How to Solve a Trigonometry Equation Using Identities and Alternative Methods?

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Homework Help Overview

The discussion revolves around solving a trigonometry equation using various identities and methods. Participants explore the application of trigonometric identities and alternative approaches to find solutions within a specified domain.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of trigonometric identities, including the sum and difference of angles and half-angle identities. There are attempts to verify the correctness of these identities and the solutions derived from them. Some participants express interest in alternative methods for solving the equation.

Discussion Status

The discussion is ongoing, with participants sharing different approaches and questioning the accuracy of each other's identities. There is a mix of agreement and differing answers, with some participants expressing a desire for further exploration of alternative methods.

Contextual Notes

Some participants mention a lack of practice in trigonometry, which may influence their understanding and application of the identities involved. The original poster seeks solutions within the domain of 0° to 180°.

chwala
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Homework Statement
Solve the equation##sin (∅+45^0)=2 cos (∅-30^0)##
giving all solutions in the interval ##0^0< ∅<180^0##
Relevant Equations
Trigonometric identities
Find the Mark scheme solution here;

1640338799514.png


Now find my approach;
Using the trig. identities It follows that,
##\frac {1}{\sqrt 2}##⋅ ##sin ∅##+##\frac {1}{\sqrt 2}##⋅ ##cos ∅##=##{\sqrt 3}##⋅ ##cos ∅##+##sin ∅##
→##sin ∅##[##\frac {1}{\sqrt 2}##-##1]##=##cos ∅##[##\frac {-1}{\sqrt 2}##+##{\sqrt 3}##]
→##sin ∅##⋅[##\frac {1-{\sqrt 2}}{\sqrt 2}]##=##cos ∅##⋅[##\frac {\sqrt 6 -1}{\sqrt 2}]##
→##tan ∅##=##[\frac {\sqrt 6 -1}{1-{\sqrt 2}}]##
## ∅##=##-74.051^0##, but we want our solutions to be in the domain, ##0^0< ∅<180^0##,
therefore, ## ∅##=##-74.051^0 + 180^0##=##105.9^0##

I would definitely be interested in another approach...
 
Last edited:
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check your trig identities, I get a different answer...
 
Dr Transport said:
check your trig identities, I get a different answer...
Which answer are you getting? The markscheme answer is attached on this post...let me know which part of my identity isn't clear. Cheers...
 
No worries, I also get irked when I don't get some things right...mostly due to a lack of practice...I want to dedicate 2022 to doing more hour practise in all mathematical areas ...pure, applied and stats..and even operations research...I just have to be a little more serious... :cool: :cool:
 
chwala said:
I would definitely be interested in another approach...
The approach you took was the most obvious one; i.e., using the sum and difference of angles identities. Any other approach would likely be longer and more obtuse.
 
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We can also use the half- angle approach by letting ##t##=tan ##\frac {1}{2}####∅##, using the knowledge and understanding of half-angles, then it follows that,

##\frac {1-t^2}{2t}##=##[\frac {1-\sqrt 2 }{{\sqrt 6}-1}]##

##1-t^2=-0.57153t##

##t^2-0.57153t-1=0##

##t_1=1.3257## and ##t_2=-0.75426##

taking, ##1.3257##=##tan## ##\frac {1}{2}####∅##

→##\frac {1}{2}####∅##=##tan^{-1}####1.3257##

→##\frac {1}{2}####∅##=##52.972074##

→##∅##=##2×52.972074=105.9^0##
 
Last edited:
There seems to be no essential alternative to your answer.
A calculation by hand
-\tan\theta=(\sqrt{6}-1)(\sqrt{2}+1) \simeq (1.414*1.732-1)(1.414+1) \simeq 3.5
\tan(\theta-\frac{\pi}{2})=-\cot \theta \simeq \frac{1}{3.5}
\theta-\frac{\pi}{2} \simeq arctan \frac{1}{3.5} \simeq \frac{1}{3.5}-\frac{1}{3*(3.5)^3}\simeq 0.278 \simeq 15.9 degree
 
Last edited:
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