Question

If Tn=sinnθ+cosnθ prove that T3-T5/T1=T5-T7/T3

Answer 1

The given equation $\frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_3}$ is proved

Step-by-step explanation:

Given that $T_n=sin^n\theta+cos^n\theta$

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

That is to prove that LHS=RHS

• From the given $T_n=sin^n\theta+cos^n\theta$
• We want $T_1,T_3,T_5and T_7$ values

Put n=1 in $T_n=sin^n\theta+cos^n\theta$ we get

• $T_1=sin^1\theta+cos^1\theta$
• $T_1=sin\theta+cos\theta$

Put n=3 in $T_n=sin^n\theta+cos^n\theta$ we get

• $T_3=sin^3\theta+cos^3\theta$

Put n=5 in $T_n=sin^n\theta+cos^n\theta$ we get

• $T_5=sin^5\theta+cos^5\theta$

Put n=7 in $T_n=sin^n\theta+cos^n\theta$ we get

• $T_7=sin^7\theta+cos^7\theta$

Taking LHS

$\frac{T_3-T_5}{T_1}$

• Now substitute the values we get
• $\frac{T_3-T_5}{T_1}=\frac{sin^3\theta+cos^3\theta-(sin^5\theta+cos^5\theta)}{sin\theta+cos\theta}$
• $\frac{T_3-T_5}{T_1}=\frac{sin^3\theta+cos^3\theta-sin^5\theta-cos^5\theta}{sin\theta+cos\theta}$
• $\frac{T_3-T_5}{T_1}=\frac{sin^3\theta-sin^5\theta+cos^3\theta-cos^5\theta}{sin\theta+cos\theta}$
• $=\frac{sin^3\theta(1-sin^2\theta)+cos^3\theta(1-cos^2\theta)}{sin\theta+cos\theta}$

• $=\frac{sin^3\theta(cos^2\theta)+cos^3\theta(sin^2\theta)}{sin\theta+cos\theta}$ ( by using the identity $sin^2x=1-cos^2x$ )

• $=\frac{sin^2\theta cos^2\theta(sin\theta+cos\theta)}{sin\theta+cos\theta}$
• $=sin^2\theta cos^2\theta$

Therefore $\frac{T_3-T_5}{T_1}=sin^2\theta cos^2\theta\hfill (1)$

Now taking RHS

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

• Substituting the values we get
• $\frac{T_5-T_7}{T_3}$
• $=\frac{sin^5\theta+cos^5\theta-(sin^7\theta+cos^7\theta)}{sin^3\theta+cos^3\theta}$
• $=\frac{sin^5\theta+cos^5\theta-sin^7\theta-cos^7\theta}{sin^3\theta+cos^3\theta}$
• $=\frac{sin^5\theta-sin^7\theta+cos^5\theta-cos^7\theta}{sin^3\theta+cos^3\theta}$

• $=\frac{sin^5\theta(1-sin^2\theta)+cos^5\theta(1-cos^2\theta)}{sin^3\theta+cos^3\theta}$
• $=\frac{sin^5\theta(cos^2\theta)+cos^5\theta(sin^2\theta)}{sin^3\theta+cos^3\theta}$ ( by using the identity $sin^2x=1-cos^2x$ )
• $=\frac{sin^2\theta cos^2\theta(sin^3\theta+cos^3\theta)}{sin^3\theta+cos^3\theta}$
• $=sin^2\theta cos^2\theta$

Therefore $\frac{T_5-T_7}{T_3}=sin^2\theta cos^2\theta\hfill (2)$

• Comparing the equations (1) and (2) we get
• LHS=RHS

Therefore $\frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_3}$ is proved