# Solve this trigonometry equation

• chwala
In summary, the trig identities state that,##\frac {1}{\sqrt 2}##⋅ ##sin ∅##+##\frac {1}{\sqrt 2}##⋅ ##cos ∅##=##{\sqrt 3}##⋅ ##cos ∅##+##sin ∅##tan ∅=##[\frac {\sqrt 6 -1}{1-{\sqrt 2}}]##f

#### chwala

Gold Member
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;

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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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...

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.

chwala
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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$$-\tan\theta=(\sqrt{6}-1)(\sqrt{2}+1) \simeq (1.414*1.732-1)(1.414+1) \simeq 3.5$$
$$\tan(\theta-\frac{\pi}{2})=-\cot \theta \simeq \frac{1}{3.5}$$
$$\theta-\frac{\pi}{2} \simeq arctan \frac{1}{3.5} \simeq \frac{1}{3.5}-\frac{1}{3*(3.5)^3}\simeq 0.278 \simeq 15.9 degree$$