My Math Forum Trig identities

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 November 6th, 2011, 10:43 AM #1 Guest   Joined: Posts: n/a Thanks: Trig identities 1. Prove: 1) $\frac{cot(\beta ) +cot(\gamma) }{cot(\alpha)+cot(\gamma) }=\frac{sin2\alpha }{sin2\beta }$ 2) $\frac{tan\alpha +tan \beta }{tan\beta +tan\gamma }=\frac{cos^{2 }\gamma }{cos^{2}\alpha }$ and $\alpha +\beta +\gamma=\frac{\pi }{2}$ 2. If $tan\alpha=1 , tan\beta =2 , tan\gamma =3$ prove that $\alpha +\beta +\gamma=\frac{\pi }{2}$
 November 6th, 2011, 03:47 PM #2 Global Moderator   Joined: Dec 2006 Posts: 17,508 Thanks: 1318 2. You need to know that the angles are in the first quadrant, so that $\alpha\,=\,\frac\pi4$ and $\tan(\beta\,+\,\gamma)\,=\,\frac{\tan\beta\,+\,\ta n\gamma}{1\,-\,\tan\beta\tan\gamma}\,=\,\frac{2\,+\,3}{1\,-\,2(3)}\,=\,-1,$ so that $\beta\,+\,\gamma\,=\,3\pi/4.$ It follows that $\alpha\,+\,\beta\,+\,\gamma\,=\,\pi.$
November 6th, 2011, 05:49 PM   #3
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Re: urgent help

Hello, noki!

Quote:
 $\text{1. Given: }\:A\,+\,B\,+\,C\:=\:\frac{\pi}{2}$

$\text{W\!e have: }\:A \:=\:\frac{\pi}{2}\,-\,(B\,+\,C) \;\;\;\Rightarrow\;\;\;\cos A \:=\:\cos\big[\frac{\pi}{2}\,-\,(B\,+\,C)\big] \:=\:\sin(B\,+\,C)$
[color=beige]. . [/color]$\text{Hence: }\:\sin(B\,+\,C) \:=\:\cos A \;\;[1]$

$\text{W\!e have: }\:B \:=\:\frac{\pi}{2}\,-\,(A\,+\,C) \;\;\;\Rightarrow\;\;\;\cos B \:=\:\cos\big[\frac{\pi}{2}\,-\,(A\,+\,C)\big] \;=\;\sin(A\,+\,C)$
[color=beige]. . [/color]$\text{Hence: }\:\sin(A\,+\,C) \:=\:\cos B\;\;[2]$

$\text{W\!e have: }\:C \:=\:\frac{\pi}{2}\,-\,(A\,+\,B) \;\;\;\Rightarrow\;\;\;\cos C \:=\:\cos\big[\frac{\pi}{2}\,-\,(A\,+\,B)\big] \:=\:\sin(A\,+\,B)$
[color=beige]. . [/color]$\text{Hence: }\:\sin(A\,+\,B) \:=\:\cos C\;\;[3]$

Quote:
 $\text{(a) Prove: }\:\frac{\cot B\,+\,\cot C}{\cot A\,+\,\cot C} \:=\:\frac{\sin2A}{\sin2B}$

$\frac{\cot B\,+\,\cot C}{\cot A\,+\,\cot C} \;=\;\frac{\frac{\cos B}{\sin B} \,+\,\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}\,+\,\frac{\cos C}{\sin C}}$

$\text{Multiply top and bottom by }\,\sin A\,\!\sin B\,\!\sin C:$

[color=beige]. . [/color]$=\; \frac{\sin A\,\!\cos B\,\!\sin C\,+\,\sin A\,\!\sin B\,\!\cos C}{\cos A\,\!\sin B\,\!\sin C\,+\,\sin A\,\!\sin B\,\!\cos C} \;=\; \frac{\sin A(\cos B\,\!\sin C\,+\,\sin B\,\!\cos C)}{\sin B(\cos A\,\!\sin C\,+\,\sin A\,\!\cos C)} \;=\;\frac{\sin A\,\!\sin(B\,+\,C)}{\sin B\,\!\sin(A\,+\,C)}$

$\text{Substitute [1] and [2]:}$

[color=beige]. . [/color]$=\;\frac{\sin A\,\!\cos A}{\sin B\,\!\cos B} \;=\;\frac{2\,\!\sin A\,\!\cos A}{2\,\!\sin B\,\!\cos B} \;=\;\frac{\sin 2A}{\sin 2B}$

Quote:
 $\text{(b) Prove: }\:\frac{\tan A\,+\,\tan B}{\tan B\,+\,\tan C} \:=\:\frac{\cos^2C}{\cos^2A}$

$\text{W\!e have: }\:\frac{\tan A\,+\,\tan B}{\tan B\,+\,\tan C} \;=\;\frac{\frac{\sin A}{\cos A}\,+\,\frac{\sin B}{\cos B}}{\frac{\sin B}{\cos B} \,+\,\frac{\sin C}{\cos C}}$

$\text{Multiply top and bottom by }\cos A\,\!\cos B\,\!\cos C:$

[color=beige]. . [/color]$=\;\frac{\sin A\,\!\cos B\,\!\cos C\,+\,\cos A\,\!\sin B\,\!\cos C}{\cos A\,\!\sin B\,\!\cos C\,+\,\cos A\,\!\cos B\,\!\sin C} \;=\;\frac{\cos C(\sin A\,\!\cos B\,+\,\cos A\,\!\sin B)}{\cos A(\sin B\,\!\cos C\,+\,\cos B\,\!\sin C)} \;=\;\frac{\cos C\,\!\sin(A\,+\,B)}{\cos A(\,\!\sin(B\,+\,C)}$

$\text{Substitute [3] and [1]:}$

[color=beige]. . [/color]$=\;\frac{\cos C\,\!\cos C}{\cos A\,\!\cos A} \;=\;\frac{\cos^2C}{\cos^2A}$

Quote:
 $\text{2. If }\tan A=1,\;\tan B=2,\;\tan C=3,\,\text{ prove that: }\:A\,+\,B\,+\,C \:=\:\frac{\pi}{2}$ [color=beige]. . . [/color][color=blue]This is not true![/color]

$\text{W\!e have: }\:\tan A \,=\,1 \;\;\;\Rightarrow\;\;\;A \,=\,\frac{\pi}{4}$

$\text{Formula: }\:\tan(B\,+\,C) \:=\:\frac{\tan B\,+\,\tan C}{1\,-\,\tan B\,\!\tan C}$

$\text{Then: }\:\tan(B\,+\,C) \:=\:\frac{2\,+\,3}{1\,-\,2\cdot3} \:=\:\frac{5}{-5} \:=\:-1$

$\text{Hence: }\:\tan(B\,+\,C) \:=\:-1 \;\;\;\Rightarrow\;\;\; B\,+\,C \:=\:\frac{3\pi}{4}$

$\text{Therefore: }\:A\,+\,B\,+\,C \;=\;\frac{\pi}{4} \,+\,\frac{3\pi}{4} \;=\;\pi$

 November 7th, 2011, 01:41 AM #4 Guest   Joined: Posts: n/a Thanks: Re: urgent help Woow! thanks a lot.

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