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

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The discussion centers on solving a trigonometry equation using identities, with one participant providing a detailed solution that leads to an angle of approximately 105.9 degrees. There is a request for alternative methods, and another participant suggests using the half-angle approach, which yields the same result. Some participants express confusion over differing answers and the importance of verifying trigonometric identities. Ultimately, the conversation emphasizes the need for practice in trigonometry to avoid mistakes and improve understanding.
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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...
 
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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##
 
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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
 
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