Prove the following trigonometric identities. If a ^3 +3a ^2 =m, a ^3 +3a ^2 =n, prove that (m+n)^2 3 +(m-n)^2 3 =2a^2 3 .

Given that, a ^3 +3a ^2 =m, a ^3 +3a ^2 =n We have to prove (m+n)^2 3 +(m-n)^2 3 =2a^2 3 Adding both the equations, we get m+n=a ^3 +3a ^2 +a ^3 +3a ^2 =a( ^3 +3 ^2 +3 ^2+ ^3 ) =a( + )^3 Also. m-n=a ^3 +3a ^2 -(a ^3 +3a ^2 ) =a( ^3 -3 ^2 +3 ^2- ^3 ) =a( - )^3 Therefore, we have L.H.S=(m+n)^2 3 +(m-n)^2 3 =a^2 3 ( + )^2+a^2 3 ( - )^2 =a^2 3 ( + )^2+( - )^2 =a^2 3 ( ^2 +2 + ^2 )+( ^2 -2 + ^2 ) =a^2 3 ( ^2 + ^2 +2 )+( ^2 + ^2 -2 ) =a^2 3 (1+2 )+(1-2 ) =a^2 3 (1+2 +1-2 ) =2a^2 3 =R.H.S L.H.S=R.H.S

Prove the following trigonometric identities. If a ^3 +3a ^2 =m, …

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Question

Prove the following trigonometric identities.
If $\text{a }\cos^3\theta+3\text{a}\cos\theta\sin^2\theta=\text{m, a }\sin^3\theta+3\text{a}\cos^2\theta\sin\theta=\text{n},$ prove that $(\text{m}+\text{n})^\frac{2}{3}+(\text{m}-\text{n})^\frac{2}{3}=2\text{a}^\frac{2}{3}.$

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Answer

Given that,
$\text{a}\cos^3\theta+3\text{a}\cos\theta\sin^2\theta=\text{m},$
$\text{a}\sin^3\theta+3\text{a}\cos^2\theta\sin\theta=\text{n}$
We have to prove $(\text{m}+\text{n})^\frac{2}{3}+(\text{m}-\text{n})^\frac{2}{3}=2\text{a}^\frac{2}{3}$
Adding both the equations, we get
$\text{m}+\text{n}=\text{a}\cos^3\theta+3\text{a}\cos\theta\sin^2\theta+\text{a}\sin^3\theta+3\text{a}\cos^2\theta\sin\theta$
$=\text{a}(\cos^3\theta+3\cos^2\theta\sin\theta+3\cos\theta\sin^2+\sin^3\theta)$
$=\text{a}(\cos\theta+\sin\theta)^3$
Also.
$\text{m}-\text{n}=\text{a}\cos^3\theta+3\text{a}\cos\theta\sin^2\theta-(\text{a}\sin^3\theta+3\text{a}\cos^2\theta\sin\theta)$
$=\text{a}(\cos^3\theta-3\cos^2\theta\sin\theta+3\cos\theta\sin^2-\sin^3\theta)$
$=\text{a}(\cos\theta-\sin\theta)^3$
Therefore, we have
$\text{L.H.S}=(\text{m}+\text{n})^\frac{2}{3}+(\text{m}-\text{n})^\frac{2}{3}$
$=\text{a}^\frac{2}{3}(\cos\theta+\sin\theta)^2+\text{a}^\frac{2}{3}(\cos\theta-\sin\theta)^2$
$=\text{a}^\frac{2}{3}\{(\cos\theta+\sin\theta)^2+(\cos\theta-\sin\theta)^2\}$
$=\text{a}^\frac{2}{3}\{(\cos^2\theta+2\cos\theta\sin\theta+\sin^2\theta)+(\cos^2\theta-2\cos\theta\sin\theta+\sin^2\theta)\}$
$=\text{a}^\frac{2}{3}\{(\cos^2\theta+\sin^2\theta+2\cos\theta\sin\theta)+(\cos^2\theta+\sin^2\theta-2\cos\theta\sin\theta)\}$
$=\text{a}^\frac{2}{3}\{(1+2\cos\theta\sin\theta)+(1-2\cos\theta\sin\theta)\}$
$=\text{a}^\frac{2}{3}(1+2\cos\theta\sin\theta+1-2\cos\theta\sin\theta)$
$=2\text{a}^\frac{2}{3}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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