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

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SUMMARY

The discussion focuses on solving a trigonometry equation using identities and alternative methods. The primary solution involves the equation ##\frac {1}{\sqrt 2} \cdot \sin ∅ + \frac {1}{\sqrt 2} \cdot \cos ∅ = \sqrt{3} \cdot \cos ∅ + \sin ∅##, leading to the result of ##∅ = 105.9^0##. Participants also explore the half-angle approach, utilizing the substitution ##t = \tan \frac{1}{2} ∅##, which confirms the same solution. The conversation highlights the importance of verifying trigonometric identities to avoid errors in calculations.

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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
 
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