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secondary 4 | A Maths
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Can someone help me proving trigonometric identities for qn (j) and (k)?
Date Posted: 4 years ago
Views: 269
tan (π/4 + A/2)
=
(tan π/4 + tan A/2)
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
(1 - tan π/4 tan A/2)
=
(1 + tan A/2)
‾‾‾‾‾‾‾‾‾‾‾‾‾
(1 - tan A/2)
=
(1 + tan A/2)(1 + tan A/2)
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
[(1 - tan A/2)(1 + tan A/2)]
=
1 + 2tan A/2 + tan² A/2
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
(1² - tan² A/2)
=
(1 + tan² A/2) □□□□□□ 2tan A
‾‾‾‾‾‾‾‾‾‾‾‾‾‾□□+□□ ‾‾‾‾‾‾‾‾‾‾‾
(1 - tan² A/2)□□□□□□ (1 - tan² A/2)
=
cos A/2 / cos A/2 + sin² A/2 / cos² A/2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ + tanA
cos A/2 / cos A/2 - sin² A/2 / cos² A/2
=
(cos² A/2 + sin² A/2)/(cos² A/2)
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ + tanA
(cos² A/2 - sin² A/2)/(cos² A/2)
=
1 / cos² A/2
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ + tan A
cos A / cos A/2
= 1/cos A + tan A
= tan A + sec A
(Proved)
=
(tan π/4 + tan A/2)
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
(1 - tan π/4 tan A/2)
=
(1 + tan A/2)
‾‾‾‾‾‾‾‾‾‾‾‾‾
(1 - tan A/2)
=
(1 + tan A/2)(1 + tan A/2)
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
[(1 - tan A/2)(1 + tan A/2)]
=
1 + 2tan A/2 + tan² A/2
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
(1² - tan² A/2)
=
(1 + tan² A/2) □□□□□□ 2tan A
‾‾‾‾‾‾‾‾‾‾‾‾‾‾□□+□□ ‾‾‾‾‾‾‾‾‾‾‾
(1 - tan² A/2)□□□□□□ (1 - tan² A/2)
=
cos A/2 / cos A/2 + sin² A/2 / cos² A/2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ + tanA
cos A/2 / cos A/2 - sin² A/2 / cos² A/2
=
(cos² A/2 + sin² A/2)/(cos² A/2)
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ + tanA
(cos² A/2 - sin² A/2)/(cos² A/2)
=
1 / cos² A/2
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ + tan A
cos A / cos A/2
= 1/cos A + tan A
= tan A + sec A
(Proved)
Ironically the “fraction line” shrank in size after posting. I am not sure if the app is capable of allowing lots of space bars between any two characters typed.
Doesn't work. It will all revert to single space
k)
Let x = π/4 + A
Let y = π/4 - A
tan (π/4 + A) + tan (π/4 - A)
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
tan (π/4 + A) - tan (π/4 - A)
= (tan x + tan y)/(tan x - tan y)
= (sin x / cos x + sin y / cos y) / (sin x / cos x - sin y / cos y)
= (sin x cos y + cos x sin y)/(cos x cos y) ÷
(sin x cos y - cos x sin y)/(cos x cos y)
= sin(x + y) / sin(x - y)
= sin(π/4 + A + π/4 - A) / sin(π/4 + A - (π/4 - A))
= sin(π/2)/sin(2A)
= 1/sin2A
= cosec 2A
(Proved)
Let x = π/4 + A
Let y = π/4 - A
tan (π/4 + A) + tan (π/4 - A)
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
tan (π/4 + A) - tan (π/4 - A)
= (tan x + tan y)/(tan x - tan y)
= (sin x / cos x + sin y / cos y) / (sin x / cos x - sin y / cos y)
= (sin x cos y + cos x sin y)/(cos x cos y) ÷
(sin x cos y - cos x sin y)/(cos x cos y)
= sin(x + y) / sin(x - y)
= sin(π/4 + A + π/4 - A) / sin(π/4 + A - (π/4 - A))
= sin(π/2)/sin(2A)
= 1/sin2A
= cosec 2A
(Proved)
See 2 Answers
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Good afternoon Annela! Here are my workings for part j. I will look at part k at a later time.
Date Posted:
4 years ago
Hello can i ask why did you multiply by cos(A-pi/2) to both numerator and denominator
The number which I multiplied to both sides is cos A/2, which is introduced because tan A/2 = sin A/2 / cos A/2 is a fraction in itself. Multiplying both numerator and denominator by cos A/2 ensures I will not have to deal with a “fraction in the numerator and fraction in the denominator” scenario.
Multiplying cos A/2 by tan A/2 is akin to multiplying cos A/2 by sin A/2 / cos A/2 which simplifies to sin A/2.
On top of this, leaving my trigonometric expressions in sines and cosines simplifies matters a lot when it comes to the double angle formula.
Multiplying cos A/2 by tan A/2 is akin to multiplying cos A/2 by sin A/2 / cos A/2 which simplifies to sin A/2.
On top of this, leaving my trigonometric expressions in sines and cosines simplifies matters a lot when it comes to the double angle formula.
done
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Hi Annela! Here are my workings for part k. At any point if you do not understand my workings or you find my handwriting too small, let me know accordingly and I will rewrite them again!
Date Posted:
4 years ago
Hello thank you for your help but may i know why did you multiply cos A/2 to numerator and denominator for part j and multiply cos A to numerator and denominator for part k?
For this part is the same reason; it’s more of anticipatory than a “must” to do this. Simplifying that tangent expression is tough, but converting it into sine divided by cosine makes the “fraction in fraction” situation difficult to simplify as well, hence my decision to multiply both numerator and denominator by cos in advance.
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