Solve the given trigonometry equation

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The discussion focuses on clarifying the use of cycles in trigonometric equations, particularly why tangent is considered with a factor of 2 instead of just π radians. The user expresses a preference for using π radians for simplicity. They also provide a known value for the inverse tangent function, which leads to solutions in the second and fourth quadrants. The derived equations for x incorporate a factor of 1/2 in the cycle, reflecting the periodic nature of the tangent function. The final solutions are constrained within the range of -10 to 0, yielding specific values for x.
chwala
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Homework Statement
see attached
Relevant Equations
Trigonometry
This is the problem. The question is simple i just need some clarification as indicated on the part highlighted below in red.

1662037532264.png
Now from my understanding tangent repeats on a cycle of ##π## radians...why do we have 2 the part circled in red below i.e ##2##? This is the part that i need clarity. Why consider cycles for sine and cosine here?
It's straightforward and less stressful to just use ##π##radians in my opinion.

1662037959983.png


The other form given below is the one that i know and am very much conversant with;

1662037901886.png
 
Last edited:
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We know that ##\tan^{-1}####\left[-\dfrac{5}{9}\right]=-0.5071##.

Tangent is negative in the second and fourth quadrant respectively. This will realize the given values: ##-0.5071+π## and ##2π-0.5071## respectively... giving us the desired ##2.6345## and ##5.7761##
i.e
##2x=-0.5071+nπ⇒x=-0.25355+\dfrac{1}{2}nπ##

##2x=2.6345+nπ⇒x=1.31725+\dfrac{1}{2}nπ##

##2x=5.7761+nπ⇒x=2.88805+\dfrac{1}{2}nπ##

where ##n=0, ±1, ±2...## and given that the solution lies on ##-10≤x≤0## then we shall get the answers as shown on the text.

Cheers.
 
Last edited:
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.

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