How to prove?trigo

sanikui · #1 $Old$ 09-04-2008, 07:48 AM

1. 1+cos2A+cosA=cotA
sin2A+sinA

2. sin2A = tanA
1+cos2A

3. sin2A = cotA
1-cos2A

4. sin2A-cos2A=secA
sinA cosA

5. 1+sinA-cosA =tan A
1+sinA+cosA 2

THANK YOU!!

ADARSH · #2 $Old$ 09-04-2008, 08:13 AM

Hello !
cos(2A)=cos^2(A)-sin^2(A)
sin(2A)=2sin(A)cos(A)
put in the above to get
[2cos^2 (A)+cos(A)] / [2sin(A)cos(A)+sin(A)]
Take cos(A) from numerator
& sin(A) from denominatorcommon you get cot(A)

ADARSH · #3 $Old$ 09-04-2008, 08:19 AM

put
1+cos(2A)=2cos^2(A) for 2nd denom.

1-cos(2A)=2sin^2(A) for 3rd denom.

sin(2A)=2sin(A)cos(A) for numerator

and cancel common terms.

sanikui · #4 $Old$ 09-04-2008, 07:27 PM

k...i got it....thx
how about 4 n 5?

Soroban · #5 $Old$ 09-04-2008, 10:22 PM

Hello, sanikui!

Here's #4 . . .

Quote:

$4)\;\;\frac{\sin2A}{\sin A} - \frac{\cos2A}{\cos A} \:=\:\sec A$

We have: . $\frac{\overbrace{2\sin A\cos A}^{\sin2A}}{\sin A} - \frac{\overbrace{2\cos^2\!A-1}^{\cos2A}}{\cos x} \;\;=\;\;\frac{2\sin A\cos A}{\sin A} - \frac{2\cos^2A}{\cos A} + \frac{1}{\cos A}$

. . $= \;\;2\cos A - 2\cos A + \frac{1}{\cos A}\;\;=\;\;\frac{1}{\cos A}\;\;=\;\;\sec A$

ADARSH · #6 $Old$ 09-05-2008, 06:56 AM

multiply denominator with both numer. and denom.
u will get
RHS=
$[1+1+2sin(A)-2cos(A) -2cos(A)sin(A)]$ / $[1+sin^2(A)+2sin(A)-cos^2(A)]$
numerator $=[2(1+sin(A))(1-cos(A)]$ denominator $=[2sin(A)(1+sin(A)]$
cancel $2(1+sin(A)$
=>RHS $= [1-cos(A)] / [sin(A)]$
Now u remember 3rd Question
u will get $1/cot(A/2)= tan(A/2)$

dere may be shortcut for this!
sOrRy for I went offline ys'trday after 3rd

sanikui · #7 $Old$ 09-05-2008, 07:35 AM

Thx ADARSH and Soroban,i try my best~!

sanikui · #8 $Old$ 09-08-2008, 10:25 PM

Question 5
i think no nid multiply denominator with both numer and denom.
Just use half-angles formulas...then can do it.