Prove that

**Sanjana Das** · July 6th 2009, 08:24 AM

If Tn = sin^nA+cos^nA
Prove that
T3-T5 = T5-T7
...T1.......T3

plzzzz explain by covering all the steps as i belong to 10th standard ......

**malaygoel** · July 6th 2009, 08:42 AM

Originally Posted by Sanjana Das

If Tn = sin^nA+cos^nA
Prove that
T3-T5 = T5-T7
...T1.......T3

plzzzz explain by covering all the steps as i belong to 10th standard ......

= $\frac{T_5-T_7}{T_3-T_5}$

$\frac{sin^5A+cos^5A-sin^7A-cos^7A}{sin^3A+cos^3A-sin^5A-cos^5A}$

= $\frac{sin^5A-sin^7A+cos^5Asin^2A}{sin^3A-sin^5A+cos^3Asin^2A}$

= $\frac{sin^3A-sin^5A+cos^5A}{sinA-sin^3A+cos^3A}$

= $\frac{sin^3Acos^2A+cos^5A}{sinAcos^2A+cos^3A}$

= $\frac{sin^3A+cos^3A}{sinA+cosA}$

= $\frac{T_3}{T_1}$

**sa-ri-ga-ma** · July 6th 2009, 08:46 AM

(T3 - T5) = sin^3A + cos^3A -sin^5A - cos^5A
= sin^3A - sin^5A + cos^3A - cos^5A
= sin^3A( 1 - sin^2A) + cos^3A ( 1- cos^2A)
= sin^3A*cos^2A + cos^3A*sin^2A
= sin^2A*cos^2A(sinA + cosA)
If you divide this by T1 you get sin^2A*cos^2A.
Proceed in the same manner in right hand side.

**Soroban** · July 6th 2009, 10:13 AM

He3llo, Sanjana Das!

Another approach . . .

If $T_n \:=\:\sin^n\!A+\cos^n\!A$

prove that: . $\frac{T_3-T_5}{T_1}\:=\:\frac{T_5 - T_7}{T_3}$

Left side:

. . $\frac{T_3 - T_5}{T_1} \;=\;\frac{(\sin^3\!A + \cos^3\!A) - (\sin^5\!A + \cos^5\!A)}{\sin A + \cos A} \;=$ $\frac{(\sin^3\!A - \sin^5\!A) - (\cos^3\!A - \cos^5\!A)}{\sin A + \cos A}$

. . . . . . . $= \;\frac{\sin^3\!A(1 - \sin^2\!A) + \cos^3\!A(1 - \cos^2\!A)}{\sin A+\cos A} \;=\;\frac{\sin^3\!A\cos^2\!A + \cos^3\!A\sin^2\!A}{\sin A+\cos A}$

. . . . . . . $=\;\frac{\sin^2\!A\cos^2\!A\cdot(\sin A + \cos A)}{\sin A + \cos A} \;=\;\sin^2\!A\cos^2\!A$

Right side:

. . $\frac{T_5-T_7}{T_3} \;=\;\frac{(\sin^5\!A+\cos^5\!A) - (\sin^7\!A + \cos^7\!A)}{\sin^3\!A + \cos^3\!A} \;= \;\frac{(\sin^5\! - \sin^7\!A) + (\cos^5\!A - \cos^7\!A)}{\sin^3\!A + \cos^3\!A}$

. . . . . . . $= \;\frac{\sin^5\!A(1 - \sin^2\!A) + \cos^5\!A( 1 - \cos^2\!A)}{\sin^3\!A + \cos^3\!A} \;= \;\frac{\sin^5\!A\cos^2\!A + \cos^5\!A\sin^2\!A}{\sin^3\!A + \cos^3\!A}$

. . . . . . . $= \;\frac{\sin^2\!A\cos^2\!A\cdot(\sin^3\!A + \cos^3\!A)}{\sin^3\!A + \cos^3\!A} \;=\;\sin^2\!A\cos^2\!A$

Therefore: . $\frac{T_3-T_5}{T_1} \;=\;\frac{T_5-T_7}{T_3}$

Edit: Too slow again . . . Sa-ri-ga-ma beat me to it!
.

**malaygoel** · July 6th 2009, 06:31 PM

Noticed this:

. . $\frac{T_n - T_{n+2}}{T_{n-2}} \;=\;\frac{(\sin^n\!A + \cos^n\!A) - (\sin^{n+2}\!A + \cos^{n+2}\!A)}{\sin^{n-2}\!A + \cos^{n-2}\!A} \;=$ $\frac{(\sin^n\!A - \sin^{n+2}\!A) - (\cos^n\!A - \cos^{n+2}\!A)}{\sin^{n-2}\!A + \cos^{n-2}\!A}$

. . . . . . . $= \;\frac{\sin^n\!A(1 - \sin^2\!A) + \cos^n\!A(1 - \cos^2\!A)}{\sin^{n-2}\!A + \cos^{n-2}\!A} \;=\;\frac{\sin^n\!A\cos^2\!A + \cos^n\!A\sin^2\!A}{\sin^{n-2}\!A + \cos^{n-2}\!A}$

. . . . . . . $=\;\frac{\sin^2\!A\cos^2\!A\cdot(\sin^{n-2}\!A + \cos^{n-2}\!A)}{\sin^{n-2}\!A + \cos^{n-2}\!A} \;=\;\sin^2\!A\cos^2\!A$

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