maths challenge!!!
let's see who's answer is detailed and correct and fastly answered.
and u will be marked the brainliest.
Answers
Exρlαnαtion:
a) cos2A = sin3A
Let's make the first trigonometric ratios equal
sinA can be written in terms of cosA is cos(90°-A)
So,
sin3A can be written as cos(90°-3A)
cos2A = cos(90°-3A)
Removing "cos" on both sides
↠2A = 90°-3A
↠2A + 3A = 90°
↠5A = 90°
↠A = 90°/5
↠A = 18°
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b) cos3A = sin7A
↠cos3A = cos(90°-7A)
sinA can be written in terms of cosA is cos(90°-A)
Removing "cos" on both sides.
➼3A = 90°-7A
➼3A + 7A = 90°
➼10A = 90°
➼A = 90°/10
➼A = 9°
______________________________
c) tanA = cot3A
cot3A can be expressed in form tanA as
cotA = tan(90°-A)
tanA = tan(90°-3A)
Removing "tan" on both sides.
➻A = 90°-3A
➻A + 3A = 90°
➻4A = 90°
➻A = 90°/4
➻A = 45/2
➻A = 22 1/2°
_____________________________
d) cotA = tan2A
tan2A can be expressed in terms of cotA are
tanA = cot(90°-A)
cotA = cot(90°-2A)
Removing "cot" on both sides
➳A = 90°-2A
➳A + 2A = 90°
➳3A = 90°
➳A = 90°/3
➳A = 30°
____________________________
So, the answers are :-
a) s
b) r
c) p
d) q[tex][/tex]
____________________________
Lєαrn Morє :
sin (90°-θ) = cos θ
cos (90°-θ) = sin θ
tan (90°-θ) = cot θ
csc (90°-θ) = sec θ
sec (90°-θ) = csc θ
cot (90°-θ) = tan θ
sin (90°+θ) = cos θ
cos (90°+θ) = -sin θ
tan (90°+θ) = -cot θ
csc (90°+θ) = sec θ
sec (90°+θ) = -csc θ
cot (90°+θ) = -tan θ
sin (180°-θ) = sin θ
cos (180°-θ) = -cos θ
tan (180°-θ) = -tan θ
csc (180°-θ) = csc θ
sec (180°-θ) = -sec θ
cot (180°-θ) = -cot θ
sin (180°+θ) = -sin θ
cos (180°+θ) = -cos θ
tan (180°+θ) = tan θ
csc (180°+θ) = -csc θ
sec (180°+θ) = -sec θ
cot (180°+θ) = cot θ
sin (270°-θ) = -cos θ
cos (270°-θ) = -sin θ
tan (270°-θ) = cot θ
csc (270°-θ) = -sec θ
sec (270°-θ) = -csc θ
cot (270°-θ) = tan θ
sin (270°+θ) = -cos θ
cos (270°+θ) = sin θ
tan (270°+θ) = -cot θ
csc (270°+θ) = -sec θ
sec (270°+θ) = cos θ
cot (270°+θ) = -tan θ