a “rule” which shows how to simplify expressions in the form of a \cos \theta + b \sin \theta but I do not understand how this “rule” works.
The simplify rule:

a \cos \theta + b \sin \theta = \sqrt{a^2+b^2} (\frac{a}{\sqrt{ a^2+b^2}} \cos \theta + \frac{b}{ \sqrt{ a^2+b^2}} \sin \theta )

a \cos \theta + b \sin \theta = \sqrt{a^2+b^2}(\cos \alpha \cos \theta + \sin \alpha \sin \theta)

a \cos \theta + b \sin \theta = \sqrt{a^2+b^2} \cos (\theta - \alpha)

where \alpha is an acute angle

In the textbook this is explained using a right angle triangle where \alpha is the unknown angle being measured, side a is the adjacent side and side b is the opposite, therefore the hypotenuse is equal to sqrt{a^2+b^2}. However I do not understand this method which the book uses to explain, and was wondering if somebody out there knew how to explain/prove how/where this “rule” has come from. Thank you for any legitimate reply,
Pavadrin

EDIT: those latex code isn't wokring properly so I've attached this image link:
PF1.JPG
 
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okay thanks for the link J77
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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